Algebraic Generalizations of Discrete Grs: A Path to Combinatorial Gr Theory Through One-Relator Products

forp=2k+l>_3 (1.6) G= be a two-generator Fuchsian group. (1) If G is of type (1.1) or (1.5) then (U,V) (A,B) (2) IfG is of type (1.6) ~hen (U,V) 2 (AB,

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(3)If (3)If

G is G is holds (a) (b)

of type (1.2) or (1.3) then (U, V) ~N (A,B) of type (1.4) then one and only one of the following cases (U, V) ~g (A, G is a (2, 3, r) triangle group with gcd(r, 6) = 1 and (U, V) ~ (ABAB2, B2ABA)

(c) G is a (2, 4, r) triangle group with gcd(r, 2) = 1 and (U, V) ~N (AB~’ B3AB3)

(d) G is a (3, 3, r) triangle group with gcd(r, 3) = 1 and (U, V) ~N (AB2’ B2A)

(e) G is a (2, 3, 7) triangle group and (U,

V) ~ (AB2ABAB~AB2AB,

B~ABAB~ABABA)

Again given a generating pair U, V for a ~mn-elementary Fuchsian group G the next two results classify the generating pairs and the corresponding types of Fuchsian groups by the trace of the commutator [U, V]. THEOREM 4.5.3. Suppose U, V ¯ PSL2(R) with tr[U, V] > 2 and suppose G =< U, V > is non-elementary. Then G is discrete if a.nd only if there is an extended Nielsen transformation from (U, V) to a pair (R, which satisfies {a~ter a suitable choice of signs} (1) 0 _< tr(R) <_ tr(S) <_ [tr(RS)l (2) tr(R) = 2cos(~r/p) or tr(R) (3) tr(S) = 2cos(~r/q) or tr(S) >_ (4) tr(RS) = -2cos(~r/r) or tr(RS) <_ -2 wherep, ¯ N \ {1 Moreover,if G is discrete then G is of type (1.1),(1.2),(1.3) or THEOREM 4.5.4. Suppose U,V ¯ PSL2(R) with 0 <_ tr(U),O ~_ tr(V) and tr([U, V]) < 2 and suppose G =< U, V >. Then G is discrete if and only i[ one of the following cases holds: (a) tr([U, V]) _< -2 (b) tr([U, V]) = -2cosOr/p),p ¯ N \ {1} (c) tr([U, V]) = -2cos(2~r/p),p ¯ N \ {1} and gcd(p, 2) =

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109

(d) tr([U,V]) = -2cos(6~r/r),r e N,r >_ 7 with gcd(r, 6) = (U, V) is Nielsen equivalent in a trace minimizing manner to a pair (R, S) which satisfies tr (R) = tr (S) -= tr (e) tr([V,V]) = -2cos(4~r/r),r e N,r > 5 with gcd(r, 2) = 1 and (U, V) is Nielsen equivalent in a trace minimizing manner to a pair (R, S) which satisfies tr (R) = tr (S) and tr (RS) 1 (f) tr([U,V]) = -2cos(3~/r),r N,r > a with gc d(r, 3) -= 1 a (U, V) is Nielsen equivalent in a trace minimizing manner to a pair (R, S) which satisfies tr (R) -~ tr (S) = tr (g) tr([U, V]) = -2 cos(4~r/7) and (U, V) is Nielsen equivalent in a trace minimizing manner to a pair (R, S) which satisfies tr (S) = tr

= tr + 1. Moreover, if G is discrete then G is of type (1.1) in case (a), of type(1.5) in case (b), of type (1.6) in case (c), a (2, 3, r) triangle groupin case (d),a (2, 4, r) triangle groupin case (e),a (3, 3, r) triangle groupin case (f), and a (2, 3, 7) triangle group in case (g). COROLLARY 4.5.1. Suppose G ~ U, V > C PSL2(R) is a Fuchsian group. Then [tr([U, V]) - 21 >_ 2- 2cos(~r/7). Weclose this section with a result based on a theorem of Majeed [Mail, (See Rosenbcrger [R 3]), concerning free subgroups in discrete groups. Notice the similarity of this result with the Ree-MendelsohnTheorem (Theorem 3.4.3). Wewill be using this result in our analysis of free. subgroups in generalized triangle groups. Recall that a subgroup of PSL2 (C) is elementary if any two elements of i~ffinite order (regarded as linear fractional transformations) have at least one commonfixed point. A subgroup of PSL2(C) is called elliptic if it consists of solely elliptic maps. THEOREM4.5.5. Suppose A,B ~ PSL,2(C) with = an d assume that G is non-elementary and non-elliptic. Then there is a generating pair (U, V) of G which is Nielsen equivalent to {A, B) such that < U~, V*~ > is a discrete free group of rank 2 for some large integer n. Wemention that a corresponding classification for two generator NEC groups has been done by E.Klymenko and M.Sakuma [K1- S].

CHAPTER V ONE-RELATOR

5.1 One-Relator

PRODUCTS

Products

As we saw in Chapter 3 there is a vast theory of one-relator groups generalizing the theory of free groups. Further, one-relator groups were tied to linear groups via their linearity properties. In this chapter we generalize one-relator groups to an extended construction called one-relator products. This construction also arises from the theory of discrete groups via the Poincare presentation. As we will see, many of the linearity properties shared by discrete groups and one-relator groups, will carry over, under appropriate conditions, to this more general class. Let {Ai}, i in some index set I, be a family of groups. Then a onerelator product is the quotient, G = A/N(R), of the free product A = ¯ iAi by the normal closure N(R) of a single non-trivial word R in the free product. Weassume that R is cyclically reduced and of syllable length at least two. The groups Ai are called the factors while R is the relator. In analogy with the one-relator group case we say R involves Ai if R has a non-trivial syllable from A~. If R = S"~ with S a non-trivial cyclically reduced word in the free product and m _> 2, then R is a proper power. Wethen also call S a relator. As in the one-relator group case the proper power case is somewhateasier to work with. In this context a one-relator group is just a one-relator product of free groups. From the Freiheitssatz a one-relator group with at least two generators in the given presentation is never trivial. On the other hand a one-relator product of non-trivial groups maycompletely collapse. For example, consider A =< a > and B =< b > to be finite cyclic groups of relatively prime order. Then the one-relator product G = A * B/N(ab) is a trivial group. Because of examples such as this, a natural question to ask is under what conditions the factors actually inject into a one-relator product. Wesay that a Freiheitssatz holds for a one-relator product G if each factor injects into G via the identity map. More generally let X, Y be disjoint sets of generators and suppose that the group A has the presentation A -- < X; Rel(X) > and that the group G has the presentation G =< X, Y; Rel (X), Rel(X, Y) >. G satisfies a Freiheitssatz, abbrevi111

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ated FHS, {relative to A} if < X >G= A, that is the subgroup of G generated by X is isomorphic to A. Alternatively this means that A injects into G under the identity map taking X to X. In coarser language this says that the complete set of relations on X in G is the "obvious"one from the presentation of G. If G satisfies a FHSrelative to A we say that A is a FHSfactor of G. In this language a Freiheitssatz holds for a one-relator product if each factor is a FHSfactor. As with one-relator groups, the starting off point for a study of onerelator products is to determine a Freiheitssatz. The example above shows that there is no such result in general and therefore some restrictions must be imposed. There are two approaches. The first is to impose conditions on the factors while the second is to impose conditions on the relator. Wewill discuss conditions on the factors first. Recall that a group H is locally indicable if every finitely generated subgroup has an infinite cyclic quotient. It turns out that local indicability of the factors is a strong enough condition to allow most of the results on one-relator groups to be carried over to one-relator products. In the next section we will give a proof of the FHSfor one-relator products of locally indicable factors (a result proven independently by Brodskii, Howie and Short). Wewill also mention some further results on local indicability and one-relator products with locally indicable factors. Local indicability is a strong form of torsion-freeness and them have been some succesful attempts to prove a FHS for one-relator products of torsion-free factors. Wediscuss these also. The second approach is to impose restrictions on the relator. The most commonrelator condition is that R is a proper power of suitably high order, that is R = S"~ with m _> 2. Wewill discuss this approach in section 5.3. If m > 7 then the relator satisfies the small cancellation condition C1(1/6) {see section 3.8} and a FHScan be deduced from small cancellation theory. A FHSdoes hold in the cases m --- 4, 5, 6 (m -- 6 due to GonzalezAcuna and Short [Go-S], m = 4, 5 due to Howie [H-3,4]) but the proofs are tremendously difficult. The cases m = 2, 3 are still open in general although specific cases where a FHSdoes hold have been proved. In particular if the factors admit representations into a suitable linear group, a FHScan be given. This will be done in Chapter 6. The technique for handling these proper power situations is combinatorial geometric and closely tied to small cancellation diagrams. Wewill only summarize these geometric results and a more complete treatement of the geometric techniques can be found in the excellent survey articles by Howie[H5] and Duncan and Howie [D-H 1] as well as the original papers. One-relator products and the Freiheitssatz are closely tied to the solution of equations over groups, also called the adjunction problem. Basically the adjunction problem is the following. Let G be a group and

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113

W(Xl,.. ¯ , x,~) be an equation (or system of equations) with coefficients G. This question is whether or not this can be solved in some overgroup of G. In the case of a single equation W(x) = 1 in a single unknownx a result of F.Levin [Le] shows that there exists a solution over G if x appears in Wonly with positive exponents. To see the connection with the FHS consider the one-relator product G = A, < t > IN(R) where R is a cyclically reduced word in the free product A* < t > of syllable length at least 2. Consider R then as a word R(t) in the variable t with coefficients in A. The equation R(t) = has a solution over A in some overgroup containing A if and only if the natural map from A into G is injective. More generally Baumslag and Pride [B.B.- P 1] have shownthat if X is a class of groups which contains the infinite cyclic group and is closed under free products with finitely many factors, then the existence of a FHSfor one-relator products of X-groups is equivalent to the fact that any equation is solvable over an X-group. By this we mean that if G = (.~H~)/N(R) where the Hi are arbitrary X-groups and R is a cyclically reduced word in the free product on the Hi of syllable length at least 2, then each Hi injects into G if and only if any equation is solvable over an X-group. In connection with the adjunction problem we mention the Kervaire Conjecture (also called the Kervaire-Laudenbach Conjecture). This says that if G = A* < t > IN(R) is trivial, then A is trivial. From the FHS this is clearly true if A is a free group. Recently A.Klyachko [K1] proved that the Kervaire conjecture is true wheneverA is a torsion-free group. In this paper he also proved a Freiheitssatz. Specifically if A is torsion-free and the exponent sum of t in R E A, < t > is :i:1, then A injects into A* < t > IN(R). Klyachko’s result is a strenghtening of the FHS for locally indicable factors. Wewill discuss this and give Klyachko’s solution in section 5.4. 5.2 The Freiheitssatz

and Locally

Indicable

Groups

As with one-relator groups the starting off point for the study of onerelator products is to provide a FHS. As indicated in the last section there are two ways to proceed: to place restrictions on the factors or to place restrictions on the relator. The factor condition which is not only strong enough to give a FHSbut also to allow the extension of most of the theory of one-relator groups is local indicability. Recall that a group G is locally indicable if every non-trivial finitely generated subgroup has an epimorphism onto Z. In particular locally indicable groups are torsion-free and hence local indicability can be considered as a very strong form of torsionfreeness. Locally indicable groups were introduced by Higman in work on group rings and the zero divisor question. Specifically, in his thesis Higman

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[Hi 2] proved that if K is an integral domain and G is locally indicable, then the group ring KGhas no zero divisors, no idempotents other than 0 or 1 and no units other than those of the form ug with u a unit in K and g E G. The following theorem was discovered independently by Brodskii[Br 1,2], J.Howie [H 7] and H.Short[Sho]. It is interesting that all three proofs are entirely different. THEOREM 5.2.1. A one-relator isfies a Freiheitssa~z. That is, if G = *~A~/N(R), is a cyclically reduced word in least two, ~hen each A~ injects factor.

product of locally indicable factors satwhere each A~ is locally indicable, and R the free product *~A{ of syllable length at into G under the identiky map, i.e. is a FHS

Before giving a proof of this major result we briefly describe the evolution of the theorem. Magnusproved the classical FHSfor one-relator groups in 1929 using the algebraic techniques as described in Chapter Three. There were simplifications of this proof by Moldavanski[Mo], McCooland Schupp [Mc-S] and others once the theory of group amalgams was better understood. In 1972 Lyndon [L 4] gave a proof of the FHS using combinatorial geometry. This proof arose out of ideas and diagrams in small cancellation theory. As a by-product of this proof, Lyndon showed that the FHS held for one-relator products whose factors were additive subgroups of the real numbers R. S.Pride [P 1] then showed that the FHS holds for one-relator products of locally fully residually free factors while this was extended to locally residually free factors by B. Baumslagand S. Pride [B.B. - P 1]. The above results axe all examples of locally indicable factors and pointed to the main result. In 1980 Brodskii [Br 1] announced Theorem 5.2.1, but it was not published until 1984. Brodskii’s method was algebraic and mimicked Magnus’ original treatment. B.Baumslag[B.B. 1] rediscovered this algebraic proof. J. Howie independently proved Theorem 5.2.1 by giving a straightforward modification of his tower proof of the classical FHSwhich was presented in section 3.3. Finally H.Short [Sho], adapting Lyndon’s combinatorial geometric arguments to the context of one-relator products of locally indicable groups, gave a third independent proof of Theorem5.2.1. Short’s technique was important because it introduced to the study of one-relator products the concept of pictures over a group. These pictures are the duals of Lyndon - Van Kampendiagrams for one-relator products and proved instrumental in handling the proper power case. We will return to these ideas in the next section. Wenow give a proof of Theorem5.2.1 along the algebraic lines of Brodskii and B. Baumslag.

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115

PROOF.(Theorem 5.2.1) The proof essentially mimics Magnus’ original treatment. Weprove the case of two factors, the general result then follows easily. Therefore assume that G = (A * B)/N(R) with A and B locally indicable and R a cyclically reduced element in the free product A * B of syllable length at least two. By symmetryit suffices to showthat A injects into G. Let Ao be the subgroup of A generated by those elements which appear in R and similarly let B0 be the subgroup of B generated by those elements which appear in R. Let Go -= (Ao * Bo)/N(R). If Ao and B0 inject into Go, it would follow that G is the amalgamated product G = A *Ao Go *Bo *B and hence A and B inject into G. Hence we can assume without loss of generality that A = A0 and B = B0. Recall that A and B now are finitely generated, and hence each has an infinite cyclic image. The proof now proceeds by an induction on the syllable length L of the relator R. If L = 2 then both A and B are infinite cyclic groups and the result holds since G is just a two-generator one-relator group. Thus we can assume that R has syllable length L greater than 2. Weconsider first the case where B =< b > is infinite cyclic. Then since A is locally indicable there is an epimorphismof A onto an infinite cyclic group C =< c > with kernel N. Suppose further that A/N =< aN >. This induces a homomorphismof A ¯ B onto C ¯ B and suppose that under this homomorphismthe relator R is mapped to R1. Wenow have the group G1 = (C * B)/N(R1) and suppose here first that c has exponent sum zero in R1. Thus R lies in the normal closure of N tO B in A, B. Let b~ = a-~ba~ and write R in terms of the b~ and the elements of N. At least two of the b~ must be involved in writing R since R is of length at least 4 (cyclically reduced of length greater than 2) and some am, with r E Z, r ~ 0, n E N, must occur in R since we are assuming that A -- A0. Let s be the minimumindex of a b~ appearing in R and let t be the maximum. Let K = N, < bs > * < bs+l > * ¯ ¯ ¯ * < bt >, Ko = N* < b~ > * < b~+~ > ,...*

< bt-~ > and

K1 = N* < bs+l > *.." * < bt > ¯ R is of length smaller, than L as a word in the free product K0* < bt >. Hence by the inductive hypothesis K0 is embedded in K = KIN(R). Similarly, R as a word in the free product < b~ > ,K1 is of length less than L and so K1 is embedded in K. Thus the original one-relator product G

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is the HNNextension of K with stable letter a and associated subgroups Ko, K1 with a-lbia = b~+l,S <_ i < t, and acting on the elements of N by conjugation in A. Therefore N and hence A is embedded in G. This handles the case where B is infinite cyclic and c has exponent sum zero in R1. Nowsuppose that c has non-zero exponent sum $ in R1 and b has nonzero expolmnt sum ~ in R~. As in the proof of the classical FHSadjoin a ~-th root ~ of a to obtain the group A and a ~3-th root ~ of c to obtain the infinite cyclic group C. Then from the subgroup theorem for free products with amalgamation A is still locally indicable and there is a homomorphism a : ~ ~ ~ such that ~ -~ E, a --* c. Let -~ = ba-~ and ~ =< ~ >. From the homomorphismdefinition of free products it follows that A * B = A * B. Further in the image of R under the homonmrphism induced by a from A * B into C * B the exponent sum of E is zero. Therefore the result follows from the argument above when there is zero exponent stun since, as in the case of a one-relator group, G embeds into G = (A ¯ B)/N(R). Finally, for the case B = B0 ---< b >, suppose that b has zero exponent sum in R1. Then b has also zero exponent sum in R. R lies in the normal closure of A which has the form M= .iA~, i E Z, where A~ = b-*Ab~. Since B = B0 and R is cyclically reduced of length at least four, at least two of the Ai are involved in R. Suppose s is the minimumindex such that Ai is involved in R and t is the maximumindex. Define Ko -- A~+I *-.- * At and K1 = A~ *... * At-~. Then R belongs to Mo ---- A~ * Ko = At * K1, and the A~, K0, K~ and M0are all locally indicable. Let L1 denote the length of some cyclically reduced conjugate of R, regarded as a word in A~ * K0. We have 2 _< L1 < L since s ~ t and 2 <_ L. By the inductive hypothesis, the canonical maps of A~ and K0 into G~ = Mo/N(R) are injective. The result now follows since G is the HNNextension of (]2 with stable letter b aand associated subgroups Ko and K~ where b-~Klb -- Ko. This completes the case whenB is infinite cyclic. We now consider the case where B is arbitrary. Then there is an epimorphism ¢ from B onto an infinite cyclic group D =< d > which further induces a homomorphsim from A ¯ B onto A * D. Let R2 be the image of R under this map. If d appears with non-zero exponent sum in R2 then if the length of R~ is one then (A * D)/N(R2) is the free product of A and a finite cyclic group and hence A is embeddedin it which implies the required result. Otherwise if R2 has length at least 2 then we can use the first case, where B is infinite cyclic to establish that A is embeddedin (A * D)/N(R2) and hence in (A * B)/N(R) which gives the result. As alluded to, local indicability of the factors is strong enoughto not only allow the FHSas above but to extend muchof one-relator group theory. We

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117

close this section by mentioning someof these extensions. Further material in this area can be found in the articles by Howie[H 1,5,6]. Theorem 5.2.1 can be iterated and extended in several ways to obtain the FHSfor groups formed by more general constructions than one-relator products. These extensions are generally of the form (*Ai)/N(Rj) where the A~ are locally indicable and the relators Rj are "staggered" [H 5]. We present one result of this type. This involves a straightforward iteration using a result of Howie (see Theorem5.2.3 below). Wemention that there are many other results along these lines especially by Howie [H 5], Howie and Pride [H-P] and M. Edjvet [E 1,2,3]. THEOREM 5.2.2. Let A1,... , An be locally indicable groups, and/’or each i = 1,... , n - 1, let R~ be a non-trivial word in the [ree product G~ = AI*A2*... ,A~+I o[ syllable length at least ~vo and involving A~+~. Assume fureher that [or i = 1,... , n - 2, R~ not a proper power. Then each A~ is a FHSfactor for G ~- (*iAi)/N(R1,... , P~-l). PROOF. Suppose n = 3 so that G -- (A,B,C)/N(R~, R2) with A, B, C locally indicable and R~ involving A and B not a proper power and R2 involving C. From a result of Howie [HI(see Theorem 5.2.3 below) one-relator product of locally indicable factors is locally indicable if the relator is not a proper power. Therefore GI = (A*B)/N(R~) is locally indicable. Now G = (GI*C)/N(R2) is then a one-relator product with locally indicable factors so both G~ and C inject. Further A and B inject into G1 and thus into G. The general result then follows easily by induction. This type of iteration will be used in Chapter 9 in the study of groups of special NECtype. The next result, due to Howie[H 6], generalizes a result of Brodskii, and answers a question posed by Pride. This is that torsion-free one-relator groups are locally indicable. THEOREM 5.2.3. Let G = (A * B)/N(R) be a one-re/ator product o[ locally indicable groups with the relator R being cyclically reduced of syllable length at least 2 in the free product A * B. Then the [ollowing are equiwlent: (i) is locally ind icable. (ii) ~ is torsion-free. (iii) is not~ pr oper power in A * B. COROLLARY 5.2.1. Aft torsion-free cable.

one-relator groups are locally indi-

Theorem 3.2.4 was the Spelling Theorem of Newman. This said in essence that in a one-relator group ~ with relator R = S~, m ~_ 2, any

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non-empty word which represents the identity in G must contain a cyclic subword of R or R-1 longer than S"~-1. There have been improvements and more precise versions of this result for one-relator groups given by Gurevich [Gu], Schupp [Sc] and Pride [P 1]. The following, due to Howie and Pride [H-P] extends this to one-relator products of locally indicable groups. Further this result in turn allows for the solution of the word problem in one-relator products of locally indicable groups when the relator is a proper power. THEOREM 5.2.4. Let G = (A * B)/N(R) where A and B are locally indicable and R = S"~, m >_ 2. Let W be a non-empty word in A. B which lies in N(R). Then either W is conjugate to R or -1 or Wcontains tw o almost disjoint cyclic subwords, each o/: which is a cyclic subword oze R or R-~ "-~. longer than S COROLLARY 5.2.2. I/: m > 1 and A and B have solvable word pro.blems in the above theorem ~hen G also has a solvable word problem. Brodskii and Mazurovskii have also solved the word problem for G in the case where rn -- 1 but under the stronger hypothesis that A and B are effeetlvely locally indicable. By this they mean that there exists an algorithm which, given a finite set of generators for a subgroup H, will decide whether or not H is trivial and if not will exhibit an epimorphism of H onto Z. Corollary 5.2.2 and the extension cited above lead to a consideration of the general decision problems in one-relator products with locally indicable factors. Recall that for one-relator groups, the solvability of both the word problem and the generalized word problem was obtained by Magnus. The conjugacy problem is also solvable and was handled in the torsion case by Newmanand in general by Juhasz. These standard decision problems are subsumed in a class of decision problems called the genus problems. (see Duncan and Howie [D-H 1]). The genus problem GP(g, n) is the following algorithmic problem: given a group G, a set of generators for G and n words W1, ..., W,, in these generators decide whether or not the equation XlW1XF1...XnWnX~1[y1,

Z1]...[Y9,

Zg]

= 1

can be solved for Xi, Y~, Zj in G and if so to find an explicit solution in terms of the given generators. GP(0, 1) is then the word problem while GP(O, 2) is the conjugacy problem. Pride [Pr] termed the class GP(0, n) the dependence problems. In general we say the genus problem is solvable for a given group G if GP(g, n) is solvable for all g, n. For one-relator products with locally indicable factors the main result is the following.

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119

THEOREM 5.2.5. Let G ~- (A * B)/N(S "~) where A and B -axe locally indicable, m >_ 2, and S is cyclically reduced in A * B of length at least 2. Suppose that A and B are g/yen by recursive presentations for which the genus problem GP(g’, n’) is solvable for all pairs of integers (g’, n’) that 0 <_ gl <_ g and 1 <_ n’ <_ n ÷ 2(g - g’) and S is given explicitly in terms of the generators for A and B. Then GP(g, n) is solvable for There are extensions of this result to one-relator products where the factors are not locally indicable with however higher power relators. We will return to this is the next section. Another stand~d result of one-relator group theory which can be extended directly to one-relator products of locally indicable factors is Theorem 3.7.1 also due to Magnus. Recall, this says that in a free group F, if two elements R1 and R2 have the same normal closure in F then R1 is conjugate to R2 or to R~1. In the context of one-relator products we have the following (see [Br 3],[H 6,7],[E 1]) THEOREM 5.2.6. Suppose RI,R2 are cycfically reduced non-trivial words of length at least 2 in the free product A ¯ B where A and B are locally indicable and suppose that ml, m2 are positive integers. Then if R~1 and R’~2 have the same normal closure in A * B, then ml = m~. and 1. R1 is a cyclic permutation of R2 or R~ Weclose this section by mentioning that Lyndon was able to determine the cohomology of one-relator groups. While we will not discuss group cohomologyin any depth at this time we note that Lyndon’s determination was based on the following algebraic formulation in terms of the relation module N(R)/[N(R), N(R)]. The corresponding result (Theorem 5.2.7) called Lyndon’s Identity Theorem and is based on the following ideas. Let G =< X; R > be a one-relator group with relator R. Then there is a well-defined action of G on the relation module N(R)/[N(R), N(R)] where N(R) is the normal closure of R in the free group F(X) on X. This action is defined as follows: if g E G is represented by the word U in F(X), then for any P e N(R),gP[N(R),N(R)] = UPU-I[N(R),N(R)]. Lyndon’s Identity Theoremis then: THEOREM 5.2.7.

Let G =< X : R >.

(1) If G is torsion-free, then N(R)/[N(R), N(R)] -- ZG as ZG - modules where ZGis the integral group ring over G. (2) If R = S’* with m >_ 2 and S not a proper power, then there is a short exact sequence of ZG-modules 0 --* ZG(s - 1) --* ZG -, N(R)/[N(R),N(R)] --~ 0 where s is the element of G represented byS.

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A strong form of the identity

theorem was given by Cohen and Lyndon

THEOREM 5.2.8. Let G =< X; R > where R = S "~ and S is not a proper power. Suppose T is a transversal for the subgroup < S > N(R) F(X). Then {tRt -1, t ¯ T} is a free basis for N(R). COROLLARY 5.2.2. Let G =< X; R >. then the cohomological dimension of G is 2 (1) H G is torsion-free, unless G is free. (2) H R = Sr~ with m >_ 2 and S not a proper power, then for n >_ 3 and any ZG-module M, H~(G, M) ~ H~(W, where W i s the cyclic subgroup of G generated by the element s represented by S. These ideas can be extended somewhat to one-relator indicable factors. In particulaz we have.

products of locally

THEOREM 5.2.9. Let G = (A * B)/N(R) where A and B are locaJly indicable, and R -- S"~ with m >_ 2, S cyclically reduced in A * B of length a~ leas$ 2 and S is not a proper power. Then N(R)/[N(R), N(R)] is isomorphic as a ZG-module to the cyclic module ZG/(1 - s)ZG generated by R[N(R), N(R)] and where s is the element of G represented by S. COROLLARY 5.2.3. Let G, A, B be as in Theorem 5.2.9 and suppose C, the subgroup generated by s is a cyclic group of order m. Then for each Ha(G;-)

= Ha(A;-)

+ Ha(B;-)

+ Hq(C;-)

Ha(G; -) = Ha(A; -) + Ha(B; -) + Hq(C;-). As a consequence of these cohomological results, an extension of the theorem of Bagherzadeh [Bag] on Magnus subgroups in one-relator groups (see Theorem3.7.6) is obtained [H 2]. THEOREM 5.2.10. indicable, and R is any subgroup H of or B ~ gBg-~ with g any subgroup of this

Let G = A * B/N(R) where A and ar e lo cally cyclically reduced in A * B of length at least 2. Then the form AAgBg-l, g ¯ G or AAgAg-~ with g ¯ G\ ¯ G \ B is cyclic. Further if R is a proper power then form is trivial.

There is also an analog of the strong Cohen-Lyndon Theorem (Theorem 5.2.8) for one-relator products of locally indicable groups due to Edjvet and Howie [E-HI. THEOREM 5.2.11. Let G = (A * B)/N(R "n) where A and B are locally indicable, and R is cyclically reduced in A * B of length at leas~ 2. Let C be the cyclic group generated by R. Then there is a set U of double coset

5.3 THE FREIHEITSSATZ FOR HIGH POWERED RELATORS represeneatives

121

of N(R’~)A * B/C such Chat N(R"~) is freely generated by

{uR’~u-1;u ~ U}. Finally, locally indicable groups are torsion-free. in the theory of one-relator products is: CONJECTURE. The Freiheiessaez eorsion-free faceors.

A standard conjecture

holds for one-relator

products

of

Brodskii and Howie [Br-H] and Klyachko [K1] have proved some results on special cases supporting this conjecture. Let x = (xl,... ,x~) be sequence of elements in a group G. x~ is isolated in x if no xj belongs to the cyclic subgroup generated by xi for i ~ j. THEOREM 5.2.12. ([Br-H]) Lee G = (A.B)/N(W) where A and eorsion-free groups and W -- a~b~.., a~bk is a cyclically reduced word in A.B such that some ai is isolated in (ax,... , ak) and some bj is isolated in (b~,... ,bk). Then A and B naturally inject into G. If W= al Zml ¯ ¯ ¯ akx rn~ is a word in A* < x >, then the sign-index a is the number of sign changes in the cyclic sequence (ml,... , mk). THEOREM 5.2.13. ([Br-H]) Let G = A* < x > IN(W) where eorsion-free and W = a~xTM ... a~x"*~ is a cyclically reduced word in A* < x > of length at least 2 and sign-index a <_ 2. Then A naturally injects into G. (In particular the Kervaire conjecture is satisfied). Pur~her suppose that in addition one of the following holds: (i) The sign-index a > 0 (ii) a~...a~ is in A\ {1} (iii) a = rnk q- mla~ + m2(a~a2)+... + mk_~(al . . . ak_~) is not a in QG

(iv) ~ _< Then < z > naturally in~ects into G, that is z has infinite

order in G.

COROLLARY 5.2.4. Let G = (A*B)/N(W) where A and B are torsionfree groups and W= a~bl . . . akbk is a cyclically reduced word in A* B such that ai 7~ 1 ~ b~ and k <_ 3. Then A and B naturally inject into G. In section 5.4 we return to this conjecture relative to Klyachko’s solution of the Kervaire conjecture over torsion-free groups. 5.3 The l~reiheitssatz

for High Powered Relators

The second set of conditions imposed on one-relator products are conditions on the relator. Most commonlyconsidered are one-relator products G = A* BIN(R) where R = S"~ with S cyclically reduced and not a proper

122

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

power in the free product A ¯ B and m _> 2. As a general rule the larger the value of m the more results that can be obtained and the easier the proofs. This mirrors the situation in one-relator group theory where onerelator groups with torsion are easier to handle than torsion-free one-relator groups. In the study of one-relator products with high powered relator the techniques become much more technicial and geometrical in nature. These techniques are related to the geometric methods used in small cancellation theory which we introduced in section 3.8. Webriefly review some of these ideas and then show how they are tied to the present situation. Suppose F is free on a set of generators X. Let R be a syrnrnetrized set of words in F. By this we mean that all elements of R are cyclically reduced and for each r in R all cyclically reduced conjugates of both r and r -1 are in R. If rl and r2 are distinct elements of R with rl = bCl and r~ = bc2, then b is called a piece. Pieces represent those subwords of elements of R which can be cancelled by multiplying two non-inverse elements of R. The small cancellation hypotheses state that pieces must be relatively small parts of elements of R. Wedefine three small cancellation conditions. The first is the most commonand is a metric condition denoted C~(A) where/k is a positive real number. This condition asserts that if r is an element of R with r = bc and b a piece, then Ibl < ),lcl. If G is a group with a presentation < X; R > where R is symmetrized and satisfies C~(A), then G is called a ,~- group. So for example, if A = 1/6, G is a sixth group, etc. Ifp is a natural number, the second small cancellation condition is a nonmetric one denoted C(p). This asserts that no element of R is a product fewer than p pieces. Notice that C’(A) implies C(p) for :>1/( p - 1). The final small cancellation condition is also a non-metric condition denoted T(q) for q a natural number. This asserts the following: Suppose rl,... , rh with 3 <_ h < q are elements of R with no successive pair inverses. Then at least one of the products rlru,... , rh-lrh, rhrl is reduced without cancellation. T(q) is dual to C(p) for (l/p) + (l/q) = suitable geometric context [L-S]. As mentioned in Chapter 3 Greendlinger [Gre 1] proved purely algebraically that sixth-groups satisfy a Dehn Algorithm while Schiek showed the same for fourth groups also satisfying T(4) [Sch]. Lyndon [L 2] placed the study of small cancellation theory in a geometric context and this is the way it is most often looked at. The geometric interpretation arises in the following manner. Suppose that the group G has a finite presentation < X; R > with R a symmetrized set of words. As before let F be free on X and let N be the normal closure in F of R so that G = FIN. If w is a word in G, then w = 1 if and only if w as a word in F is a product of conjugates of elements

5.3 THE FREIHEITSSATZ FOR HIGH POWEREDRELATORS

123

of R, that is, w = ulrlu-~ 1... u,~r,~u~ ~ where the u~ are words in F and 1 is called an the r~ are elements of R. The sequence ulr~u-~l,... , umrrnu~n R -sequence of length m for w. A minimal R-sequence for w is an R-sequence of minimumlength. Wewill associate to any R-sequence for w a connected, simply-connected diagram in the Euclidean plane called an R-diagram. The small cancellation hypotheses are analyzed by analyzing these diagrams. A Lyndon-Van Kampen diagram for a group F consists of a collection Mof pairwise disjoint vertices, oriented edges and regions in the Euclidean plane together with a labelling function f assigning to each oriented edge e an element f(e) of F. This labelling function must satisfy f(e -~) = f(e) -~ where e-1 is the oppositely oriented edge of e. Further if a is a path in Mwith a = c~ ... e,~, then f(a) is defined as f(el).., f(en). If D is a region in M, a label of D is an element f(a) for any boundary cycle of D. The following results (Theorems 5.3.1 and 5.3.2) summarize manyof the geometric properties and existence of these diagrams: THEOREM 5.3.1. (see [L-S]) Let F be a free group and el,... ,Cm a sequence of non-trivial elements of F. Then there exists a diagram M = M(cl,... , cm) over F satsifying the following properties. (i) Ire is an edge of M, f(e) ~ (ii) M is connected and simply connected with a distinguished vertex 0 on the boundary of M. There is a boundary cycle e~,... , en of M ( a cycle in M of minimal leng~h which contains all the edges in the boundary of M) beginning at 0 such that the product f(e~) . .. f(en) is reduced without cancellation f(e~). ¯ ¯ f( c, ~) -C1" " " Crr~

(iii) IT D is a region of M and el,... ,cj is a boundary cycle olD, then f(el).., f(ej) is reduced without cancellation and is a cyclically reduced conjugate of some c~. The next provides a converse to the above theorem and also allows us to relate this to R-sequences. T~EOR~.M 5.3.2. (see [L-S]) Let M be a connected, simply connected diagram over a group F with regions D~,... , D,~. Let a be a boundary cycle of M beginning at a vertex vo on the boundary of M and let w = f(a). Then there exists boundary/abels ri of D~ and elements ui ofF, 1 < i < m, -1-.. 1. such that w = UXrlU~ umrmu~ Nowsuppose that R is a symmetrized subset of words in a fi:ee group F. An R-diagram is a diagram Mover F such that if 0 is any boundary cycle of any region D of M, then f(O) is in R. If G = F/N as before, then from the two theorems we obtain the following fact. A word w in F is in N if and

124

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

only if there exists a connected, simply connected R-diagram M such that the label on the boundary of Mis w. Thus connected, simply connected diagrams provide an adequate tool for studying membership in normal subgroups of free groups. The analysis of the small cancellation conditions lies in analyzing the structure of R-diagrams under these conditions. There have been many extensions of the basic small cancellation conditions by Rips, Olshanskii, Juhasz, Collins and Perraud and others (see the book by Olshanskii [O]). Of particular interest is the W(6) condition. If group G has a presentation satisfying C(6), then the corresponding VanKampendiagram has the property that every inner region has at least six neighbors (see [L-S] or [J 1,2]). This corresponds to the regular tesselation of the plane by hexagons. Similarly the conditions C(4) and T(4) imply that each inner region has at least four neighbors and no inner vertex has valency three. These correspond to regular tesselations by squares. W(6) a more flexible condition which sometimes looks like C(6), sometimes like C(4) or T(4) and sometimeslooks like a pentagonal tesselation. Specifically a Van-Kampendiagram satisfies W(6) if for each inner region D one of the following alternatives holds: (a) D has 4 neighbors and no vertices with valency 3 (b) D has 5 neighbors and at most 3 vertices with valency 3 (c) D has at least 6 neighbors Juhasz has proved that groups satisfying the W(6) condition have solvable conjugacy problem [J 2] and has used this condition extensively in his study of the solvability of the conjugacy problem for one-relator groups. The small cancellation conditions can be extended to develop a small cancellation theory over amalgams. First suppose that F -- ,F~ is a free product of non-trivial groups F~. A word w in F has reduced form w = uv if there is no cancellation or consolidation in forming, the concatenation of the free product normal forms for u and v. This is equivalent to the fact that the last syllable of u is in a different factor than the first syllable of v. w is in semi-reduced form if w -= uv where there is no cancellation in forming uv but there may be consolidation - the last syllable of u maybe in the same factor as the first syllable of v but they don’t cancel completely. Suppose R is a symmetrized subset of F. Then b is a piece if there exist distinct elements rl and r2 in R with semi-reduced forms rl = bcl and r2 -~ bc2. To extend these concepts to free products with amalgamation and HNN groups we need slightly more general definitions of normal forms than those given previously. Suppose F is the free product of non-trivial groups F~ amalgamated over a common(isomorphically embedded) subgroup A. w ~ 1 is in F, we say that a normal form for w consists of any sequence

5.3 THE FREIHEITSSATZ FOR HIGH POWERED RELATORS

125

w = yl""yn such that y~ is in a factor of F, successive y~ come from different factors of F and no yi is in the amalgamated subgroup unless n = 1. Under this definition there will be many normal forms for the same element but their lengths will all be the same. If u has normal form YI’"Y,~ and v has normal form xl...xm, then there is cancellation in forming uv if y~x~ is in the amalgamated subgroup. If y~x~ is not in the amalgamated subgroup but y,~ and x~ are in the same factor of F, then there is consolidation. With these ideas the definition of semi-reduced forms and a piece is the same as for a free product. Finally suppose F is an HNNgroup with base H so that F =--< H, t : t-]At = B >. If w is in F, we consider a normal form for w to be any sequence w -= hotelhl ... h~te~h,~+~, e~ -- +1, where no t-reductions are possible. Any two normal forms for w will have the same number of t symbols. If u and v have the respective normal forms hot~1 h l ... h,~t ~" hn+l . ,VmV and h~otAh~¯ . ~1 ÷IMP! then there is cancellation in forming uv if either e~ = -1 , h~+lh~ois in A and I1 = 1 or if e,~ -- 1, h~+lh~ois in B and f~ = -1. If there is no cancellation in forming uv, we say that uv is in reduced form. If R is a symmetrized set in F, then the definition of a piece is as in the previous two cases except here we require the factorizations to be reduced forms. Wenow state our metric small cancellation condition for amalgams. If F is either a free product, free product with amalgamation or HNNgroup and R is a symmetrized set of words in F, then the metric condition C~(A) for A a positive real numberis defined as follows where I is the appropirate length function: CONDITION Ct(/~). If r is in R with semi-reduced form (reduced form for HNNgroup) r = bc and b a piece, then L(b) < AL(r). Also L(r) > 1/~ for all r in R. There are similar analogues for C(p) and T(q) but these are not necessary for our FHSresult. Wedefine a small cancellation product to be a group G = FIN where F is either a free product, free product with amalgamation or HNNgroup and N =- N(R) is the normal closure of a symmetrized set of words in F satisfying a small cancellation condition C~(A) for some Our major result is the following which is actually a summaryof several results (see [L-S]). T~EOaEM 5.3.3. (The FHS for Small Cancellation Products) ( see[L-S] and the references there) Let G be a small cancellation product satisfying C’(A) for some A < 1/6. Then G satisfies a FHS relative to any of the amalgamfactors, that is, any factor of the underlying area/gain injects into G under the identity homomorphism. The proof of these results follows the same general outlines as for small

126

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

cancellation theory over free groups, that is, the analysis of appropriate diagrams. The approach given in Lyndon and Schupp[L-S] is to treat each amalgam construction separately. J. Corson [Cor] has considered Small Cancellation Theory over graphs of groups. His approach both unifies and extends this classical treatment. He defines non-spherical sets of words which are the analogues of symmetrized sets of words satisfying small cancellation conditions. He then proves that if F is a graph of groups and N is the normal closure in F of a non-spherical set of words, then the group natural map F --* FIN embeds the vertex and edge groups of F. Similar results have been done independently by R.M.S. Mahmud[Ma]. Wemention also that Collins and Perraud have proven a FHS for small cancellation products (over a free product) satisfying different small cancellation conditions [C-P]. The relevance of the above discussion to the case of one-relator products is the following. Suppose G = (A * B)/N(S "~) with A, B arbitrm’y groups and m _> 2. If m>_ 7 then the resulting group is a 1/6-th small cancellation product and the FHS holds. Further the geometric pictures and analyses used in small cancellation theory can be utilized in this case to extend most of the results on one-relator products of locally indicable groups. The case m = 6 was handled by Gonzalez-Acuna and Short [Go-S] in conjunction with a topological problem. Their technique provided an adaptation of Lyndon’s combinatorial geometric proof of the FHSto the situation of one-relator products. In doing this they adapted the concept of pictures which axe duals of Lyndon Van-Kampendiagrams. We now briefly describe these objects. Pictures were introduced as a technique in group theory by Rourke[Ro]. We mention also the FHS for m = 6 was done independently by Collins and Perraud [C-P] who showed that in this case the relator m S satisfies the non-metric small cancellation condition C(6). A picture F of a one-relator product G = (A * B)/N(I~), m _> 2, on a compact surface E consists of: (i) A disjoint union of (small) discs Vl,... ,V n in int(~), called the vertices of F. (ii) A properly embedded 1-submanifold ~ of To = ~,, \ int(Uv~). components of this submanifold are called the arcs of F. (iii) An orientation of the boundary 0~0 and a labelling function that associates to each componentof 0~0 \ ~ a label which is an element of AUB. A picture over 82 is called a spherical picture. The data above is also required to satisfy a numberof properties reflecting the fact that it is to represent the one-relator product. (a) In any region A of F--componentof ~ \ (t3v~ (~ ~), either all labels

5.3 THE FREIHEITSSATZFOR HIGH POWEREDRELATORS

127

belong to an A-region or a B-region accordingly. (b) Each arc separates an A-region from a B-region. (c) The vertex label of any vertex v~-the word consisting of the labels of Ovi read in the direction of orientation from somestarting pointis identically equal to Rm in the free monoid on A U B up to cyclic permutation. (d) Suppose A is an orientable region of F of genus g with k boundary components. Then each boundary component has a boundary 1label a~ ¯ ..a~~ , ~ = =t=1, where al,... ,a~ are the labels of that boundary component in the cyclic order induced from some fixed orientation of A and e~ -- ÷ 1 if the orientation of the segmentof 0~0 labelled a~ agrees with that of the boundary component of A, and -1 otherwise. If al,..., ak are the boundary labels of A, then the equation XlalX~"1... XkakX[~[Y~, Z1]’" [Y~, Zg] --- 1 is solvable for (X~, Y~, Zj) in A if A is an A-region or in B if A is a B-region. Under the conditions (a) through (d) above, pictures can be constructed over G. Further there are certain allowable operations on pictures which, when applied, yield an essentially equivalent picture. These arc called bridge moves and insertion or deletion of floating dipoles or exceptional spherical pictures. We refer to [D-H 1] or [Re] for a precise definition of bridge moves and floating dipoles but describe exceptional presentations which are crucial to some further discussions. As will be seen these axe related to triangle groups. Wesay a relator R is exceptional and the corresponding picture is exceptional if R = xUyU-1 for some word U and letters x,y up to cyclic permutation. If p,q are the orders of x, y respectively, then we say that G is of type E(p, q, m) and we then call R"~ of the form E(p, q, m). In the case where U is trivial, A =< x >, B =< y >, and then G is the triangle group T(p, q, m) (see Chapter 4). If liP-t- 1/q ÷ 1/m - 1 = s > 0 then T(p, q, m) is finite order 2Is. In this case, that is, G -~ T(p,q, m), s > 0, there is a canonical spherical picture F(p, q, m) arising form the action of G on 2. In general, if G is exceptional of type E(p, q, m) with s > 0, then there is a natural homomorphismfrom T(p, q, m) to G and F(p, q, m) induces a spherical picture over G. Wenote that a one-relator product can be exceptional in more than one way and so great care must be taken in handling such exceptional presentations. (see [D-H1]). Two pictures over G on the same surface are equivalent if each can be obtained from the other by a sequence of these allowable operations. Pictures can be associated to maps from the surface E to a certain space with fundamental group G (see [D-H 1]) and equivalent pictures have maps which differ only by a certain homotopy.A picture on ~. over G is efficient if it has the least numberof vertices in its equivalence class.

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ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

For the proof of the FHS,the crucial condition on pictures is the following: CONJECTURE F. Suppose G = (A * B)/N(R "~) is a one-relator product where S is a cyclically reduced word in the free product A ¯ B of syllable length at least 2 and suppose m _> 2. Let F be an e//icient picture on the ~ by disc D~ over G such that at most 3 vertices of F are connected to OD 2 by arcs. arcs. Then every vertex of F is connected to OD TI~EOREM5.3.4. ([H 3,4])Suppose that G = (A* B)/N(R) and that Conjecture F holds for pictures over G on the disk. Then the FHS holds for G. The techniques in handling the FHSfor high powered relators then reduce to showing that Conjecture F holds. The geometric arguments however become tremendously complex. The main results obtained by Howie are given in the next theorem. An analysis of pictures yielded the FHSfor m = 5 [H 3] while a much deeper analysis did the same for m = 4 [H 4] and certain cases when m = 3 [D-H 2] (The last result joint with A. Duncan). TItEOREM 5.3.5. Suppose G ---- (A * B)/N(Rm), m >_ 2, is a one-relator product with R cyclically reduced of length at least two. If at least one of the following conditions hold, then G satisfies conjecture F:

(i) > 4; (ii) m ---- 3 and no letter oeeuring in R has order 2. THEOREM 5.3.6. Suppose G = (A, B)/N(R m) is a one-relator product where R/s a cyclically reduced word in the free product (,Ai) of syllable length at least 2 and suppose m >_ 4. Then the FHSholds, that is, each factor Ai natura/ly injects into G. Further if m -- 3 and the relator R contains no letters o[ order 2, then the FHSholds. There are no known examples for which the FHS fails therefore we have the following conjecture.

for m _> 2 and

CONJECTURE. The FHS holds for any one-relator product G = (A * B)/N(R’~), where m >_ 2 and where R is a cyclically reduced word in the free product A * B of syllable lengeh at least 2.

COROLLARY 5.3.1. Let G = (A* B)/N(R~), m >_, where R is a cyclically reduced word of leng’~h at least two in A. B and is not a proper power. If conjecture F holds for G, the no proper cyclic subword of R"~ represents the identity in G. In particular, r, the element of G represented by R, has order m in G. The technique of pictures and geometric analysis has also been used to further extend many of the results on one-relator products of locally

5.3 THE FREIHEITSSATZ FOR HIGH POWERED RELATORS

129

indicable factors to the situation of high powered relators. However in most of these results, special care must be taken in the case of exceptional relators. There is first a generalization of the LyndonIdentity Theorem and the Cohen-Lyndon Theorem. THEOREM 5.3.7. Let G = (A*B)/N(Rm), m >_ 2, where R is a cychcally reduced word in A * B of length at least 2 which is non-exceptional. conjecture F holds for G, then N(R)/[N(R), N(R)] is isomorphic as a module to ZG/(1 - r)ZG where is the element of G represented by R. COROLLARY 5.3.2. Let G, A, B, R, r and m be as in Theorem 5.3.7. Let conjecture F hold for G. Then for q >_ 3 the restriction induced maps Hq(G;-)

-,

Ha(A;-)

+ Hq(B;-)

+ Hq(C;-)

are naturaJ isomorphisms of functors on ZG-modules, and for q = 2 a natural epimorphism, where C =< r > is the cyclic group of order m generated by r. Dually the maps Hq(A; -) + Hq(B; -) + Hq(C; -) --~ Hq(G; are natural isomorphisms for q _> 3, and it is a natural monomorphismfor q=2 THEOREM 5.3.8. Let G = ( A* B) /N ( ~) where R is cyclicedly red uced in A * B of length at least 2. Suppose that m >_ 6 or m >_ 4 and letter of R has order 2. Let C be the cyclic group generated by r, r the element represented by R, except in the case where R is exceptional of type E(2,2,m), that is R = xWyW-l,x 2 = y2 = 1, in which case C is -1. Then there is a set U of the diheclral group generated by x and WyW double coset representatives of N ( R’* ) A * B / C such ~ha$N ( m) i s f reely generated by {uRmu-x; u E U}. Using the cohomological results together with a result of Serre [Se] a classification of the finite subgroups of one-relator products can be obtained. TI~EOREM 5.3.9. Let (7, A, B, R, m, C be as in Theorem 5.3.8. Let conjecture F hold. Suppose K ~ {1} is a finite subgroup of G. Then K C gag -1, K C gBg-1 or K C gCg-1 for some g E G. Moreover precisely one of these occurs and the left coset gA (respectively gB, gC) uniquely determined by K. In particular if A, B are locally indicable and m >_ 2, then any element of ~nite order is conjugate to a power of R. Note that the second part of the theorem is the direct analog of the torsion theorem for one-relator groups. The theorem of Magnus on normal closures of elements can also be extended.

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ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

THEOREM 5.3.10. Suppose R1,R2 are cyclically reduced non-trivial words of length at least 2 in the free product A * B where ml, rn~ are positive integers. Suppose that for each i -- 1, 2 at leas~ one of the foflowing conditions holds: (i) m~_> and R~is notof t he formE(2, 3, 4) or E(2, 3 , 5); (ii) ml _> and nolet ter of R~ hasorder 2. If R~1 and R’~2 have the same normal closure in A. B, then m.~ = m2 and 1. R2 is a cyclic permutation of R2 or R~ There are also various versions of the Spelling Theorem. Theorem5.3.8 gives the result when m _> 4. There is a corresponding technical result for m >_ 3 (see [D-H1]). THEOREM 5.3.11. Let R be a cyclically reduced word of length I >_ 2 in the free product A * B. Assume that m >_ 4 and that R is not of the form E(2, 3, 4) or E(2, 3, 5). Let W be a non-empty, cyclically reduced word in the normal closure of R~. Then either (1) W is a cyclic permutation of R~"~; or (2) W has two disjoint cyclic subwords U1, U2 such that each Ui is identical to a cyclic subword Vi of R±’~, L(UI) = L(U2) _> (m 1)/- 1 and W has a cyclic permutation xtU~x2U2 for some elements Xl,X2 of the pregroup A U B; or (3) W has k disjoint cyclic subwords U1,...,Uk for some k, 3 _< k _< 6 such that each Ui is identical to a cyclic subword Vi of R±’~, and Vi has length at least (m - 2)/- 1 for i < 6- k and at least (m - 3)/fori>6-k. Weclose this section with the next result which mirrors Theorem 5.2.5 on the genus problem for one-relator products of locally indicable groups. TtiEOREM5.3.12. Let G = (A * B)/N(R"~) where R is cyclially reduced in A ¯ B of length at least 2. Suppose that A and B are g/yen by recursive presentations for which the genus problem GP(g~, n’) is solvable for all pairs of integers (g’, n’) such that < g’< gand 1 < n’< n +2(g - g’) and R is given explicitly in terms of the generators for A and B. Suppose further that one of the following conditions holds (i) m > 5 and G is not of type E(2, 3, 5) or E(2, 3, (ii) > 4 and nolet ter in R hasorder 2. Then GP(g, n) is solvable for 5.4 The Kervaire

Conjecture

and Klyachko~s

Solution

As described in section 5.1, one-relator products and the Freiheitssatz are closely tied to the solution of equations over groups, also called the

5.4 THE KERVAIRECONJECTUREAND KLYACHKO’SSOLUTION131 adjunction problem. The adjunction problem is the following. Let G be a group and W(xl,... ,x,~) = 1 be an equation (or system of equations) with coefficients in G. This question is whether or not this can be solved in some overgroup of G. To see the connection with the FHS consider the one-relator product G = (A. < t >)/N(R) where R is a cyclically reduced word in the free product A* < t > of syllable length at least 2. Consider R then as a word R(t) in the variable t with coefficients in A. The equation R(t) = 1 has solution over A in some overgroup contaaining A if and only if the natural map from A into G is injective. More generally B.Baumslag and Pride [B.B.-P 1] have shown that if X is a class of groups which contains the infinite cyclic group and is closed under free products with finitely many factors, then the existence of a FHSfor one-relator products of X-groups is equivalent to the fact that any equation is solvable over an X-group. By this we mean that ifG = (*iH~)/N(R) where the Hi are arbitrary X-groups and R is a cyclically reduced word in the free product on the Hi of syllable length at least 2, then each Hi injects into G if and only if any equation is solvable over an X-group. There are three standard conjectures concerning equations over groups. CONJECTURE 5.4.1. (The Kervaire Conjecture) - Given any non-trivial group A, then the one-relator product (A* < t >)IN(R) is non-~rivial. CONJECTURE5.4.2.

Any single equation over a torsion-free

group A is

solvable. CONJECTURE 5.4.3. Any single power equation, that is an equation of the form (W(t)) k = g, is solvable over an arbitrary group A. Fromthe Freiheitssatz for one-relator products of locally indicable factors it follows that all three conjectures are true whenA is a locally indicable group. The obvious question then is can this be extended to torison-free groups. The main result of this section is to give the solution, due to Klyachko[K1], of the Kervaire conjecture when A is a torsion-free group. THEOREM 5.4.1. Let G = (A, < t >)IN(R) where each letter in R from A has infinite order. Then G is non-trivial. In particular G is non-trivial if A is torsion-free. Before giving the proof of Theorem 5.4.1 we make some brief comments on the history of the adjunction problem and equations in general. For more information we refer to the articles by Howie [H 1,5], Duncan and Howie [D-H 1], the older survey article by Lyndon [L 5] and the relevant sections in [L-S]. The study of the adjunction problem was initiated by B.H. Neumann [Ne] who proved that given a group G and g E G the equation x"~ = g

132

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

for m E Z can always be solved over G. Gerstenhaber and Rothaus [G-R] showed that any equation, or more generally any system of equations, over a compact connected Lie group A is solvable in some overgroup of A. As a corollary they obtain that any equation over a finite group is solvable in some finite overgroup. F. Levin proved that any equation W(x) = is solvable over an arbitrary group if x appears in Wo~fly with positive exponents. We can also consider systems of equations. To any finite system of equations Z : r~(tl,... ,tn) ..... rm(tl,... ,t,~) = 1 over a group, in variables {t~}, we can associate the integer matrix M(E) (#~j) where #~j is the exponent sum of t~ in the word (equation) rj. If the rank of M(E) is equal to the number of equations in E, then E is called independent. Howie [H 7] as a consequence of a result of Gersten [Ge] has proved that every independent system of equations over a locally indicable group has a solution in some overgroup. Welist and summarize some of these results and then give the proof of Klyachko’s Theorem. THEOREM 5.4.2. (Gerstenhaber and Rothaus) Every independent system of equations over a i~ni~e group has a solution in some [inite overgroup. Tn~,OREM 5.4.3. (Levin) Any single equation W(x) = i is solvable an arbitrary group G if x appears in W with only positive exponents. TH~,OR~,M5.4.4. (Howie) Every independent system of equations over a locally indicable group has a solution in some overgroup, In connection with this last result we mention another result, due to Howieand Short [H-S] along the same lines but yielding a Freiheitssatz. THEOREM 5.4.5. Let G = (A, < tl,... ,tn >)/N(rx,... ,r~). Then injects naturally into G under either of the following two conditions: (i) A is locally indicable and the system (rx,... , rn)/s independent (ii) A is locally indicable, n = 1 and rl is not conjugate to an element of A. Wenow turn to the proof of Klyachko’s Theorem. The proof depends on the following non-evident but relatively elementary topological observations. Suppose we have a simply connected domain D on an oriented surface and a moving point - Klyachko refers to this as a car - on the boundary. The car is said to move properly around the domain if it moves along the boundary continuously in the positive direction with no stops, no reverses and passing through every point on the boundary infinitely often. If two or more such moving points meet, it is called a collision. The observations are then:

5.4 THE KERVAIRECONJECTUREAND KLYACHKO’SSOLUTION 133 LEMMA 5.4.1. Suppose there is an oriented 2-sphere and a finite connected graph on it which divides the sphere into a finite number of simplyconnected domains. Suppose, for each domain, there is a car moving properly on the boundary of each domain. Then there exist at least two points of the sphere where there is a complete collision. A collision is complete if at a point of multiplicity k, k cars collide simultaneously. LEMMA 5.4.2. Suppose a11 the conditions of Lemma5.4.1 are satisfied but stops are allowed. Then there will also exist at least two points of complete collisions provided the following condition holds: Suppose in a vertex V, n cars Cl, ..., c~, listed in counter-clockwise order, stop. Then cars ci, ci+l rood (n) are never simultaneously situated in V. The proofs of these lemmas are somewhattricky but otherwise relatively straightforward. Weleave them as exercises. Wealso need the following result of Howie [H 8]. LEMMA 5.4.3. (/H 8]) Suppose W~(t), i E I, is a system of equations over a group A and suppose that the natural map A-~< A,t : W~(t) = 1,i ~ I is not that (i) (ii) (iii) (iv)

a monomorphism.Then there exists a tesselation

of a 2-sphere such

The edges are oriented. The corners axe labelled with elements of A. The exterior vertex has non-trivial label. Each interior vertex has trivial label. "(v)" The/abe/of each face is equal to some Wi up to permutation and inverse.

From these lemmasthe following two additional results can be obtained: LEMMA 5.4.4. Let H be a group with P, P1 isomorphic subgroups of H with P1 = ¢(P) where ¢ is a given isomorphism. Suppose that ai, bi are elements of infinite order in H for i = O, 1, ..., k with the property that < ai,P >= *P,< hi,P1 7=< bi > *PI fori -- 0,...,k andsuppose c ~ H. Then the sy~em of equations k (H bit-Xa~t)ct -i~o {t-lpt

= ¢(p),p ~

is solvable over H. PROOF.If this system was not solvable over H then from Lemma5.4.3 we could construct a tesselation of the 2-sphere satisfying those conditions.

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ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Weshow that on an arbitray tesselation of this form we can determine a motion of cars satsifying the conditions of Lemma5.4.2 with fewer than two complete collision points. This contradiction will prove the result. A tesselation of the form of Lemma5.4:3 has three types of faces. We picture these in figure 5.4.1. Suppose each car moving around a face of type 1 (2) is situated at some initial time period at corner b0 (b~1) and move with velocity one edge per unit of time away from bo (b~1) but stay at corner c(c-1), the corner immediately prior to b0 for 2k - 1 time periods. For cars moving around faces of type 3 let them move without stops and with the same constant velocity and pass the corner ¢(p) at the inital time period.

Figure 5.4.1 Tesselation Face Types

It is easy to see that cars of different kinds never stop simultaneously and it is evident also that during the time intervals (0, 1) + 2/, 1 E Z, no car movesin the positive direction with respect to orientation of the edge and

5.4 THE KERVAIRE CONJECTUREAND KLYACHKO’SSOLUTION135

Figure 5.4.2 Source and Sink Vertices

Figure 5.4.3 Corners c and c-1 alternate - no stops in other corners during the intervals (1, 2)+2/, l E Z, no car movesin the negative direction. In particular there can be no collision outside vertices. If there is a collision in an interior source (or sink) vertex (see figure 5.4.2), it means a relation in < a~, P > (or < b~, ¢(P) >). This relation must be trivial and the tesselation is reducible. For vertices of other kinds it is clear that the schedule of cars stopping in such a vertex satisfies the conditions of Lemma 5.4.2. (figure 5.4.3) Therefore the only point which can be a point of complete collision is the exterior vertex. This contradiction then proves the lemma. For any group A consider B -- A* < t >. We use the notation A~ -t-~At ~ C B, P,~ the subgroup of B generated by (A~, 0 < i < m}, that is P,~ =< {A~},0 < i < m >B, R,~ =< (A~}, 1 < i < m >B and ¢ : P,~-I -~ R,~ the natural isomorphism p -~ t-lpt. LEMMA 5.4.5.

Let W ~ A * t and suppose the exponent sum oft in W

136

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

is one. Then W is qonjugate to a product k i~-0

where ~or some m each a~, b~, c e P,~ (1) each ai¢ Pro-1 (2) each b~ ~ R,~, PROOF.Clearly Wis conjugate to a product vt, v ¯ P~, v ~ Rs for some s. Choose the conjugate of Win this form such that s is minimal and the length of v as an element of Ps-~ *riB-1 Rs = P~ is minimal for this s. If we change each R~ factor h in the normal form of v by t-l¢-~(h)t, we will obtain a product satisfying the required conditions for m-- s - 1. PROOF.(Theorem 5.4.1) We now give the proof of the main result that is that the Kervaire conjecture is true for A a torsion-free group. In particular we assume G = A. < t > IN(R) and assume that each letter of R has infinite order in A. Consider R then as an equation W-- 1 with each coefficient having infinite order in A. Wecan assume that t has exponent sum 1 in Wfor otherwise G has a non-trivial cyclic quotient. Further from Lemma5.4.5 we can assume that W has form 5.4.1. The equation W-- 1 given in form 5.4.1 can be considered as an equation over H = P,~ = Ao * ... * Am the free product of m ÷ 1 copies of A. Let P = P,~_~ and ¢(P) -- R,~. Then from Lemma5.4.4 the system equations over H (5.4.2) W = 1, (t-~pt = ¢(p),p has a solution t in an overgroup H of H. The conditions of Lemma5.4.4 hold since if u ¯ (B ¯ C) \ B and u has infinite order, then < u, u>*B. Clearly W(t) = 1. Hence H is an overgroup of A containing a solution of the sytem 5.4.2 and therefore the map A -~ G is a monomorphism. It follows that G is non-trivial. Notice that the proof contained the following Freiheitssatz. COROLLARY 5.4.2. Suppose A is torsion-free and the exponent sum of t in R ¯ A, < t > is 4-1, then A is a FHS factor ofG = (A* < t >)/N(R) Weclose by noting that Klyachko’s paper contained several other results on equations over torsion-free groups. In particular., THEOREM 5.4.6. The sy~em of equations Iv(t), G] -- over a group G is solvable over G if every coefficient of v(t) has infinite order - in particular if G is torsion-free.

CHAPTER VI ONE-RELATOR

PRODUCTS

6.1 One-Relator

OF CYCLICS

Products of Cyclics

Wenow turn to the main focus of these notes - one-relator products of cyclics, their linearity properties and their generalizations. A one-relator product of cyclics is a one-relator product where each factor is a cyclic (possibly finite) group. Thus these are groups with presentations of the form (6.1.1)

V ~-~

Xl,...~xn,

.x ~1 -~- .... 1

x~,~=S m~ 1>

where n >_ 2, e~ -~ 0 or ei >_ 2 for i = 1, ..., n, S is a non-trivial cyclically reduced word, not a proper powerin the free product of cyclics on Xl, ..., xn and m _> 1. If m _> 2 then the relator R = S"~ is a proper power. As was the situation with one-relator groups and general one-relator products, the proper power case is somewhat easier to handle. If each e~ = 0, then G is just a one-relator group and hence the class of one-relator products of cyclics clearly generalizes the class of one-relator groups. Further each finitely generated co-compact Fuchsian group falls into this class via the Poincare presentation. Therefore the class of onerelator products of cyclics provides a natural algebraic generalization of Fuchsian groups. The next few chapters are primarily concerned with the following general question: General Program on One-Relator Product of Cyclics: Given an algebraic property true in all Fuchsian groups, must it be true in all onerelator products of cyclics. Further, if not, what specific conditions on the relator (if any ) are sufficient so that it will hold. In regard to this general program we concentrate on the linearity properties exhibited also by one-relator groups. In particular, the Tits Alternative, the virtual torsion free property, SQ-universality, amalgamdecompositions and separability properties. These will be taken up in subsequent chapters. In this chapter we consider the representation theoretic methods used in handling the class of one-relator products of cyclics. 137

138

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Since we allow torsion, the results on local indicability of factors do not apply. Further we do not restrict the relator to have a power higher than 4. Thus we go beyond what we have previously discussed. The basic technique to handle these groups is to consider certain special representations (defined in the next section) into some suitably chosen matrix group. For most applications this object group is PSL2(C_.). In the next section we examine these special representations, which are in a sense "close" to faithful. It is from the properties of the image group coupled with the special properties of these representations that our results will be derived. Of special interest is the case when n -- 2. The groups then have the form (6.1.2)

G =< a, b; an = bq .= Srn = 1 >

where as before S is cyclically reduced, not a proper power in the free product of cyclics on a, b involving both a and b. These groups are called the generalized triangle groups and clearly generalize the ordinary triangle groups (see Chapter 4). Wewill discuss them in detail in the next chapter. Historically the interest in one-relator products of cyclics, as separate from general one-relator products, arose out of a result of G.Baumslag, Morgan and Shalen [B-M-S] showing that if m _> 2 the resulting generalized triangle groups are always non-trivial. Their result grew out of a topological question which reduced to the group theoretical question of whether it is possible to kill a free product of two cyclic groups with a single high-powered relation. If the relator is a proper power, m _> 2, then almost all the linearity questions have a positive solution. Howeverif m = 1 and there is a generator of finite order { not all ei = 0} then almost nothing is true in general. (see [FH-R 1],[F-L-R 1,2]). Therefore specific restrictions must be imposed on the relator. The strongest algebraic analog (via presentations) of a Fuchsian group seems to be a group of F-type. This is a group with a presentation of the form (6.1.3)

e--< Xl, ...,xn;x~ ~" .....

Xenn = 1, UV = 1 >

where n _> 2,e~ = 0 or e~ _> 2, i = 1,...,n,1 _< p _< n-1,U = V(Xl, ...,Xp), V = V(Xp+l,..., Xn) with U a cyclically reduced word in the free product on xl, ...,xp which is of infinite order and V a cyclically reduced word in the free product on Xp+l,...,x,~ which is of infinite order. These groups are the analog for the class of one-relator products of cyclics of the cyclically pinched one-relator groups and satisfy manyof the algebraic properties of Fuchsian groups. Welook at these in detail in Chapter 8.

6.2 ESSENTIAL ANDESSENTIALLYFAITHFULREPRESENTATIONS 139 6.2 Essential

and Essentially

Faithful

Representations

The special representations needed in the study of one-relator of cyclics are defined in the following way. Suppose G is a group with the presentation (6.2.1)

G =< al,

..-,

an; a~ 1 .....

a~’~ ---- R~1 .....

products

R~~ = 1 >

where ei = 0 or e~ _> 2 for i = 1, ..,n, mj >_ 1 for j = 1,...,k and each R~ is a cyclically reduced word in the free product of the cyclic groups < al > ,..., < a,~ > of syllable length at least two. A representation p : G -~ Linear Group over a field of characteristic zero is an essential representation if for each i = 1, ..., n, p(a~) has infinite order if e~ = 0 or exact order e~ if ei >_ 2 and for each j = 1, ..., k, p(Rj) has order mj. For any group G a linear representation p over a field of characteristic zero is an essentially faithful representation if p is finite dimensional with torsion-free kernel. In much of the work on one-relator products of cyclics and related groups certain essential representations will be shownto be essentially faithful. For a finite presentation < X; R > the deficiency d is the difference between the number of generators and number of relators. For presentations of the form 6.2.1 we further define the extended deficiency d* to be n-k. Note that if the orders of the generators (al, ...,aN} and the additional relators {R~,..., R~}are exactly as they appear in the presentation, then an essentially faithful representation must be essential. On the other hand if the conjugacy classes of torsion elements are precisely given by the powers of the generators (if of finite order) or the powersof the other relators, then an essential representation is essentially faithful. In this section we consider some basic results on groups which admit essential and essentially faithful representations. Our techniques then will evolve into showing that a particular class of groups of interest do admit such representations. The first results are straightforward and handle the situation where the group G is virtually torsion-free. THEOREM 6.2.1. Let G be a finitely generated group. Then G admits an essentially faithful representation if and only if G is virtually torsion-free. PROOF.Suppose G is finitely generated and p : G --~ Linear Group is an essentially faithful representation. Since G is finitely generated, p(G) is a finitely generated linear group. From a result of Selberg [Se] p(G) is then virtually torsion-free. Let H be a torsion-free normal subgroup of p(G) of finite index and let H* be the pull-back in G. H* has finite index in G. If g ~ 1 has finite order, then p(g) has exactly the same order since p is an essentially faithful representation. Therefore g cannot be in H* since its

140

ALGEBRA|CGENERALIZATIONSOF DISCRETE GROUPS

image would then be an element of finite order in the torsion-free group H. Therefore H* must be torsion-free and G is virtually torsion-free. Conversely suppose G is virtually torsion-free. As explained above G must then contain a torsion-free normal subgroup H* of finite index. Choose a faithful finite dimensional representation p* of the finite group G/H*. The composition of this with the natural homomorphism from G to G/H*will give the desired representation. A modification of the proof of the Ree-Mendelsohn Theorem (Theorem 3.4.3) shows that any one-relator group with torsion must admit an essential representation. Coupling this with the torsion theorem for one-relator groups yields a proof of the following corollary due to Fischer, Karrass and Solitar (Theorem 3.4.4.) COROLLARY 6.2.1.

Any one-relator

group with torsion

is virtually

torsion-free. PROOF.Let G =< X; R~ > be a one-relator group with torsion so that R is not a proper power in the free group on X and n _> 2. From the ReeMendelsohn proof G admits an essential representation p : G ---* PSL2(C). Hence p(R) has exact order n. By the torsion theorem in one-relator groups each element of finite order in G is conjugate to a power of R. Since the image of R has exact order n the image of any power of R has exactly the same order in p(G) as it had in G and hence any element of finite order has its order preserved under p. Therefore no non-trivial element of finite order can be in the kernel of p. Therefore p is essentially faithful and G is virtually torsion-free.. For groups with presentation 6.2.1 and where each mj _> 2 and where the extended deficiency _> 3, the next, result shows that admitting an essential representation implies the existence of a subgroup of finite index mapping onto a non-abelian free group. This in turn implies SQ-universality. This result extends certain deficiency results of B.Baumslagand Pride [B.B.-P 2]. (G.Baumslag,Morgan and Shalen [B-M-S] and R.Thomas IT 1] proved similar results using different methods. Wepoint out that although this conclusion extends these other results, the hypotheses are somewhatstronger. Recall that a group G is SQ-universal if every countable group can be embedded in a quotient of G. If G has a SQ-universal quotient, then G is itself SQ-universal and similarly if G has a SQ-universal subgroup of finite index then G is itself SQ-universal (see[L-S]). From the work Higman,Neumannand Neumanna free group of rank 2 and hence any nonabelian free subgroup is SQ-universal. Weneed the following result due to B.Baumslag and S. Pride [B.B.-P 2]

6.2 ESSENTIALANDESSENTIALLYFAITHFULREPRESENTATIONS141 THEOREM 6.2.2. Let G be a finitely presented group of deficiency greater than or equal to 2. Then there exists a subgroup of finite index mapping onto a Tree group of rank 2. In particular G is SQ-universal. THEOREM 6.2.3. Suppose G is a finitely tation of the form (6.2.1)

G =<: Ial,...,an,

. a el .....

presented group with presen-

a~n = R~I

..

Rkm~= 1 >

with each Ri cyclically reduced in the Tree product of the cyclic groups < al >, ..., < a,~ >, of syllable length at least 2 and for i = 1, ...,n,e~ = 0 or e~ > 2. Suppose each m~ > 2 and that the extended deficiency d* = n - k > 3. Then if G admits an essential representation, G must contain a subgroup of finite index which maps onto a non-abelian free subgroup. In particular G is SQ-universal and contains a non-abelian Tree subgroup. The result is still true if d* ---- 2 but mj > 2 for somej or ei = 0 for somei or ei > 2 ]:or somei. PROOF.Suppose G has a presentation of form 6.2.1 as in the statement of the theorem. If a quotient of G contains a subgroup of finite index which maps onto a non-abelian free subgroup, then the group G does also. Hence we can, without loss of generality, assumethat ei _> 2 for i = 1, ..., n passing to a quotient if necessary. Suppose that mj _> 2 for j -- 1, ..., k and that G admits an essential representation p into a linear group V (over a field of characteristic zero) Therefore from Selberg’s theorem on finitely generated subgroups of linear groups [Se] there exists a normal torsion-free subgroup H of finite index in p(G). Thus the composition of maps ( where r is the canonical map) G --~P p(a) -~ p(a)/H gives a ¢ of G onto a finite group. 1 ..... a,~ * -- 1 > be the free product of the Let X =< al,...,a~;a~ cyclic groups < al >, ..., < an >. There is a canonical epimorphism/~ from X onto G. Wetherefore have the sequence X --~ G -~¢ p(G)/H Let Y -- ker(¢ o fl). Then Y is a normal subgroup of finite index X and Y is torsion-free. Since X is a free product of cyclics and Y is torsion-free, it follows that Y is a free group of finite rank r. Suppose [X : Y[ = j. Regard X as a Fuchsian group with finite hyperbolic area #(X). Every finitely generated free product of two or more cyclics can

142

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

faithfully represented as such a Fuchsian group if it is not isomorphic to the infinite dihedral group. From the Riemann- Hurwitz formula we have that j#(X) = #(r) where #(Y) -- (2~r)(r and #(X) -- (2~r)((n - 1) - (~

~- ~)).

Equating these expressions we obtain r=l-j(---+

.....

+-- - (n- 1)). en

G is obtained from X by adjoining the relations R~I, ..., R~ X/K, where K is the normal closure of R~ ..., Rk . Since K is contained in Y, the quotient Y/K can be considered as a subgroup of finite index in G. Applying the Reidemeister- Schreier process or repeated applications of Corollary 3 in [B-M-S] Y/K can be defined on r generators subject to (j/m1) + .... ÷ (j/mk) relations. The deficiency d of this presentation for Y/K is then J d--r J Since each ei >_ 2 and each mj >_ 2 we have the inequality d_>

l÷j(

n-k

2 1)

By assumption the extended deficiency n-k _> 3 and hence the deficiency d of the above presentation is at least 2. From the result of Baumslagand Pride, Theorem 6.2.2, Y/K, and hence G, has a subgroup of finite index mapping onto a free group of rank 2. Therefore Y/K is SQ-universal and since this has finite index in G, G is also SQ-universal. Further an SQuniversal group must contain a non-abelian free subgroup. This completes the main parts of the theorem. If the extended deficiency is 2 but not all mj -- 2 or not all e~ -- 2, then the inequality above becomes proper - that is: n-k d>l÷( ~ 1)=1. Hence d > 1 and since d is an integer d _> 2 and the proof goes through as before.

6.2 ESSENTIALANDESSENTIALLYFAITHFULREPRESENTATIONS143 This type of argument using deficiency and the Baumslag-Pride result was used by G.Baumslag, Morgan and Shalen in their study of generalized triangle groups. Wereturn to this in the next chapter. Theorem6.2.3 is closely related to the following result of R.Thomas[T1] proven by looking at the Cayley graph. THEOREM 6.2.4. (6.2.3)

Let G be a group with a presentation

G --(

Xl,...,xn,

W~ .....

Wrmr-~ l >

where W~axe non-trivial words in the generators xl,...,x~ and ml,..,mr are integers _> 2. Suppose that W~, ..., Wr have exact orders m~, ..., mr respectively in G. Let a = (l/m1) + ... q- (1/mr). H G is finite then a - n÷ 1 > 0 and IGI >_ 1/(a - nq- 1). In particular this is true fig admits an essential representation. PROOF.If F is a graph, a walk in F consists of an alternating sequence Po, 11,p1,12, ...,p,~-~,In,p,~ of points pi and edges li such that li is incident with p~_l and pi. The walk is closed if P0 = P~ and open otherwise. If there is a closed walk with n > 2 distinct points Po, ...,Pn, then the walk is called a cycle. The cycles of F under the operation of symmetric difference span a vector space over the field of two elements knownas the cycle space of F and if F is a connected graph with p points and q edges then this vector space has dimension q - p + 1. Werefer to the book by Harary. [Har] for these results and terminology. Nowlet G be a group with finite presentation < X; R > where X and R are finite sets of respective orders n and r. Let Y denote the elements of X together with their inverses, elements of order 2 occurring twice so that [Y[ = 2n. The Cayley graph of G is defined as follows. The vertices of F are the elements of G and there is an edge {a, b} in F if and only if the elements a, b E G satisfy ay = b for some y E Y. The edges of F are undirected and each pair of points a, b ~ F is connected by as many edges as there are elements y ~ Y with ay = b; thus there may well be multiple edges in F. A trivial generator would give rise to a pair of loops at each point of F. Nowsuppose that G is a finite group of order m. The Cayley graph as defined above then has m points and mn edges and is regular of degree 2n. If g ~ G and w is a word in the elements of Y which equals the identity in G, then the edges of F starting at the point g and corresponding to the elements of w form a closed walk in F. Thus each relator gives rise to m closed walks in F although there may well be repetitions amongthese. From [Har] we have that since F is connected with m points and mn edges, the cycle rank of F equals mn - m + 1 {[Har], corollary 4.5}. Since

144

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

each relator in G can be defined from R, we have that the r closed walks at each point of F corresponding to the elements of R span the cycle space of Fandhencethatmn-m+l<_mrorn-l<_r-1/n
1 ..... x~; W~

WF~ = 1 >.

If each W~ has exact order m~ in G, then of the m closed walks in ~ starting at each point of F, each F corresponding to the relator W~ is repeated at least mi times, so that we now have a set of at most m/ml + m/m~ + ... + m/m~ = rn~ cycles spanning F. It follows that ms > mn-m+l or ce >_ n-l+l/m. Therefore c~-n+l > 0 and m > 1/(a - n + 1) as required. If G admits an essential representation, then the orders of the elements Wi are exactly mi. Wenote that it is not always possible to find an essential representation. Consider the group G defined by < a, b; a3 = b12 = (ab) 2 = (a-lb) 13 = 1 >. If G had an essential representation~ it would be infinite from the last theorem. However a-~b = aab = ab-la -1 so that (a-lb) 12 = ab-12a-1 = 1. It follows that a-Xb = 1 so that a = b and hence a3 = aa = 1 and so a = 1. Thus G is the trivial group. 6.3 Essential

Representations

of One-Relator Products

Wenow show that every one-relator product of cyclics with proper power relator admits an essential representation into PSLg.(C). From this we can deduce easily that these groups are never trivial and in most cases infinite. Our main result is the following concerning one-relator products whose factors admit faithful representations into PSL~(C). THEOREM 6.3.1. Let G = (A * B)/N(R "~) where A and B are groups admitting faithful representations into PSL2 (C) and where R is a cyclically reduced word in the free product A * B of leng~h at least 2 and m >_ 2. Then G admits a representation p : G --* PSL2(C) such tha~ PlA and PlB are faithful and p(R) has order m. In particular A -~ G and B -~ G are injective, that is, the b-~eiheitssatz holds. Before proving this result we will prove a special case of it (Theorem 6.3.2) where each factor is a free product of cyclics. The proof of this theorem is based on the technique of Ree and Mendelsohn (Theorem 3.4.3)

6.3 ESSENTIAL REPRESENTATIONS OF ONE-RELATORPRODUCTS 145 and is a refinement of a technique used by Baumslag, Morgan and Shalen [B-M-S] who used it to prove that all generalized triangle groups are nontrivial. Wewill return to the Baumslag, Morgan and Shalen results in the next chapter when we discuss the generalized triangle groups. Wefirst note that Theorem 6.3.1 has the following corollary which greatly extends the classes of factors for which a Freiheitssatz holds in a one-relator product. COROLLARY 6.3.1¯ Let G = (A * B)/N(R "~) where R is a cyclically reduced word in the flee product A * B of length at least 2 and m >_ 2. Then the t~eiheitssatz holds whenever both A and B are in any of the following classes of groups: (1) Fuchsian groups (2) Kleinian groups (3) Cyclic groups (4) Surface groups (5) Free abelian groups of rank 2 (6) Free metabelian groups of rank 2 (7) Cyclically pinched one-relator groups with malnormal cyclic amalgamated subgroups in both factors (8) Mixed combinations of any of the above The corollary is a consequence of the fact that any group in the classes cited admit faithful representations in PSL2 (C). THEOREM 6.3.2. Suppose G is a one-relator product of cyclics proper power relator. That is, G has a presentation V -~< al,...,an;a~’

= a~2 .....

with

a~"~ = Rm(al,...,an) -~- 1

with n >_2, m >_ 2,ei = 0 or e~ >_2 for i -- 1,...,n and R(al,...,a,~) is cyclically reduced word in the ~ree product on al,..., a,, which involves all the generators. Then G admits an essenital representation into PSL2(C) which is faithful on the free product on a~, ..., a,~-l. In particular ~ 1 al~...~an-1 that is < al, ...,a,~-i

~--~ al,a ¯ el -- 1 > *..-* ( an-l; ae"-I -- 1 > > is the free product ofcyclics of the obvious orders.

PROOF.(Theorem 6.3.2) From our results on Fuchsian groups (Chapter 4 ) we can choose projective matrices A1, A2, ..., A,_I E PSL2(C)such that the subgroup generated by these is the appropriate free product of cyclics. That is < A1,...,A,~_~

>-~< A1, Ai = 1 > *...*

< A,~_I;An_1 -~ 1 >

146

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

so that < A1, ...,An-1 > faithfully represents the free product of cyclics e~-~ ~ 1 >. ~ al, ...~ an-l; a1 ~ a2 e~~- ... = an-1 We will determine a projective matrix A,~ E PSL2(C) of order e~ so that the subgroup generated by A1, ...,A,~ provides a representation p : G -~ PSL~(C) with p(ai) = Ai for i = 1, ...,n and p(R) having order m. The representation will thus be essential. Further, since the image of < al, ..., a,~-i > will then be a free product of cyclics with each of maximal possible order, it follows that p restricted to the subgroup generated by al, ..., a,_l is faithful and this subgroup is the appropriate free product of cyclics. Choose

where t = 2cos(~/e,~) and w is to be determined. Since tr(A~) = t 2cos(~/en) we have A~- = 1 in PSL2(C). Consider the relator word R(al, ..., as). Substituting A~, A2, ..., A~ into R we obtain the projective matrix

wheref~, f2, f3, f4 are polynomialsin the coefficients of A1, As, ..., A,~. Considering w as the only unknown,f~, f~, fa, f4 are then polynomials in w. If tr(R(A~, ..., A~)) = f~ + is not a co nstant poly nomial in w , t hen the polynomial equation

+

= cos( /,n)

can be solved for w by the Fundamental Theorem of Algebra. For a particular solution wo in A~ we would then have R’~(A~, ..., A~) 1 because of the trace. Therefore the subgroup generated by A1,..., A,~ provides an essential representation p: G -~ PSLu(C~) under p(a~) = A~,i 1, ..., n. F~her by the choice of A~, ..., A~_~we have that al, .., a~_~ is the appropriate ~ product of cyclics and p restricted to the subgroup < a~..., a~-i > is faithful. What is le~ in order to complete the proof is to show that there is a choice of A~, ..., A~_~such that tr(R(A~, ..., ~)) is nonco~tant in Since R(a~, a:, ..., a~) is cyc~cally reduced, without loss of generafity we can ~ite the relator matrix R(A~, ..., A~) R(A~, A2,...,

A~) = B~A~~ B:...BkA~ ~

with B~ non-trivial words in A~, A2, ..., andl~t~<e~.

A~_~and hence non-trivial ~trices

6.3 ESSENTIAL REPRESENTATIONS OF ONE-RELATORPRODUCTS147 1

has lower left entry c + ya - yd - y2b which is non-zero for all but finitely many choices of y. Similarly for B2, ...,Bk. Then by conjugating all the B1,...,Bkbyasuitable(ly ~) if necessary, we can assume that the lower left entries of B1,..., Bk are non-zero. By considering the diagonalization of An we see that A~ has the form

where gl,g2,ga, g4 are polynomials in w of respective degrees 1, 2, 0, 1. It follows that each factor B~A~in the expression for the relator R must have the form

where hi, h2, ha, h4 are also polynomials in w. Since each B~ has non-zero lower left entry, we must have that the degree of h4 is exactly 2 while the degrees of h~, h2, ha are <_ 1, <_ 2, _< 1 respectively. Using an induction on k _> 1 we can conclude that in R, fl and f3 are of degree < 2k, f2 is of degree _< 2k and f4 is of degree exactly 2k. This guarantees that tr(R(A1, ..., An)) is nonconstant in w completing the proof. As a consequence of the theorem we get the following corollary which was proved by Baumslag, Morgan and Shalen in response to a topological question (see [B-M-S]). COROLLARY 6.3.2. Le$ G =< a,b;a p = bq -- R’~(a,b) generalized triangle group. Then G is non-trivial.

-- 1 > be

Since m >_ 2 in the definition of a generahzed triangle group, the corollary follows directly since the non-trivial cyclic group on a injects into G. Baumslag, Morgan and Shalen proved more, depending on the values of p, q, m. Wewill return to this in the next chapter. Wenote that Corollary 6.3.2 and a version of Theorem 6.3.2 were proved independently by S.Boyer /By], who used essential representations into SU(2) rather than into PSL~(C). By extending the proof of Theorem 6.3.2 we obtain a proof of Theorem 6.3.1.

148

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

PROOF.(of Theorem 6.3.1) Let G = (A * B)/N(R "~) where A and B are groups admitting faithful representations into PSL~(C,) and where R is a cyclically reduced word in the free product A * B of length at least 2 and m _> 2. Write the relator as R = albl...akbk

with ai E A, bi ~ B.

Since R is cyclically reduced of length _> 2, we may assume that ai ~ 1 and bi ~ 1 for all i. Choose faithful representations PA : A --} PSL~(C) and PB : B PSLu(C,) such that (after a suitable conjugation if necessary), pA(ai)

= ( : x~ wit hxi~0

pB(bi) ---- ( ** Yi wit hyi~0 for all i -- 1,..., k. w E C let p] denote the representation

/ PA conjugated by (~

That is

Considering w as a variable define T(w) = tr(p~4 (al)pB (bl)...p~4 (ak)pB (bk)). Then T(w) is a polynomial in w. As in the proof of Theorem 6.3.2 the coefficient of wek is equal to ±xlyl...xkyk and is therefore non-zero because of the choices of #A and pB. Then as in the proof of Theorem 6.3.2 there exists a w0 with T(Wo) = 2cos(r/m). Now define the representation p : G -~ PSL:(C) by PlA = #A and PiB = pB. Then p(R) has trace T(wo) - 2 cos(r/m) and hence p(R) has order m. Furt~h~er PlA = p]O is faithful on A and PIB = PB is faithful on B and therefore p is the desired representation. Since a one-relator product of cyclics with proper power relator has now been shown to admit an essential representation, the results, appropriately phrased, of section 6.2 can be applied. This will be the focus of much of the remainder of these notes. For the rest of this chapter we look at several other results on essential representations.

6.4 FREIHEITSSATZ FOR ONF~-RELATORAMALGAMATED PRODUCTS149 6.4 Freiheitssatz

for

One-Relator

Amalgamated Products

In analogy with the one-relator products we define a one-relator gamated product as a group of the form

amal-

G = (A *c B)/N(R) where R is a non-trivial, cyclically reduced element of length at least two in the amalgamated product A *c B. In this section we generalize the results and techniques of the previous two sections to obtain a Freiheitssatz for a certain class of one-relator amalgamated products. In particular suppose that G = (A *c B)/N(S "~) with C cyclic and m >_ 2. If (A *c B) admits a complex two-dimensional representation which is faithful on A and B, then under certain conditions, this representation can be extended to a representation p : G -~ PSL2(C)which is also faithful on A and B. It will then follow that A a.nd B both inject into G. Our proof is related to ~ result of Vinberg [V]. Once we establish this Freiheitssatz, we can prove a series of results about these one-relator amalgamated products mirroring some of the previous material on one-relator products. If A, B E PSL2(C), then we say that the pair {A, B} is irreducible if A, B, regarded as linear fractional transforlnations; have no commonfixed point, that is, tr [A, B] ¢ 2. The main result of this section is the following which gives a Freiheitssatz for certain one-relator amalgamatedproducts. THEOREM 6.4.1. { The Freiheitssatz } Suppose H = H1 *A H2 with A =< a > cyctic. Let R ~ H\A bc given in a reduced form R = alb~...a~b~ with k >_ 1 and ai ~ Hi \ A, bi ~ 2 \ A for i = 1, ...,k. As sume that th ere exists a representation ~ : H -~ PSL2(C) such ~hat ¢IH~ and ¢lt~ are faithful and the pairs {¢(a~), ¢(a)} and (¢(b~), ¢(a)} irre ducible for i =1, ..., k. Then the group G = H/N(R~), m >_ 2, admits a representation p : G -~ PSL2(C) such that Ht -~ G ~-~ PSL2(C) and H2 -~ G ~-~ PSL2(C) are faithE~1 and p(R) has order m. In particular Ht -~ G nnd H2 -~ G are injective. PROOF.The proof generalizes somewhat the proofs of Theorem 6.3.1 and Theorem 6.3.2. Let ¢ : H -~ PSL2(C) be the given representation of H such that ¢1H1and ¢IH~ are faithful and the pairs {¢(a~), ¢(a)} {¢(b~), ¢(a)} are irreducible for i = 1, ..., We may assume that ¢(a) has the form ¢(a)=

0)

s_1 or

150

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS ¢(a) ---- (~

Suppose first that ¢(a)

1

(: 0)

s_ 1 and let

-1 T---

0 t

with t a variable whose value in C is to be determined. Define (1) p(hl) ¢(hl) fo r hi e H1 and (2) p(h2) --- T¢(h2)T-~ for h2 E H2. Since T commutes with ¢(a) for any t, the map p : G -~ PSL~(C) will define a representation with the desired properties if there exists a value of t such that p(R) has order m. Recall, as in the last section, that a complex projective matrix A in PSLu(C) will have finite order m >_ 2 if trA = :]=2cos(~r / m). As in the statement of the theorem assume R = albl...akbk with k _> 1 and a~ E H1 \A,b~ E H2 \A for i = 1,...,k, and assume that the pairs {¢(a,), ¢(a)} and {¢(b~), ¢(a)} are irreducible for i = 1, ..., k. f(t) = tr(¢(a~)T¢(bi)T -1 .... ¢(ak)T¢(bk)T-~). f(t) is a Laurent polynomial in t of degree 2k in both t and t -1. The coefficients of t 2k and t -2k are non-zero because the pairs {¢(a~), ¢(a)} {¢(b~), ¢(a)} are irreducible for i -- 1, ..., k. Therefore by the fundamental theorem of algebra there exists a to with f(to) = 2cos(n/m). With this choice of to we have tr(p(R)) -- 2cos(~r/m) and thus p(R) has order m. Therefore p is a representation with the desired properties. Now assume ¢(a)

. In this

case define

T=0

again t a variable. Again t commuteswith ¢(a) and the proof goes through as above with tr(p(R)) polynomial in t o f degree 2k (se e the proof of Theorem6.3.1) giving the desired representation. Exactly the same proof as that used for Theorem 6.4.1 can be employed to prove the following two generalizations. THEOREM 6.4.2. Le$ H = H1 *A H2 with A = H~ AH~. Let R E H\ be given in reduced form R = albl ...akbk with k ~_ 1 and a~ ~ H1 \ A, b~ H2 \ A for i = 1, .., k. Assume that ~here exists a representation ¢ : PSL2(C) with the properties that ¢IH~and ¢lu~ are faithful, for each a ~ A ~here exists

% ~ C, such that

¢(a) 0= (% ~ ~ or for

eachaeAthere

6.4 FREIHEITSSATZ FOR ONE-RELATORAMALGAMATED PRODUCTS151

each a E A, a ¢ 1, the pairs (¢(ai), ¢(a)} and (¢(bi), ¢(a)} irre ducible. Then the group G = H/N(R’*), ra >_ 2, admits a representation p : PSL2(C) such that H1 --* G ~-L PSL2(C) and H2 -~ G P-~ PSL2(C) are faith[ul and p(R) has order m. In particular H1 --~ G and H~ --* G are injective. THEOREM 6.4.3. Let H -- HI *A H2 with A = H~ NH~. Let R ~ H\ be g~ven in reduced [orm R = alb~...akbk with k >_ 1 and H2 \ A [or i -- 1, ..., k. Assumethat there exists a representation PSL2 (C) with the properties that ¢]H1 and ¢1H2are [aith[ul, [or each a ~ A

there

existsx~O,y~O,

such that

¢(a~) ----

xi

’

Yi * "

Then the group G -= H/N(R’~), m >_ 2, admits a representation p : PSL2(C) such that H~ -~ G P-~ PSL2(C) and H2 --+ G P-~ PSL2(C) are [aithful and p(R) has order m. In particular H1 -~ G and H2 -+ G are injective. The proof of Theorem6.4.2 is exactly the same as the proof of Theorem 6.4.1. The conditions on the images of the elements of the amalgamated subgroup again force the trace polynomial to be non-constant. In Theorem 6.4.3 the form of the images of the elements which appear in the relator force these images to be irreducible in pairs with the images of the nontrivial elements of the amalgamated subgroup. Again this forces the trace polynomial to be non-constant and the proof goes through as in theorem 6.4.1. Weclose this section with a general question related to the above. QUESTION.Suppose H = H~ * a H2 with A = H~ ~ H2. Suppose there exist [aithf~l representations ¢i : Hi -~ PSL2(C) [or i 1, 2. Under what conditions on H~, H2, A may we construct a representation ¢ : H --~ PSL2(C) such that ¢IH~ and ¢1~ are Wenote that it is possible to construct such a representation in the following cases: (1) H~,H~are free and A is cyclic. (2) H1, H2 are free products of cyclics and A is cyclic. (3) Each Hi, i ---- 1,2 is either free, a free product of cyclics or a nonelementary Fuchsian group with positive dimensional Teichmuller space and A is cyclic. (4) A is finite cyclic

152

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

A general criterion to answer the question is given by P. Shalen in [Sh] and generalized by S.Litz to GL2((2) [Lit]. Finally we note that if H1, H2 embedinto PSL2 (C) then the free product H1 * H~ also embeds into PSL~.(C). In later sections we will return to the representation of one-relator amalgamated products to consider certain linearity properties in such groups. 6.5 Faithful

Representations

of One-Relator

Products

Suppose that G = (A * B)/N(R) is a one-relator product where the relator R is a non-trivial word of syllable length at least two in the free product A * B. Wehave seen that if A and B admit faithful representations into PSL2 (C) and R is a proper power, then G admits an essential representation into PSL2 (C). This result and the techniques of the previous sections in this chapter then suggest the following two questions. (1) Under what conditions on the relator R will the one-relator product G admit a faithful two-dimensional complex representation? (2) Under what conditions on the relator R will the one-relator product G admit a two-dimensional complex representation which is faithful with a discrete image? In this section we describe some partial answers to these questions. Wefirst survey what is knownabout faithful discrete representations into PSL~ ((2). Wethen prove a necessary condition on the relator, based on a classical result of Magnus, for a two generator one-relator group to admit a faithful discrete representation into PSL~(C). Using cancellation arguments in free groups we then describe conditions on words R(x, y) in a free group on two generators x, y which allow them to meet the above condition. Recall from Chapter 4 that a finitely generated non-elementary Fuchsian group F has a Poincare presentation of the form (2.1) F -----< el,..,ep,

h~,..,h~,a~,b~,...,ag,bg;e~’ = 1, i ----- 1,..,p,R = 1 >

where R -- e~..%hl..h~[al,bl]...[ag,b~] andp> O,t > O,g > O, pq-t q-g > O, and m~_> 2 for i ~ 1, ...p. Hence as mentioned Fuchsian groups fall into the class of one-relator products of cyclics. The Euler Characteristic of F is given by x(F) = -I~(F) where p(F) -- 2g - 2 -b t q- ~ (1 1/ mi). If p(F) > 0 then 2~r#(F) represents the hyperbolic area of a fundamental polygon for F and a presentation of the above form can be represented by an actual Fuchsian group and hence admits a faithful~ discrete representation. Recall that a group of F-Type is a one-relator product of cyclics,with a presentation of the form (2.2)

G =< al,. .... ,an, ¯ ael1 .........

a~n -~ 1, UV= 1 >

6.5 FAITHFUL REPRESENTATIONSOF ONE-RELATORPRODUCTS 153 wheren _> 2,ei --- 0 or e~ _> 2, 1 _< p _< n- 1, U = U(al, .., ap) is a cyclically reduced word in the free product on al, ..., ap which is of infinite order and V -~ V(ap+l, ...,an) is a cyclically reduced word in the free product on av+l, ..., a,~ whichis of infinite order. In Chapter 8 we will prove that if neither U nor V is a proper power in the free product on the generators which they involve, then the group G admits a faithful two-dimensional complex representation. This uses a result of P.Shalen [Sh] from which we can deduce that cyclically pinched one-relator groups with malnormal cyclic amalgamated subgroups in both factors admit faithful two-dimensional complex representations. Whether the image group is discrete or not depends on the further exact structure of R. If both U and V axe proper powers then we will see that there is no faithful representation. In general a faithful, discrete representation p : G --~ PSL2(C)of a group G is said to be of finite volttme if Ha/p(G) has finite volume where Ha is hyperbolic 3-space. Helling, Kim and Mermicke [H-K-M]have shown that if m _> 4 the group G =< a, b; am = b2 = ((a-lb)2(ab)a) 2 = 1 > has a faithful, discrete image in PSL2(C). Further Helling, Mennicke and Vinberg [H-K-V] show that the groups G :-<: a, b; a~ = bI = (aba-lbab-~) ’n = 1 > with at least one of k, l, m-- 0 or k, l, m>_ 2 and with k <_ l have a faithful, discrete representation if at most one of k, l, mis 2 and if (k, l, m) ~ (2, 3, 3). Moreover group G has a faithful discrete representation of finite volume if 2 < k < l and (l/k) + (1//) + (l/m) >_ 1. In connection with these results it shown that the groups G --< a,b;a a = b3 = (aba-lbab-~) ~ = 1 > and G =< a, b; a3 = b4 = (aba-~bab-1) 2 = 1 > axe arithmetic. In a similar manner Hagelberg [Ha] and Hagelberg, Maclachlan and Rosenberger [HaMc-R] showed that the groups G =< a, b; aa = b~ = [a, b] "~ = 1 > with at least one of k, t, m= 0 or k, t, m_> 2, and k <_ t have faithful, discrete representations of finite volumeprecisely when(k, t, m) --- (3, 3, 3), (3, 4, (4, 4, 2) and that the groups G -~< a, b; ak = b~ = (a-~bab-~ab-~a-lb)m 1 > with k,t,m >_ 2 and k <_ t have faithful discrete representations of finite volume if (l/k) + (l/k) + (l/m) _> 1 and (l/t) + (l/t) + except for (k, t, ra) = (2, 2, m) and (2,3,2). In addition Hagelberg, Maclachlan and Rosenberger proved that for (l/k) + (l/k) + (l/m) _> (1/t)+(1/t)+(1/m) _> 1 the groups G = with (k,t,m) (3, 3, 3)~(3, 4, 2 ) o r ( 4,4,2) are axit hmetic. If (k,t ,m) = (3,3 ,3) the group G is a subgroup of index four in the Bianchi group PSLu(O3) while if (k, t, m) -- (3, 4, 2) or (4, 2), G is comme nsurable with the Picar d group PSL2(O~). On the other hand if G --< a, b; ak = b~ -- [a, b]"~ 1 > with at least one o~ k,t,m ---- 0 or k,t,m >_ 2, and k _< t, G has a faithful, discrete representation into PSL~(C) if and only if (k, t, m) (~, 3, ~), (~, ~, ~), (3, 3, ~), (~, The above examples seem to lead to a general necessary condition for

154

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

a generalized triangle group G to have a faithful, discrete representation into PSL2(C.) of finite volume. In [Ha-Mc-Ro]there is the following partial result. THEOREM 6.5.1. Let G =< a,b;a p = bq -~ Rm(a,b) = 1 > with p : or p > 2, q = 0 or q > 2 and m > 2 and R(a, b) a cyclically reduced word, not a proper power, in the free product on a, b which involves both a and b. Suppose one of the following holds: (1) m_>4 (2) m = 3 and the word R(a, b) does not involve a letter (with respect to the free product on a and b) oT order 2. Suppose/ur~her that G has a Taithful, discrete representation into PSL2 (C) of finite volume. Then p > 2,q > 2 and (l/p) + (l/q) + (l/m) > 1. A generalization of this theorem would be of great interest. In the special case of finitely generated one-relator groups we can directly obtain such a generalization using results of ChiswelI[Ch] and Ratcliffe IRa] on Euler characteristic. THEOREM 6.5.2. Let G =< al ..., n,R (al, ..., an) 1 > with n 2, m _> 1 and R(al, ..., an) a cyclically reduced word, not a proper power in the free group on al, ...,an involving all the generators. Suppose that G has a TaithIul, discrete representation into PSL2(C)oT finite volume, then n = 2 and m = 1 - that is, G is a torsion-free, two-generator, one-relator group. PROOF.From [Ch] G has an Euler characteristic given by x(G) = l-n+ (l/m). From IRa] if G has a faithful,discrete representation into PSL2(C) of finite volume then the Euler characteristic must be non-nesativ% hence n= 2 and m= 1. The result of theorem 6.5.2 leads us to consider the problem of classifying all the torsion-free two-generator one-relator groups which admit a faithful, discrete representation into PSL2(C) of finite vohtme. There are many known examples of such groups with this property, for instance G a, b; aba- l bab-1 = 1 >. Wenow consider the case of a two-generator one-relator group G with presentation (6.5.1)

G =< a, b; R’~(a, b) = 1

where R(a, b) is a non-trivial cyclically reduced word in the free group on a, b involving both a and b and m _> 1. in general there are no faithful representations into PSL2(C). For example it can be shown that the group

6.5 FAITHFUL REPRESENTATIONSOF ONE-RELATORPRODUCTS 155 H =< a, b; atbs = 1 > with s, t _> 2 has no faithful representation. However from certain special properties of complex projective matrices coupled with a property of conjugates in free groups due to Magnuswe get the following necessary condition for such a group to admit a faithful two-dimensional complex representation. This result was proved in a slightly different manner by Magnus [M 4]. THEOREM 6.5.3. Let G be a two-generator one-relator group with form (6.5.1) and suppose G is non-metabelian. If G admits a faithful representation into PSL2(C), then the relator R(a,b) must satisfy the property that the word R(a, b) is conjugate in the £ree group on a,b to the word R+l(a-l,b-1). PROOF.Assumethat there is a faithful representation p : G -~ PSL2(C) with a --* A, b -~ B. From the properties of complex projective matrices it is known that there is a projective matrix C with CAC-1 = A-~ and -~ = B-~. Therefore we must have that R’~(A-~, B-~) = 1. Because CBC p is faithful, we then must also have Rm(a-~,b -~) = 1 in G. Since G is a finitely generated linear group it is residually finite and Hopfian and therefore the mapa --~ a-~, b --~ b-1 defines an automorphismof G which is induced by a Nielsen transformation. From a result of Magnus{see Chapter 3} R(a, b) must be conjugate in the free group to R+l(a-1, b-l). Recall that in a free group F if X~n is conjugate to Y±’~, then X is conjugate to Y±I. Wenote that Theorem 6.5.3 does not hold ifG is metabelian. The group G(n) = where n >_ 2 has a faithful representation into PSL2(C) given by a-~A--

(0 0 ) (01 (v~)_ ~ ,b~

but bab-la -n is not conjugate to (b-la-lban) ±1 in the free group on a, b. From standard cancellation arguments in free groups we can further describe conditions on words R(x,y) in a free group on two generators x,y which allows them to be conjugate to R±~(x-l,y -~) and thus can be permissible relators in a two generator one-relator group which admits a faithful two dimensional complex representation. First, in what follows, we give some notation. Suppose F is a free group on a,b. If s,t -- 4-1 then U(aS,b t) means U(aS,b ~) = 1 or U(a~, bt) -~ a~’btl’...aSe’btl~ for some natural number r and with non-zero integers. If U(aS,bt) ?~ 1 then U-l(a~,b t) clearly means that -~. Analogously wc use the notation u-l(a s, bt) = b-~1~a-~...b-tl~a ~, U(b at), s, t = 4-1.

156

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

THEOREM 6.5.4. Suppose F is a free group on a, b and let 1 # U = U(a,b) be an elemen~ off or 1 ~ U = U(b,a) be an element If U =- U(a,b) assume that there is a V in F with V-1U(a,b)V U-t(a-l,b-1). If U = U(b,a) assume that there is a V in F V-1U(b,a)V = U-t(b-~,a-~). Then, possibly after replacing U by u-l(a-l,b -1) = U*(b,a) ifU = U(a,b) or U by U-l(b-l,a-1) = U*(a,b) iT U = U (b,a) - one of the following cases holds: (1) U = U(a, b) and a-*U(a, b)a* = U-l(a-l,b -1) for some non-zero integer t (2) U = U(a,b) = a~U~(b,a)bSa~U2(b,a)b q for some non-zero integers t,s,q and b-sUi(b,a)b ~ -- u~l(b-i,a -~) and b-qUu(b,a)b q = -1) u~l(b-l,a (3) U =- U(b,a) and b-tU(b,a)b ~ = U-i(b-i,a -~) for some non-zero integer t (4) U = U(b, a) = b~U~(a, b)aSb~U2(a, q for somenon-z ero integers t, s, q and a-sUl(a, b)a~ -- u~-l(a -1, 5-1) and a-qU2(a, b)aq --u~l(a-l,b-1). PROOF. Suppose, without loss of generality, that 1 ~ U = U(a,b) = a~’bA...a~rbl~,r >_ 1, with ei, fi non-zero integers. Let V = adtbkl ...ad"bk~a d’~+~with all ki non-zero integers, di a non-zero integer for i = 2, .., n and dl and d,,+l integers. Let L denote free product length in the free product decomposition F =< a > ¯ < b > of F. Cancellation arguments with respect to L give immediately that d~ ~ 0 if and only if d~+l ~ 0. If d~ = d~+t = 0 then V = b~ ...ad"b k" and since U(a, b)V = VU-~(a-~, b-~) we have U-~(a -~, b-~)V -~ = V-~U(a, b) and so bl~ a~...b l~ a~ b-k~...b -k~ -~ b-k~...b -k~ ae~ b~. ..a~b ~. Therefore here we may replace V(a,b) by U-l(a-~,b -~) = W(b,a) (and Y by -1 ) . H ence we may assume that d~ ~ 0 ~ d,~+l ¯ Then ~ = ad~bkl...ad-bk~ad~+~b$~ae,...b$~ae~ aelbl~...ae,bl~ ad~bk~...ad~bk-ad-+ with d~ ~ 0 ~ dn+l ¯ If n > r then from the above equation V = V1V2 with V1 = U(a,b) and U(a,b)V2 V2u-i(a -1, 5- 1). He re we canrepl ace V by V2. Therefore we may assume that r _> n. -~, b-~) the first of the two possibiliIfn = 0 then U(a, b)ad~ = ad~U-~(a ties. Nowassume that r _> n _> 1. If r -- n then again from the above equation V = U(a, b)a d~+~ and therefore U(a, b)a d~+~ = ad-+iU-~(a-1, b-i). Now assume that r > n _> 1. Then U(a,b) = Wl(a,b)W2(a,b) with V = W~(a,b)ad~+t,d~+~ -- e,+l and we get W2(a,b)W~(a,b)a d~+~ = ad"+~W~(a-i,b-~)W~(a-i,b-~). Let W~(a,b) = a~U~(b,a)b ~ and $. W2(a, b) ---- adU2(b, Wethen get Ul(b, a)bkd~+~-~ bk U~l(b -1, a-l)a e.

6.5 FAITHFUL REPRESENTATIONSOF ONE-RELATORPRODUCTS 157 Hence e -- dn+l and Ul(b, a)b k = baU~t(b-t, a -1) . Analagously d = d~+l and U2(b, a)bI ~- blUf l(b -~, a-t). This completes the second possibility in the theorem. After possibly exchanginga and b if U(a, b) is conjugate to U-1 (a-1, b-1) we are left with the situation 1 ~ U = U(a, b) = ~1 bIt . ..aerb Ir , r >_1, with ei, fi non-zero integers and U ( a, b) d =adU- 1 ( a-t, b- t) forsome non-zero integer d. If r = 1 then we have the equation a~lblta ~1 = a*lblta ~1 so U(a,b)a d = adU-t(a-l,b -1) with d = el. Nowlet r _> 2. Comparing the exponents in the above equation leads to the next result. THEOREM 6.5.5. Let 1 # U = U(a,b) = aelblt...a~rbI~,r _> 2, with ei, f~ non-zero integers. Suppose U(a, b)a d = adU-t(a -~, b-t) /=or some non-zero h~teger d. (1) Hr = 2s _> 2 is even then U(a, b) = ~t bit.., a ~ bI~ a~ +t bI. a~¯ bl. - ~ .. ... (2)

a~ bit

Hr=2s-l>_3isoddthen U(a, b) =~1bit.., a ~ bf. ae. bI~-1. .... a~:bI~.

The final U(a-t,b-1).

theorem handles the case where U(a,b) is conjugate

to

THEOREM6.5.6. Let 1 ~ U = U(a,b) be in F or 1 ~ U U(b,a) be in F. HU = U(a,b) assume that there is a V in F with V-1U(a,b)V = U(a-l,b-t). H U = U(b,a) assume tha~ there in F with V-tU(b, a)V = U(b-t, a-t). Then possibly after replacing U by U(a-t,b -~) = U*(b,a) i/=U = U(a,b) or U by U(b-~,a -t) = U*(a,b) U = U(b, a) one of the following cases hold: (1) U -= U(a, b) and U(a, b) = (S(a, b)S(a -1, b-l)) d [or some na~urM number d and some elemen~ S(a, b) in F. d for some natural (2) U = U(b,a) and U(b,a) = (T(b,a)T(b-t,a-~)) number d and some element T(b, a) in F. PROOF. Suppose, without loss of generality, that 1 ¢ U = U(a,b) = a~tblt...a~rbI~,r >_ 1, with ei, f~ non-zero integers. Let V = d’+t with all ki non-zero integers, di a non-zero integer adtbk~...ad’~bk’~a for i = 2, .., n and dl and d~+t integers. Again let L denote the free product length in the free product decomposition F =< a > * < b > of F. In this case cancellation arguments with respect to L give that dt ~ 0 if and only if d~+t = 0. If dl = 0 and d,~+t ¢ 0 then

158

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

V --- bkl...ad’~bk"ad’~+l and a-elb-ll...a-e"b--5"a-d’~+Ib-k’~...a-d~b -k~ -a-d,~+~ b-k,~...b k~a~Ibll ...a¢r b~r, and we may replace U(a,b) by U(a-l,b -1) = W(b,a) (and V by v-l). Therefore we may assume that dl ~ 0 and d=+l = O. Then ae~ bfl...a ~ bfr adl bkl...a d" bk" =a’~1 bk~ ...a d’~ bk’~ a-~ b-f~...a -~ b- fr with dl # 0. Ifn > r then V = V1V~with V1 = U(a,b) and U(a,b)V2 V2U(a-1, b-~) and here we may replace V by V2. Therefore we may assume that r >_ n. If r = n then V = U(a,b) = U(a-l,b -1) which is impossible. Thus we have r > n. Then U(a,b) = Ul(a,b)U2(a,b) with Ul(a,b) = and we get the equation U2(a,b)Ul(a,b) = Ul(a-~,b-~)U2(a-l,b-i). If -l) L(Ul(a,b)) = L(U2(a,b)) then U2(a,b) = Ul(a-l,b and therefore U(a, b) = Ul(a, b)Ul(a-1, 5-1). If L(UI(a, b)) < L(U2(a, thenU2(a,b) = Ul(a-~,b-~)W~(a,b) and W~(a,b)U~(a,b) = Ul(a,b)Wl(a-l,b-1). If -1) and L(U2(a,b)) < L(U~(a,b)) then Ul(a,b) = W2(a,b)U2(a-l,b Uu(a,b)W2(a,b) = W2(a-l,b-1)U2(a,b) that is W~(a,b)U2(a,b) U2(a,b)W~(a-l,b -1) with W~(a,b) = W2(a-l,b-1). In both these last two cases the desired result follows by induction.

CHAPTERVII LINEAR ONE-RELATOR

7.1 Linear Properties

PROPERTIES PRODUCTS

OF OF

CYCLICS

of One-Relator Products of Cyclics

As we saw in Chapter Six, one-relator products of cyclics with proper power relators admit essential representations into PSL2(C). Although many of these representations are far from being faithflfi their existence leads to the existence of manylinear properties in this class of groups, The existence of these linearity properties mirrors the situation for one-relator groups (see Chapter Ttn:ee). Throughout this chapter we concentrate on the case where the relator is a proper power. Therefore our groups all have the form (7.1.1)

< al, ...,a,,;a~ 1 .....

a~," = R"(al, ..,aN) ---- 1

wheren _> 2, ei = 0 or ci _> 2 for i = 1, ..., n, R is a cyclically reduced word in the free product on al, ..., a~,, involving all al, ..., a,~, and m>_ 2. For n = 2 the resulting class of groups is the class of generalized triangle groups. These all have the form

(7.1.e)

< a, b; a"p = bq = R’" (a, b) = 1

where as above R is a cyclically reduced word in the free product on a, b involving both a and b and m _> 2. For many of the linearity properties examined the generalized triangle groups must be considered separately. In section 7.2 we first consider the Tits Alternative. Weprove that if, n _> 3 then any group with a presentation of the form 7.1.1 satisfies the Tits Alternative. Related to this we consider the SQ-universality of these groups. The complete solution of the Tits Alternative and SQ-universality the generalized triangle groups is still open. In section 7.3 we discuss the general theory of this class. In particular the Tits Alternative is knownto hold for a generalized triangle group except possibly in one situation. The techniques used to study the generalized triangle groups leads to showing 159

160

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

the existence, in many cases, of Ree-Mendelsohn pairs. These are generating pairs (g, h} for which there is a sufficiently large power t such that < gt, ht > is free of rank two. Filmlly in section 7.3.3 we give the complete classification of the finite generalized triangle groups recently completed by Howie, Metafsis and Thonlas [H-M-T 1], and Levai, Rosenberger and Souvignier [L-R-S]. In section 7.4 we consider mffticient conditions for a one-relator product of cyclics with proper power relator to be virtually torsion-free. From work of Culler and Shalen, if a finitely generated group admits "enough" representations into SL2(C), it must decompose as a non-trivial amalgam. A result of Bass in a different direction shows that manyfinitely generated subgroups of GL2(C) are in fact non-trivial free products with amalgamation. These results are used in section 7.5 to examine the existence of amalgam decompositions tbr groups in the class of one-relator products of cyclics. Finally in section 7.6 we discuss the existence of an Euler characteristic for groups in this class. 7.2 The Tits

Alternative

and SQ-Universality

The first linear property we consider is the Tits alternative. By a linear group we mean a subgroup of GLn(F) for some commutative field F. For these notes we restrict F to have characteristic zero although this is not necessary in all cases. A theorem of J. Tits [Ti] states that a finitely generated linear group either contains a non-abclian free subgroup or is virtually solvable, that is contains a solvable subgroup of finite index. In general we say that a group G satisfies the Tits alternative if it either contains a non-abelian free subgroup or is virtually solvable. Theorem3.4.1 due to Karrass and Solitar shows that any one-relator group, and more generally any subgroup of a one-relator group, satisfies the Tits alternative. Our first result here is that if n _> 3 then a one-relator product of cyclics with a proper power relator must satisfy the Tits Alternative. THEOREM 7.2.1. Let G be a one-relator product of cyclics with proper power relator so that G has a presentation (7.1.1)

< a~, ...,

a~; a[~= ....

a~~

= R’~’at~,..,a~)"=1

with m >_ 2, ei = 0 or ei >_ 2 for i = 1, ...,n and R(a~, ...,a~) a cyclically reduced word in the free product on al, ..., a~ involving all a~, ..., a~. Then G satisfies the Tits alternative ifn > 3. I~ particular (1) /f n _> 4 or n _> 3 and (e~, e2, Ca) ~ (2, 2, 2), then G contains a non-abelian free subgroup.

7.2 THE TITS ALTEI~NATIVEANDSQ-UNIVERSALITY

161

(2) Ii’n = 3 and (el,e2, e3) ---- (2,2,2), then either G contains a nonabelian free sllbgrollp or G contains a free abelian s~ibgrollp of rank 2 and of index 2. PrtooF. If n _> 4 this is a direct consequence of the Freiheitssatz. Since n >_ 4 the subgroupgenerated by al, a2, a3 will be a non-trivial free product of rank 3. This clearly contains a free subgroup of rank 2. Nowsuppose n = 3 and (el, e2, e3) # (2, 2, 2). ~omthe l:~eiheitssatz, contains a non-trivial free. product of cyclics of rank 2 which is not infinite dihedral. Hence G has a free subgroup of rank 2. Nowsuppose that n = 3 and (~, e2, e3) = (2, 2, 2) so that G has form (7.2.1.)

G -~<

al,a2,a3;a’~

-~ a29 -~-

a~ =

R’~(al,a2, a.~) --

>

If m _> 3 then from Theorem6.2.3 it follows that G contains a subgroup of finite index mapping epimorphically onto a non-abelian free group. Hence it follows that G must itself contain a non-abelian free subgroup. As a consequence G must be SQ-mfiversal, a fact that we will return to after the completion of the proof. Therefore we are reduced to the case where n -- 3, m -- 2 and (ebe2,e3) = (2,2,2) and G has the form (7.2.1). In this case a = a~a2, b = ala3 and let H =< a, b >. The index ]G : HI _< 2 and we show that H either contains a free subgroup of rank 2 or that H is free abelian of rank 2 with ]G: HI = 2. First write R(a~, a2, a3) = a~S(a, b) where e -- 0 or e -- 1 and S(a, b) is freely reduced in the free group on a, b. S(a, b) involves both a and b since R(a~a2, a3) involves all al~ a2, a3. Case 1. m--2andc--0. Choose three projective matrices of order 2 in PSL2 (C) A1-- q-

1 A2 -: =t= -1 1 0 ’ 1

2 and Aa -- + -w

with w e C. A1,A2 generate an infinite dihedral group. Let A -- A1A2 and B -AIAa. Then < A, B > cannot be an infinite dihedral group because if it were then tr(B) = tr(AB) and this imples that w2 + 2 = 0 and 3+ 2w2 - 2w = (w- 1) 2 = 0 giving a contradiction.

162

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Nowas in the proof of the ~reiheitssatz for one-relator products of cyclics choose w0 so that tr(R(A1, A2, A3)) = 2 cos(~/m). Then < A1, A2, provides a representation for G in PSL2 (C). Wcclaim that A and B = B(wo) havc no commonfixcd point for if thcy did then < A, B > would be metabelian and tr([A, B]) = 2. From this would follow that Wo= 0 or w0 = -1. We then obtain that A3 --- +A1 or As = +A2. In both cases it follows that S(A, B) = ~ f or s ome k~ Z and R(AI,A2,A~,) = A~Ak. Since m = 2 and e = 0, this shows that R(AI,A2, A~) cannot have order m in < AI,A,2,Aa > contradicting the way the trace was chosen. Thereibre A and B have no commonfixed point. Further tr(A) = tr(AiA2) = 3 so tr(A) ~ If tr(B) ~ and tr (AB) ~ wemay a ssume ( aft er a suit able conjugation) that < A,B >C PSL2(R). It then follows from a result of Rosenberger [Ro] that < A, B > contains a generating pair {u, v} such that < ut, v~ > is free of ra~k 2 for a sufficiently large integer t. If both B and AB are loxodromic {tr(B) ¢ ~, and tr(AB) 6 It~} then there exists a positive integer t such that ]tr(Bt)l > 2 and Itr(AB)tl 2. From results of Majeed [Maj], < Bt, (AB)t > is free of rank 2 for a sufficently large integer t. Clearly < B, AB >=< A, B >. If tr(B) ~ but tr (AB) ~ th entr(BA B) q~ ~ sin ce tr(BAB ) = tr(B)tr(AB)-tr(A) and < BA, BAB >=< A, B >. Then from the results of Majccd thc subgroup < (BA)t, (BAB)t > will bc free of rank 2 for a suffciently large integer t. Finally iftrB ~ ~ but tr(AB) ~ th en tr (AB-1) ~ ~ si nce tr (AB-~) = tr(A)tr(B) -tr(AB) and < B, AB-~ >=< A,B >. Then a.s above < Bt, (AB-1)t > is frcc of rank 2 for a sufficicntly largc intcgcr t. So in all cases H =< a, b > contains a generating pair {u, v} such that < ut~ vt > is free. of rank 2 for a mffficiently large integer t. This completes case 1. Case 2. m=2ande=l. The relator now has the form R(al,a2,a3) = alS(a,b). H =< a, then has index 2 in G and applying the Reidemeister-Schreier process we find that H has the ~presentation H --< a, b; S(a-~, b-~)S(a, b) -~ 1 Wecan assume without loss of generality that S(a, b) is cyclically reduced in the free group on a, b and that there is no free. cancellation between S(a -~, b-~) and S(a, b). Thc cxponcnt sum on b is zcro, so wc can cxprcss H as an HNNgroup with stable letter {b}. For each i ~ Z let xi = ~. b-~ab Then S(a-~,b-~)S(a, canbe e xpressed as a fre el y reduc ed word T on the {xi}.

7.2 THETITS AUI’~RNATIVE ANDSQ-UNIVEILSALI’I’Y

163

Let M be the greatest integer such that XM appears in T and m the least integer such that Xmappears in T. Clearly IM - mI _> 1. H can then be expressed as H =< x~,~, .... x~,b;T(x,,,,. .... xaz) = 1, b-tx~b = x~+~,i = m,....M - 1 > Let K-~-<

xm,

....XM;

gl

=~ xm,

T(x,~, .....

.....

XM-1;

K2 =< Xm+l, .....

XM)=

1

>

>

XM;> ¯

Then H is an HNNgroup with base K, free part generated by b and associated subgroups K1 and K2. If ]M - m] _> 2 then K1 is free of rank at least 2 and is contained in H, so in this case H contains a free subgroup of ra~k 2. If IM - m] = 1 then d~¢ = m + 1. It follows that either x,~ = x,~+l or x,, ~ x,,+l in H. If Xm = Xr~+l then b-lab = a or ab = ba. Then H =< a,b > is free abelian of rank 2 and is of index 2 in G. Finally if x,~ ¢ x,~+l then there exists no p, q ~ Z \ {0} such that p (.T.,~X~+I)P----..T.q.,, or (x,~:r.~l+~) =x,,,+~q sincetheabelianization of His frcc abclian of rank 2. Thcn from Britton’s lcmma on HNNgroups (scc Chapter Two) < x,~x~+~, b > is a free subgroup of rank 2. This completes case 2 and the proof of the theorem. As mentioned in Chapter Three, closely related to the Tits alternative is SQ-universality. Recall that a group G is SQ-universal if every countable group can be embedded in a subgroup of a quotient of G. All non-abelian free groups are SQ-universal and from a theorem of Levin [ Le] any nontrivial free product except Z~ ¯ Z2 is SQ-universal. If IG : HI < oc then H being SQ-universal implies G is. Further if G maps onto a non-abelian free group or contains a subgroup of finite index which maps onto a non-abelian free group, then G is SQ-universal. The tie to the Tits alternative is that having manynon-abelian free subgroups is a good indicator that the group should be SQ-universal. As a direct consequence of Theorem6.2.3 and the existence of essential representations ibr one-relator products of cyclics we obtain the ibllowing result. THEOREM 7.2.2. Let G be a one-relator product of cyclics with proper power relator so that G has a presentation of the tbrm 7.1.1 (7.1.1)

< al,...,a,~;a~

x .....

a~~ = R"~(al,..,an)

=1

as in Theorem7.2.1. Suppose thag n >_4 or n >_ 3 and (e~, e2, e3) ¢ (2, 2, or n ~ 3,(e~,e2,e~) = (2,2,2) and m >_ 3, t, hen G contains a subgroup

164

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

of finite index mapping epimorphically onto a non-abelian free group. In paxticulax under these conditions G is SQ-universal. Thus ibrn _> 3, relative to SQ-universality we are reduced to the two special cases considered in the proof of Theorem7.2.1, that is when m = 2 and (el, e2, e3) = (2, 2, 2). Thus we now consider groups of the G =< al, a2, a3; a~ = a’~ = a~ = R’2(al, a2, a3) = 1 where R(at, a2, a3) is cyclically reduced in the free product on at, a2, a3 and involves al:a2,a3. A.~ before let a = ala2, b = ala3 and let, H =< a,b > in G. Thus H has index less than or equal to 2 in G and again as in the proof of Theorem7.2.1 write R(al, a2, a3) = a~S(a, b) where ~ = 0 or e = 1 and S(a, b) is freely reduced in the free group on a, b. S(a, b) involves both a and b since R(a~, a2, a3) involves all a~: a2.. a3. Notice that the element R(at, a2, a3) is in H only if e = 0. Wethen obtain the following theorem. THEOREM 7.2.3.

Suppose the group G has a presentation

of" the tbrm

G = 1 and ai,~i axe non-zero integers for i = 1, ..., k. Suppose further that one of the following holds: (i) ->2 or [fill/-~ 2 if k = 1 (ii) gcd(a~,...,crk) > 2 or gcd(3~,...,/dk) > 2 ilk >_2 (iii) S(a, b) a p roper power in thefree groupon a, b Then G contains a subgroup of finite index mapping onto a nonabelian free subgroup and hence is SQ-universal. PROOF.Suppose first H has the presentation

that e = 0 so that R(al, a2, a3) = S(a, b). Then

H =< a, b; S’2(a, b) = S2(a-~, b-1) = 1 > . ~Yoma theorem of M.Edjvet [E 3] H has a subgroup of finite index mapping onto a non-abelian free. group and hence G has also.

7.2 TtiE TITS ALTERNATIVE ANDSQ-UNIVERSALITY

165

Nowsuppose that e = 1 so that R(al, a2, az) al S(a, b) ThenH has the presentation H =< a,b;S(a,b)S(a-~,b

-1) = 1 > .

If (i) or (ii) holds, then the group H has a free product of cyclics Zr with r _> 2 as an epimorphic image, so the result follows. If (iii) holds thcn S(a, b) = T’~(a, with 0/>2. Hthcn has a s a factor group the group H with the presentation ~ =< x,y;T’~(x,y)

= T’~(x-l,y -~) >.

The theorem of Edjvet used in the first part of the proof can be applied to thissituation to get the result. Finally suppose that k ----- 2. Wemayassume, without loss of generality (passing to ~nother generating pair if necessary and after a suitable conjugation) that 0/1,a2,fl~,f12 are all greater than or equal to 1. H has a free., product of cyclics Zr * Z~ with r >_ 2, s > 3 as an cpimorphic imagc, from which the result would tbllow, except in the tbllowing cases: (a) a~ =2, a2=fl~=l,f12_>2;

(d) 0/1 = fll = f12 = 1, 0/2 = Consider first case (a). Let a~ = 2, c~2 = fl~ = 1, fl~ > 2 and let fl = Let K be the subgroup of H generated by x = a~,y = b and z = aba-1. K then has index 2 in H and has a presentation K =< x,y,z;xyz~x-~z-ly

-~ = xzxyZx-~y-lx-lz-~

=1 >

which has the free product Zfl+l * ~2 as an epimorphic image. In case (b) where a~. = 3, a~. =/~ = 1, ~2 = 2 let K be the subgroup H generated by x = b~, y -- a and z = bab-1. K then has index 2 in H and has a presentation K =< x, y, z; y3zxz-3y-lx-1

= z3xyxy-3x-~z-lx

-~ = 1 >

which has the free product Z2 * Z4 as an epimorphic image. In case (c) where 0/~ = a2 = ~ = 1, ~ = 2 let K be the subgroup of generated by x = b2,y = a and z = bah-~. K then has index 2 in H and has a presentation K =< x, y, z; yxyz-~x-~z -1 = zxzxy-~x-ly-~x-~

=1 >

166

ALGEBRAIC

GENERALIZATIONS

OF DISCRETE

GROUPS

which has the free product Z2 * Z as an epimorphic image. Finally in case (d) where al = fll --- f12 = 1, a2 = 2 then H has the group H with presentation ~ =< x,y;x’~ = y4 = (xy’2)’~ = 1 as an epimorphic image. Let K be the subgroup of H generated by u = y and v = xyx. K then has index 2 in H and has a presentation K =< u,v;u 4 = v 4 = (u2v’2) "2 = 1 > which has the free product Z2 ¯ Z4 as an epimorphic image. This completes the proof. Weconjecture that if k _> 3 then the subgroup H in the above theorem contains a subgroup of finite index mapping epimorphically onto a free group of rank 2. From a result of Sacerdote and Schupp IS-S] we know that a group H as in Theorem 7.2.3 is SQ-universal if k _> 3. Combining Sacerdote and Schupp’s result with Theorems 7.2.1 and 7.2.2 ~ then have the following theorem. THEOREM 7.2.4. Let G be a one-relator product or" cyclics with proper power relator so that G has a presentation of form 7.1.1 (7.1.1)

< a~,...,a,~;

a~ =

.

with m >_ 2, e~ = 0 or e~ >_ 2 ~or i = 1, ...,n and R(al,...,a,,) a cyclically redfaced wordin the £r~ product on a~, ..., a,~ involving all al, ..., a,~. Then i[ n _> 3 either G is SQ-univcrsal or v/rtually ~cc abclian. In the next section we consider the generalized triangle groups. We conjecture that ibr this class each group is either SQ-universal or virtually solvable. 7.3 The Generalized

Triangle

Groups

Whenn = 2 we have the case of the generalized triangle groups. Relative to the Tits alternative, the complete solution ibr these groups is still open. Essentially what is presently knownis that the Tits alternative holds except possibly when both generators have finite order p, q, the relator has order m = 2, (l/p) + (l/q) _> (1/2) and the relator has syllable length greater than eight in the free. product on the generators. This is the content of Theorem 7.3.1 given below. First ~ fix some notation. A generalized triangle group is a group G with a presentation (7.3.1)

G =< a, b; a r = bq = R’~(a, b) = 1

7.3 THE GENERALIZEDTRIANGLEGROUPS

167

where p < q, p k 2 or p = 0, q > 2 or q = 0, R(a, b) is a cyclically reduced word in the free. product on a and b involving both a and b and m _> 2. If p >_ 2 we let s(G) = (l/p) + (l/q) + (l/m). A generalized triangle clearly generalizes an ordinary triangle group T(p, q, m) which has the presentation T(p, q, m) =< a, b; p

=b q=

(ab"~ = 1 > .

Wenote that many authors restrict the class of generalized triangle groups to the case where both generators have finite order. Wedo not make this restriction. For the generalized triangle groups the major result relative to the Tits alternative is the following. THEOREM 7.3.1. tion

Let G be a generalized

triangle

group with presenta-

G =< a, b; a p = bq = R"~(a, b) = 1

where p _< q, p _> 2 or p = O, q > 2 or q = O, R(a, b) is a cyclically reduced word in the free product on a and b involving both a and b and rn > 2. Then G satisfies the Tits alternative except possibly whenp >_ 2, q > 2, rn -2, (l/p) + (l/q) > 1/2 and the relator R(a, b) has syllable length greater than eight in the ~ree product on a, b. This result is actually a summaryof several results whose precise statements and proofs will be given in sections 7.3.2 and 7.3.3. These results generally show more than just the existence of free subgroups or solvable subgroups of finite index. If G is a 2-generator group then a generating pair (u, v} is a Ree-Mendelsohn pair or RM-pair if there exists an integer t such that < ut, vt > is a free, subgroup of G of rank 2. In particular included amongthe results summarized in Theorem7.3.1 are the following: (1) If > 3, p = 0 and q _>2 t hen G c ontains an RM-pair. (2) If m = 2,p = 0 and q _> 3 then G contains an RM-pair. (3) If ra = 2,p = 0 and q = 2 and suppose R a’ ~ba’~b...a’*~b with k _> 1, then: (a) If k >_ 2 or ]’n~] > 3 for some i = 1,...,k, then G contains a non-abelian free subgroup. ~hrther if k is even or k is odd with k > 3 and ni +... + nk # O, then G contains an RM-pair. (b) If k = 1 and In~l >_ 3 then G has an RM-pair. (c) If k = 1 and ]n~] < 2 then G is infinite and solvable. The reason that there are two types of results - ones giving the existence of RM-pairs and others only guaranteeing that G satisfies the Tits alternative, comes from the method of proof. This was seen previously in the proofs in the last section. ~lb obtain an RM-pair we consider the essential

168

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

representations of G in PSLu(C). Let G =< A, B > be the image group. If we can find a representaion such that < A, B > is non-elementary and non-elliptic, we can use the results of Rosenberger [R 3,13] and Majeed [M~j] as in the proof of Theorem 7.2.1 to obtain an RM-pair. If we cammt show that the image group has this property we can in the remaining cases use combinatorial arguments to get that G either has a non-abclian free subgroup or a solvable subgroup of finite index. Before going to these results we first review and highlight some of the main features of the ordinary triangle groups T(p, q, m). 7.3.1

The Ordinary Triangle

Groups

Let p, q, m be integers greater than one. Then the ordinary triangle group or just triangle group T(p, q, m) is tim group defined by the presentation (7.3.1.1)

T(p,q,m) :< a,b;a p = bq = (ab) *’~ = 1 > .

Notice first, that as abstract groups any permutation of the triple (p, q, rn) leads to an isomorphic ordinary triangle group. This is clearly not necessarily true for the generalized triangle groups. Note also that in the definition of the generalized triangle groups wc did not require the exponents p, q to be greater than one, so that we allow generators to have infinite order, l~br the ordinary triangle groups it is generally standard to require only generators of finite order. Group theoretically, if one or more of the exponents is zero, we get a non-trivial free product of two cyclics. It is often convenient to think of these free products as triangle groups. For example the Modular group PSL~.(Z) has the structure Z2 * Z3. This can be considered as the triangle group T(2, 3, oo). The terminology triangle group comes from the following construction. Let p,q, m > 1 be integers with (l/p) + (l/q) + (l/m) < 1. Then exists a non-Euclidean hyperbolic triangle A with angles ~/p, r/q, rim. Let L, .¢¢, N respectively be the reflections of the hyperbolic plane 7~ in the sides of A and let T* be the subgroup of the isometry group of 7-/ generated by L, M, N. It can be shown (see Magnus[M 6]) that T* has the presentation T*

~=Mu=Nu=(LM) ~=(MN)q=(NL)~’=1>. =
Nowlet T be the subgroup of T* consisting of the orientation preserving isometries. T has index 2 in T* and group theoretically consists of those words of even length in the generators of T*. Using this characterization

7.3.1

THE ORDINARYTRIANGLEGROUPS

169

it can be shown that T is generated by a = LM and b = MN with a presentation T =< a, b; ap -- bq : (ab) TM = 1 >. Therefore T is isomorphic to T(p, q, m). Further since now T consists of orientation preserving isometries of 7-/, the elements of T can be considered as being in PSL2(R) (see Chapter 4) and hence T(p, q, m) has a faithful representation in PSLg.(R). The union of the original triangle A with the image L(A) under the reflection L serves as a positive area fimdamental domain for T. From tlfis it follows that T is discrete and thus a Fuchsian group. Wesummarize this by saying that ibr any p, q, m > 1 with (l/p) (l/q) + (l/m) < I the triangle group T(p, q, m) has a faithful representation as a Fuchsian subgroup of PSL2(R). (This also follows from Poincarc’s Theorem (see Chapter 4)). This construction can be mirrored in both the Euclidean plane and ill the spherical (double elliptic) plane. In both these cases the resulting triples axe restricted. For a triangle in the Euclidean plane, we must have (l/p) (l/q) + (l/m) = 1. If < p < q < m,the n the onlypossi ble tripl es are (2, 3, 6), (2, 4, 4) and (3, 3, 3). Usingreflections in the sides of tile triangle generate a subgroup of the Euclidean isometry group and then taking the subgroup of orientation preserving isometries we get the Euclidean triangle groups T(2, 3, 6), T(2, 4, 4) and T(3, 3, 3). In each case these groups to tesselatior~ or filings of the Euclidean plane. (See. Magnns[M 6] for somevery pretty pictures of these tesselations as well as tesselations of from hyperbolic triangle groups.) It can be shown that these groups have faithful (but not Vhchsian) representations in PSL2(C). Wesay that a representation of a group in PSL2 (C) is a Fuchsian representation if the image group is a Fuchsian subgroup of PSL2(C). Finally for a spherical triangle we mttst have 2 < p, q, m and (l/p) (l/q) + (l/m) > 1 and if 2 _< < q _<m t heposs ible trip les are (2, 2, m), (2, 3, 3), (2, 3, 4) and (2, 3, 5). As before we get the triangle T(2, 2, m), T(2, 3, 3), T(2, 3, 4) T(2,3, 5) which are al l fi nite groups. These groups car~ be faithfully represented ir~ PSL2(C) and describe the finite groups of symmetries in R3 of the regular solids. T(2, 2, m) axe the dihedral groups D,~ and are the symmetry groups of the regular m-gon or dihedron. T(2, 3, 3) has order 12 and is the symmetry group of a tetrahedron. It is isomorphic to A4, the alternating group on ibur symbols. T(2, 3, 4) has order 24, is the symmetrygroup of a octahedron, and it i~morphic to $4, the symmetric group on four symbols. Finally T(2, 3, 5) has ordcr 60 and corrcsponds to thc symmctry group of thc rcgular icosahedron. It is isomorphic to As. ~om this discussion we see that T(p, q~ m) is finite if and only if 2 < p,q,m and (l/p) + (l/q) + (l/m) > 1. The ~mhsian triangle

170

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

groups are clearly infinite while one can show that the Euclidean triangle groups contain free abelian subgroups of finite index. If 2 _< p, q, mand (l/p) q- (l/q) q- (l/m) < 1 then T(p, q, m) can be considered as a group. It follows from the Fenchel-Fox Theoremthat there is a torsion-free Fuchsian group of finite indcx. This must bc an oricntablc surface group of genus g _> 2. Such groups clearly map epimorphically onto non-abelian free groups and hence in this case T(p, q, m) is SQ-universal. If 2 _< p, q, and (l/p) + (l/q) + (l/m) ---- 1 the resulting three, groups are all virtually free abelian of rank two. It follows that they are all infinite and contain subgroups of finite index mapping onto Z. Finally from the concrete construction as groups of isometries it is clear that in T(p, q, m) the element a has exact order p, the element b has exact order q and the element ab has exact order m. Wesummarize all these facts below. THEOREM 7.3.1.1. Let G = T(p,q,m) =< a,b;a v = bq = (ab) "~ = 1 > with p, q, m > 1 be a ~ria~gle group. Let s(G) (l /p) + (l /q) + (l Thdn (1) G admits a faithful representation into PSL2(C). The image group can be ~bchsian if and only if s(G) < (2) G is flnife if and only if s(G) (3) If s(C) < 1 ~hen C contains an orientable surface group of genus g >_ 2 as a subgroup of finite index. In particular G contains a subgroup of finite index mapping onfo a non-abelian free group and hence (I is universal (4) H s(G) ---- 1 ~hen~ is virtually free abelian of rank (5) The element a ha~ exact order p, the element b h~ exact, order q and $he element ab has exact order m. (6) G is virtually torsion-free. (7) G satisfie~ ~he Tits al~,ernative. The hyperbohc triangle groups are of further interest amongthe Fuchsian groups as the subclass of ~hchsian groups with fundamental domains having small hyperbolic area. Of particular interest here is the group T(2, 3, 7) which has a fimdamental domain of the absolutely minimum hyperbolic area. This is a straightforward consequence of the Poincare presentation and its relation to the hyperbolic area of a fundamental domain (see Chapter 4). The group T(2, 3, 7) is of further interest in Riemannsurface theory because of the following set of results of Hurwitz (see Magnus[M 7]). The group H of conformal self-mappings of a Ricmannsurface $ of genus g _> 2 is a finite group of maximumorder 84(g - 1). If Sa is the fundamental group of S, then H is a quotient group of a l:hchsian group F by a normal subgroup isomorphic to S.q. The order of H is the maximalorder 84(g- 1) and only if F is isomorphic to T(2, 3, 7). Thus the finite groups of maximal

7.3.2

THE TITS AEI’ERNATIVE- GENERATORS OF FINITE ORDER171

order which can be isomorphic to the group of conformal self-mappings of a Riemannsurface are precisely the finite quotients of T(2, 3, 7). Such groups are called Hurwitz groups and have been studied fairly extensively (see the survey by Conder [Co 2]). We mention that Higman [Hi 3] proved that all alternating groups An for n sufficie~flty large are Hm’witzgroups. Macbeath [Mac 1,2] proved that PSL2(q) is a Hurwitz group if and only if either (i) q = 7 ; (ii) q = ±1rood 7; or (ii i) q = p3wher e p -- ±2 or -- :t:3 rood 7. 7.3.2

The Tits Alternative

- Generators of Finite

Order

Our purpose in the next two sections is to prove Theorem 7.3.1. suppose that G is a generalized triangle group with presentation (7.3.1)

We

G --< a,b;a ~ = bq -- R"~(a,b) = 1

where p _< q, p >_ 2 or p = 0, q _> 2 or q = O, R(a, b) is a cyclically reduced word in the free., product on a and b involving both a and b and m >_ 2. We will give a series of results which combinedwill showthat G satisfies the Tits alternative except possibly whenp _> 2, q >_ 2, m -- 2, (l/p) + (l/q) _> and the relator R(a, b) has syllable length greater than eight in the free product on a, b. In this section we concentrate on the case where both generators have finite order, that is we assumethat p _> 2, q _> 2 in presentation 7.3.1. The main result, due to Baumslag, Morgan and Shalen, mirrors the situation fbr the ordinary triangle groups and also reduces the problem to a finite set of triples (although this involves i~ffinitely manypossible relators). THEOREM 7.3.2.1. with presentation

[B-M-S] Suppose G is a generMized triangle

group

G =< a,b;a ~ = bq = R"~(a,b) = 1 where 2 <_ p ~_ q, R(a, b) is a cyclically reduced word in the Tree product on a and b involving both a and b, m >_ 2 and let s(G) = (l/p) + (l/q) + (l/m). (1) ITs(G) < 1 then G contains a subgroup o[ ~nite index mapping a non-abelian free subgroup. In particular G is SQ-universal and contains a free subgroupof’ rank 2. (2) If s(G) = 1 then G contains a subgroup of finite index mapping onto Z. (3) a has exact order p, b has exact order q and R(a, b) has exact order PROOF.~rom Theorem 6.3.2, G admits an essential PSL2 (C). Part (3) follows directly from this fact.

representation

into

172

ALGEBRAICGENERALIZATIONSOE DISCRETE GROUPS

The proofs of parts (1) and (2) mirror the proof of Theorem 6.2.3. s(G) _~ 1 and let p be an essential representation of G into PSL2(C). Therefore from Selberg’s theorem (Theorem 4.3.5) there exists a normal torsion-free subgroup H of finite index in p(G), the image of G. Thus the composition of maps ( where rr is the canonical map) C A p(G) -~ p(C)/H gives a ¢ of G onto a Let X =< a, b; ap < a; ap = 1 >, < b; onto G. Wetherefore

finite group. = bq -- 1 > be the free product of the cyclic groups bq = 1 >. There is a canonical epimorphism/~ from X have the sequence X ~-~ G --~¢ p(G)/H

Let Y = ker(¢ofl). Then Y is a normal subgroup of finite index in X and Y is torsion-free. Since X is a free’, product of cyclics and Y is torsion-free, it follows that Y is a free group of finite rank r. Suppose IX : YI = J- Since every finitely gmmrated free product of two or more cyclics can be faithfully represented as a Fuchsian group if it is not isomorphic to the infinite dihedral group, we may regard X as a Fuchsian group with finite hyperbolic area #(X). From the RiemannHurwitz formula we have that

jit(x) where It(Y) = (2rr)(r and #(X) = (2u)(1

- (~

Equating these expressions we obtain r = 1 -j(:

1 +- - 1). P q

G is obtained from X by adjoining the relations R’~(a, b) and so G = X/K where K is the normal clom~re of R"~. Since K is contained in Y~ the quotient Y/K can be considered as a subgroup of finite index in G. Applying the Reidemeister- Schreier process or repeated applications of

7.3.2

THE TITS AUFERNATIVE - GENERATORS OF FINITE ORDER173

Corollary 3 in [B-M-S], Y/K can be defined on r generators subject (j/m) relations. The deficiency d of this presentation for Y/K is then d=r-

jm--:l-j(~+-ql+-ml

to

_l)=l_j(s(a)_l).

If s(G) < 1 then d > 1 and since d is an integer, the deficiency of the above presentation is at least 2. From the result of Baumslag and Pride, Thcorcm 6.2.2, Y/K, and hcncc G, has a subgroup of finitc indcx mapping onto a free group of ra~k 2. Therefore Y/K is SQ-universal and since this has finite index in G, G is also SQ-universah l~lrther an SQ-universal group must contain a non-abelian free subgroup. If s(G) = 1 then the deficiency is d = 1 and the group Y/K maps onto an infinite cyclic group and is thus infinite. Therefore G is infinite. If G is a generalized triangle group with presentation G =< a, b; ap = bq -~ R’~(a, b) = 1 where 2 < p < q, m > 2, and R(a, b) is a cyclically reduced word in the free product on a and b involving both a and b, then R(a, b) can be written as a word R(a, b) = ~ bq~ . .. a ~ bq~: with k _> 1 and 1 _< p.¢ < p, 1 _< qi < q fbr j = 1,..., k. The syllable length as a word in the free product on a, b is then 2k. In the next result we show that the Tits alternative holds wheneverk = 1, that is the syllable length is two. THEOREM 7.3.2.2. sentation

Suppose G is a generalized triangle

group with pre-

G =< a, b; a v = bq = (a~bt) m = 1 > where2 <_ p <_ q,m >_ 2, 1 < s < p, 1 < t < q. Then G satisfies the Tits alternative. In particular if G is not an ordinary triangle group then C contains a free subgroup of rank 2 or G is infinite and solvable. PROOF.Without loss of generality we may assume that sip and tlq passing to another generating pair if necessary. If f = s = 1 then G is an ordinary triangle group and the result follows. If s(G) < 1 it follows that there is a free subgroup of rmrk two from Theorem 7.3.2.1. This reduces to the case where (t,s) # (1, 1) and (l/p) + (l/q) (l/m,) >_ 1. Its _> 2 andt_> 2 then the free product < a,b : a ~ = b~ ~ 1 > is a epimomorphicimage of G. If s + t _> 5 this free product is not Z2 * Z~ so it

174

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

has a free subgroup of rank 2 and therefore so does G. Therefore the only other cases to consider are s _> 2, t = 1 or s = t = 2 or s = 1,t_> 2. Consider the case where either s > 2, t = 1 or s = t = 2. Then p = st, r _> 2 and q _> 4 since p < q. Wemay then write G as a free product with amalgamation G = HI * H H2 where H1----~ a; ap = 1 > H2 =< a~, b; (aS) " = b~ = (a~bt) "~ = 1 > and

H =< a~; (a~y = 1 >. Let x = -1 abaSb-la and y = b. Standard cancellation methods in free products with amalgamation show that the subgroup generated by x, y has the presentation ~ x, y; x’*" ~- yq --~ 1 >. Since r _> 2 and q _> 4 this has a free subgroup of rank 2. Next consider the case where s -- 1, t _> 2. Then as in the previous situation, q = tr with r _> 2. As before G is a free product with amalgamation G--H1 ~:H H2 where H1 =< b;b q = 1 > H2 =< a, bt; ap -= (bt) " = (abt) "~ = 1 > and H =< bt;(bt) r = 1 >. Ifp _> 3 let x = babta-lb -~ and y = a. Then the subgroup generated by x, y has the presentation < x, y; xr = yP ---- 1 > which has a free subgroup of rank 2 since r _> 2 and p _> 3. If p ---- 2 and m _> 3 let x = babta-lb -1 and y = abt . Then < x,y >=< x,y;x""

= y"~ = 1 >

which has a free subgroup of rank 2. Finally, if p = m = 2 then the cyclic subgroup < bt > is normal in G sinceabta=abta -~ = b-t. ThenG/ < b t >--~, =< a,b;a 2-- b t = 1 >. If t ___ 3 then G has a free subgroup of rank 2 and therefore so does G. If

7".3.2

THE TITS ALTERNATIVE - GENERATORS OF FINITE ORDER175

t = 2 then G is infinite dihedral and so is infinite and solvable. Therefore G is infinite and solvable since both < bt > and G/ < bt > are solvable. These cases exhaust the possibilities for (s, t) and therefore complete the proof. It ibllows from the previous results that G contains a non-abelian free subgroup if s(G) < 1 and in general satisfies the Tits alternative if the relator has syllable length 2 in the free product on a and b. This reduces the Tits alternative to the case of syllable length greater than or equal to four and the followingset of triplex: (2, 2, ,~), (2, q, 2), (2, 3, 3), (2, 3, 4), (2, 4, 4), (2, 3, 6), (2, 6, 3), (2, 3, 5), (2, 5, 3), (3, 3, 2), (3, 4, 2),(3, 5, and (3, 3, 3). Most of these are covered in the next theorem. We~nention here that J.Howie [H 9] has completely settled the case when s(G) = and shown that these have non-abelian free subgroups unless they are equivalent to an ordinary triangle group. His theorem handles the triples (3, 3, 3),(2, 4, 4) and (3, 6, 2) which are done in a different manner next results. Someof the results of this section are u~d by Howie for the proof of his theorem. Wedefer the precise statement of Howie’s result until the end of the section. THEOREM 7.3.2.3. senta~ion

Suppose G is a generalized triangle

group with pre-

G =< a,b;a" = bq = R"~(a,b) = 1 where2 <_ p <_ q,m >_ 2 and R(a,b) = a~bq~...a~b q~ with k >_ 2 and 1 <_p~ b e th e image o f G in PSL2( C). If < A, B >= p(G) is a non-elementary subgroup of PSL2(C), then it will

176

ALGEI~RAIC GENERALIZATIONS OF DISCRETE GROUPS

contain a non-abelian free subgroup and hence G will also. If p(G) is elementary, then p(G) is finite or metabelian, since a two-generator elementary subgroup of PSL2(C) is finite, infinite dihedral or metabelian. Therefore we will describe up to isomorphism, those G, such that it is always the case that p(G) is finite of metabelian. Hence, we now assume that always p(G) is elementary. It is known that the only finite subgroups of PSL2(C) axe up to isomorphism .the cyclic groups Zn for all n ¯ N, the dihedral groups Dn for all n ¯ N, the alternating groups A4 and As and the symmetric group 5’4. Thus if the image of G is finite under the essential representation p we know the possible structures of this image . Recall from the proof of the Freiheitssatz tbr one-relator products of cyclies (Theorem 6.3.1) that we can choose A, B ¯ PSL2(C) such that tr(A) = 2cos(r/p) and tr(B) = 2 cos(~r/q). It follows that tr(R(A, B)) is a polynomial in x = tr(AB). Therefore tr(R(A, B)) = f(x). We call f(x) the trace polynomial for G. The possible values for tr(AB) are the possible roots of a polynomial f(x)=t= 2cos(rr/m) with 1 _< r < m and god(r, m) = 1. the possible finite subgroups of PSL~ (C) are known, if the image group is fi~tite this gives a fi~tite list of possible values for tr(AB) and trance a fi~dte list of possible values for the roots of the polynomial f(x) + 2 cos(rr/m) with 1 <_ r < m and god(r, m) = 1.. Wemake this more precise. If A and B are as above and R(A, B) = Arabq~...Ae~Bq~ with k _> 2, 1 _< pj < p, 1 _< q~- < q and x = tr(AB), then tr(R(A, B)) is a non-constant polynomial f(x) = a~x~ + a~_~x~-~ + ... + ao of degree k and leading coefficient

cos(r/q))

= s;, where S~(z), r ¯ N U {0} is defined recursively by

So(z) = O, S~ (z) = 1 and S~(z) zS~_l (Z ) - S~_2(z) for r_> 2. In the following we refer to A and B as the images of a and b respectively under an essential representation p : G --~ PSL,2(C) such that tr(A) 2 cos(tip) and tr(B) = 2 cos(r/q), and we assume that each such posssible image < A, B > is finite or metabelian. Casc(1): p=q--m=3: To obtain an essential representation of G with elementary image we may choose tr(R(A, B)) = e,e = =t=l. Therefore we must consider the polynomials g~(x) ---- f(x) + e,~ = -t-1 and

7.3.2

THE TITS AUI’ERNATiVE- GENERATORS OF FINITE ORDER 177 gg.(x) = f(x) --

which are both in Z[x]. We have ak = 1 and tr(AB) is a zero of gl(x) or of g2(x). The possibilities for tr(AB) are 0, 1, 2, -1, ~, 1 - ~ where ~ = 2cos(~r/5) and ~2 = ~ + 1. This is because we cannot have < A, B being dihedral and further the synnnetric group $4 cannot be generated by elements of order 3. Hence the possible finite < A, B > are the alternating groups A4 and As. Note also that no zero of g~(x) is also a zero of g2(x) because gl(x) - g2(x) = 2~ = -4-2 Further A and 1 - ~ always occur in pairs since gl(x),g2(x) Z[x] an d we have (x- A)(x- 1 + A) ~- x- 1 . Suppose first that tr(AB) = 0 occurs. Without loss of generality let gl(0) = 0. ThEn2 must be a simple zero ofg2(x) since gl(x) -g2(x) = :i:2. Therefbre we have g, (x) = xr(x 1)~(x + 1) 2 - x - 1)" xk + d~_lx~-~ + ...

+ d~x,r >_ 1 and

g2(x) = (x- 2)(x- 1)~(x + 1)t(x u - x- 1) ~ = xk + c~,_~xk-~ + ... + c~x + co. There are now manyrestrictions on the coefficients. Wemust have s = 0 if I > 1 and t = 0 if g > 1 representing the fact that gl (x) and g2 (x) do have a commonzero. Further di = c~ for i = 1, ...,k - 1 and c~) = -t-2. c~ must be an odd integer hence we get that r = 1. We cannot have l = n = 0 for otherwise dk-~ > 0 and c~¢_ ~ < 0 (recall that k _> 2). Also cannot have l = O,g > 0 and n > 0 for otherwise ca_~ = -2 - s # d~-i = g - n = 2g + n - s (recall that here t = v = 0 and k - 1 = s = g + 2n). Assume that n = 0 and l > 0,g > 0. Then l+g = 2v, and from dk-1 = ck-1 and dk-2 ---- ok-2 we get the following three equations

(a) l + g = (b)

g-l

=-2-

-(c) 2 - + From (a) and (b) we get that v = 2g + 2 and I = 3g + 4. Using this in we obtain g = -3 which contradicts g > 0. Hence, n = 0 and l > 0, g > 0 also does not occur. In any case we must have g = 0. Now assume that t = 0. Then either n = 0, s = 0 and l = 2v or v = 0, l = 0 and s = 2n. If n = 0, s = 0 and l = 2v then dk-1 = Ok-1

178

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

implies that 2n = n+2 or equivalently n -- 2 and k = 5. But then d3 -- 6 ~ 5 = c3 giving a contradiction. If v -- 0, 1 -- 0 and s -- 2n then dk-1 ---- Ok-1implies that n ---- 2n + 2 or n = --2 giving a contradiction. Hence we must have t > 0. Assume that n > 0. Then v --- 0. We cannot have s = 0 because otherwise dk-1 : Ok-1implies that l + n ---- 2t which contradicts t ---- 2n + l. Hence we have s > 0 and l ~- 0. Then 2n ---- s + t and from dk-1 ---- Ck-1 and dk-2 ~- Ck--2 we get the fbllowing three equations. (a) 2n = s + (b) -n=-s+t-2

(c) ~--~-’~ ~------ 2~-1) 2~ st + ~+ 28. From (a) and (b) we get that n = 2t - 2 and s -- 3t - 4. These together with (c) then imply that t - 3 which in tt~rn gives that n = 4, s = 5 and k--9. Indeedwchavcthat ds =cs =-4andd~ --c7-2but nowwc also have d6 -- 6 ~ 10 - c6 giving a contradiction. Therefore we must have n~0. Nowl ---- t + 2v, and dk-1 -~ ck-~ implies that -l = -2 + t - v ----2-l-2t+v-l, that is, 2t+v=2. Sincet_> lwemust havev-~0and t ~-- l ---- 1. It ibllows that k -= 2. This gives the two polynomials g~(x) = x(x- 1) and g2(x) = (x- 2)(x + 1). Up to isomorphism these polynomials can be realised only tbr the group G =< a, b; a a = ba = (abab’~) ~ = 1 >. Now let G~ be the commutator subgroup of G. Using the Reidemeister -Schreier method G~ has the pre~ntation G~ -~

al~a2~aa;aa;a~a4

-~

aaal;a2a3

~-

aaa2:a

--1 --1 --1 2 a 3 alaaa I a 4

a2a4 = 1 >.

G~ ~ has a free group of rank 2 as an epimorphic image and therefore G contains a non-abelian free subgroup and hence so does G. It follows then that if tr(AB) ~ th en G contains a non-abclian fr ee su bgroup. Nowsuppose that tr(AB) ~- does no t oc cur. Th en al so tr (AB) = does not occur because gl (x) - g2 (x) = ±2. Therefbre we g~ (x) ---- (x - 1)~(x + ~ - x - 1)"~-=x~ + d~,-~x~-~ + ... + d~x + doand gu(x) = (x 1)~(x + 1)~ - x - 1)" = x~ ÷c~._~xk-~ -t- ...

÷ c~x ÷ co.

7.3.2

THE TITS AIJFERNATIVE- GENERATORS OF FINITE ORDER179

As before we have s = 0 if l > 1 and so on as a consequence of the fact that gl(x) and g2(x) have no commonzero. bhrther d~ = c~ for i = 1, ..., k - 1. Assumethat 9 > 0. Then t = 0. Suppose n > 0. Then v = 0, l = 0, k = s = g+ 2n and from dk-1 = ck-1 we get that g- n = -s. This implies that 2s = 3n and -s = 3g which is impossible since g > 0 and s > 0. Therefbre we have n = 0. Wemust have l > 0 fbr otherwise dk-1 > 0 and ck_l < 0. Hence we have l > 0 and s = 0. Further we have l + g = 2v = k and from d~-i = ck-~ and dk-2 = Ck--2 we get the fbllowing three equations: (a) l + g --

(b) g - l = (c) ~2 Ig + ~- ’~ 2

-- 2 From (a) and (b) we get that l = 3g and v = 2g. Using these we from (c) that 2g = 3g, that is, g = 0 which gives a contradiction. Therefore we cannot have g > 0 and so g = 0. By a symmetrical argument we also have t = 0. Finally let s > 0 and n > 0. Then v = l = 0. Now, s = 2n and -s = -n from dk-~ = Ck-1 giving a contradiction. In an analogous manner we can handle the case where l > 0 and v > 0. Therefbre it is impossible to get an elementary image fbr G if tr(AB) = 0 does not occur fbr the triple (3, 3, 3). Fromthis it follows that G znust contain a non-abelian free subgroup. This complctcs case (1) whcrc p = q = m = Case (2): p = 2, q = 3,m = In general if p = 2 and k is even, then f(x) = a~,x~ + a~_~x~-’z + ... + a~x’~ + ao; while if k is odd then f(x) = a~x~ + a~,_~x~-’~ + ... + a.~x~ + a~x. Henceif k is even and x0 is a zero of f(x) + c, c e C \ {0}, then -x0 is also a zero of f(x) + c. If k is odd and x0 is a zero of f(x) + c, c ~ C \ {0}, then -x0 is also a zero of f(x) Nowsuppose that p = 2, q = 3, m = 4. ’lb get an essential representation of G in PSL2(C) with elementary image group < A, B > we must choose tr(R(A, B)) -- ex/~, e = d=l. Hence we must consider the polynomials gl(x) = f(x) ex/2, e g2(x)-=f(x) --

= +1and

180

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

We cannot get < A, B > either dihedral or metabelian and if < A, B > is then elementary, it must be finite and isomorphic to $4 since it has an element of order 4. Therefore the possibilities for tr(AB) are x/~ or -v~. Further tr(AB) is a zero of either gl(x) or g2(x). It follows without loss of generality that we can only have k odd and

al(x) = (x - k and g~_(z) = (x kx/2) which is a contradiction fbllowing from the remarks on p = 2 above. Therefbre in the case p ---- 2, q = 3, m ---- 4, G must have a non-elementary image and hence must contain a non-abelian free subgroup. Case (3): p = 2,q = 3, m---To get an essential representation of G in PSL2 (C) with image group < A, B > we must choose tr(R(A, B)) = ai, 1, 2, 3, 4 wher e a~ = e(2cos(~r/5)),

e

a2 ---- --~(2 cos0r/5)) a3 = ~(2cos(2~/5),

~=

a4 = -~(2 cos(2r/5). Hence, we must consider the polynomials gi(x) = f(x) ÷ (~, 1,2,3, 4. As beibre we cannot get < A, B > either dihedral or nmtabelian and if < A, B > is then elementary, it must be finite and isomorphic to A5 since it has an element of order five. Thereibre the possibilities fbr tr(AB) are c~,c~2, c~3~ c~4. Further tr(AB) is a zero of one of the g~(x). It follows without loss of generality that wc can only have k odd and g~(x) = (x-~)a,i

1, 2,3,4

which again is a contradiction ibllowing from the remarks on p ---- 2 above. Thereibre in the case p ---- 2, q ---- 3, m = 5, G must have a non-elementary image and hence must contain a non-abelian free subgroup. Case(4): p=2, q=3, m=6: To get an essential representation of G in PSL~(C) with image group < A, B > wc must choose tr(R(A, B)) = evf~, ~ = ±1. Hcnee, wc must consider the polynomials gl(X)

"~-

f(x) + ev~, e = ±1 and

7.3.2

THE TITS AUFERNATIVE - GENERATORS OF FINITE ORDER181

as(x)= f(x) Wecannot get < A, B > to be finite since it has an element of order 6. Thereibre if < A, B > is elementary, it must be metabelian. It follows that the possibilities for tr(AB) are v~ or -x/~. Further tr(AB) is a zero of either gl(x) or g2(x). It follows as before that without loss of generality we can only have k odd and g~(x) = (x v/ 3) k and

as(x) = (x k which again is a contradiction. Thereibre in the case p = 2, q -- 3, m = 6, G must have a non-elementary image and hence must contain a non-abelian free subgroup. Case(5): p---2, q--4, m=3: To get an essential representation of G in PSL~.(C) with image group < A, B > wc must choosc tr(R(A, B)) -- e, e = =t=1 and wc must considcr the polynomials gl(x) f( X) + e, e = :t :1 an d

This holds wheneverp ---- 2 and m = 3. If q = 4 then an elementary < A, B > must be isomorphic to Sa and so the possibilities ibr tr(AB) are +1. Again this forces k to be odd and without loss of generality the polynomials to be g~(x)--ak(x-1)

~ and

g’~(x) = ak(x 1) a. As before this is a contradiction and we can conclude that in the case p - 2, q -- 4, m = 3, G must have a non-elementary image and hence must contain a non-abelian free subgroup. Case (6): p = 2, q-- 5,m ~To get a.n essential representation of G in PSL2(C) with ima.ge group < A, B > which is elementary we must have < A, B >~- A5. Let ,\ 2 cos(~r/5). Then we have s =~ + 1 and th e po ssibilities fo r tr (AB) ar 1,-1, A-land 1-A. Suppose first that k is even, that is k --- 21 _> 2. Then without loss of generality we can only have g~ (x) = a~(x"~ - 1)~ and

182

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS g~(x) = ak(x"~ + ~ -- ~ 2)

which implies that l = 1. Howeverif I = 1 we obtain from gl(x) - g2(x) ±2 that a2(1 - ~) = ±2 or equivalently that 2 - ---- ±2~. This i s a . c ontra,diction sincc a2 = Sql (A)Sa2(A), that is, without loss of gcncrality, a2 is 1 or A or A ÷ 1. Thus k cannot be even. Nowsuppose k is odd. Then we must have gl(X) -- ak(x -- e)"(x -- ~(.~ 1)) s and g~(x) = ak(x e) "(X + ~(A -- ~, where r + s ---- k and ~, y -- ±1. This also gives a contradiction since -er-

~s(A-1)¢0.

As beibre we can conclude that in the case p -- 2, q -- 5, m = 3, G must have a non-elementary image and hence must contain a non-abelian free subgroup. Case (7): p = 2, q = 6, m-To get an essential representation of G in PSL~_(C)with image group < A, B > which is elementary we may have < A, B > dihedral or metabelian but not isomorphic to A4, $4 or A5. Therefore the possibilities for tr(AB) axeO, 1, --1. If k is odd then 0 cannot be a zero of either g~ (x) or of g2 (x) and hence without loss of generality we can only have g~(x)=a~(x-

1) a and

g (x) = a (z 1) which gives a contradiction. Nowsuppose k is even. Then without loss of generality

we must have

g~ (x) = a~x~ and gu(x) = a~(x"~ - ~ 1) where k ~-- 21 _> 2. This implies that k ---- 2 which gives the polynomials g~ (x) = a2x"2 and g2(x)=a2(x ~ -- 1).

7.3.2

THE TITS AUI’ERNATIVE- GENERATORS OF FINITE ORDER183

a2 can only be 1, ~/3, 2, 3 or 2v/3. Using gl (x) - g2(x) -- ±2 we get that a2 = 2 and hence we must have gl (x) = 2 and

g2(x)= 2(xUp to isomorphism these polynomials can only be realized ibr the group G =< a, b; a"~ = b~ = (abab’~)"~ = 1 > . Let H be the subgroup of G generated by u = b and v -- aba. H has a presentation H =< u,v;u ~ = v ~ -~ (vu~) "~ = (uv’~) ~ = 1 > . This then has the free product Z3 * Z3 as an epimorphic image and so has a non-abelian free subgroup. Thereibre in the case p -- 2, q = 6, m = 3 either the group has a nonelementary image or is up to isomorphism the group G above which has a non-abelian free subgroup. Hence in all cases the group has a non-abelian free subgroup. The final case covered by the theoremis p -- 2, q -- 4, m-- 4. Case (8): p = 2, q = 4, m= To get an essential representation of G in PSL2(C) with elementary image group < A, B > we must choose tr(R(A, B)) = ev~, e = ±1 and must consider the polynomials gl(x) = f(x) ~v/2, e -- -- ±1 and

Here g~(x)-g2(x) = 2x/2e and tr(AB) is again a zero ofgl(x) or ofg~(x). Since G contains an element of order 4, < A, B > cannot be isomorphic to either Aa or A5. Therefore the possibilities for tr(AB) axe 0, 1,-1, x/~ and - ~/~. Suppose first that k is odd. Then 0 cannot be a zero of either gl (x) of g2(x) and hence without loss of generality we can only have g~(x) = a~. (x - e)r(x ~, ~/2)~ and

=

+

+

184

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

where r ÷ s = k, 0 _< r, s and e, u = ±1. Howeverthis produces a contradiction since -er - usv~ ~ O. Therefore k must be even. Let k -- 21 _> 2 and suppose first that tr(AB) = occurs an d as sume, without lo ss of generality tha t gl( 0) --= Nowwe have the possibilities:

(a):

gl(X) ---- akxk and g2(x) = a~,(x"~ - 1)r(x u - 2) ~ with r+ s = l, r, s >_O.

(b):

gl(x) = akxr(xu - 1)~ with r + 2s -- k, r, s _> 0 and g~ (x) = a~, 2 - 2)t

(c):

g~(x) = akxr(x 2 -- 2)~ with r + 2s = k, r, s >_ 0 and g2(x) a~(z 2 - 1) t.

If case (a) holds and I _> 2 then r + 2s # 0. Hence l = 1, that is, r = or s = 1, Wecan then have only a 2 = 1, V~ or 2, Hence r = 0~ s = 1 and a.~ = "v/~2 since gl(x) - gz(z) -= d=2vf~. This gives the pol:~aaomials u and gl(X) ---- X/2X g2(x) ---- X/~(x2 - 2). Up to isomorphism these polynomials can be realized only ibr the group G =< a, b; a~ = b4 = (abab’~)~ = 1 > . Let H be the subgroup of G generated by u = b and v = aba. Then H has a presentation H--
4--v4--(vuu)4--(uv2)4--1

>.

H then has the group K =< c, d; cu = d4 -- (cd2) 4 -- 1 > as an epimorphic image. This group K decomposes as a free product with amalgamation K = K~ *A K~

7.3.2

THE TITS ALTERNATIVE - GENERATORS OF FINITE ORDER 185

where K1 --< d;d 4 = 1 >, K2 =< c,e;c 2 = e ~ = (ce) 4 -- 1 > and A --< e;e 2 = 1 > with the identification e --= ar2. Let g ~- dcd[2cd-1 and h -- cd2. Using the cancellation method in free products with amalgamtion we get that the subgroup of K generated by g and h has the presentation

This is a non-trivial free product Z2 * Z4 and hence has a free subgroup of rank 2. Now suppose that case (b) holds. Again we must have 1 -- 1 since g2(x) -gl(X) : :t:2v/-~. This implies then that k = 2 and r + 2s = 2 which contradicts r > 0, s > 0. Therefore case (b) cannot hold. Finally suppose that case (c) occurs. Then we must have > 4 and a~. -- 2V~ since gl(x) - g2(x) -- ±2v/~. This implies that we have gl(x)

= 2V~x"(x~ - 2) s = 2V~xk + d~-2x k-e ~ + ...

+ d2x

with r + 2s = k _> 4,0 < r,s, and g2(x) ---- 2v~(x: - 1) ~ =- 2v/~x~ + c~-2xk-~ + ... + c2x~ + co. Fromdk-2 ~- Ok-2 we get that 2s ----- l and hence r -- l. Therefore we must have l = 2 since if l > 3 it would follow that d2 = 0 but c2 ~ 0. This gives us the polynomials gl (x) = 2v~x2 (x ~ - 2) and

g2(x)= 245(x- 2. Up to isomorphism these polynomials can be realized only for the group 4 -- 1 > . G -= . H then has the free product Z2 * Z4 as an epimorphic image and therefore H contains a non-abelian free subgroup and hence so does G. Now assume that tr(AB) = 0 does not occur. Then without loss of generality we must have the polynomials g~(x) = atc(x ~ - 1) ~ and

186

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS g2(x) = aa(x~ - 2) ~ where k = 21.

If I _> 2 then l = 21, that is l = 0 contradicting that 1 > 0. Therefore we must have l = 1 and k = 2. This gives the polynomials gl(x) = a2(x~ - 1) and g2 (x) = a2 2 -- 2). As befbre since gl(x) - g2(x) = +2v~, we must have a2 = ±2v~. However this gives a contradiction because we can only have a2 = 1, v~ or 2. Hence wc can get no group hcrc with elementary image. This completes the case p = 2, q --- 4, m= 4 and the proof of the theorem. Whenthe generators have finite order, we have now reduced to the triples (2, 2, m),(2, m, 2),(2, 3, 3),(3, 3, 2),(3, 4, 2),(3, 5, 2), and (3, 6, 2). (2, 2, m) both generators have order 2 and the relator can be rewritten (ab) t for some t. This is easily seen to be a finite dihedral group. Thus the opencases are the triples (2, m, 2),(2, 3, 3), (3, 3, 2),(3, 4, 2),(3, 5, (3, 6, 2). The next result handles the case (2, 3, THEOREM 7.3.2.4. sentation

Suppose G is a generalized triangle

group with pre-

G =< a,b;a 2 = b3 = R3(a,b) = 1 where R(a, b) = abqiabq2 ... abq~ with k > 1 and 1 _< qj _< 2 forj = 1,..., k. (1) If k >_ 2 then one of the following holds. (a) G has a free subgroup of rank 2. (b) is fin ite of order 1440 andup t o i somorphism has the presentation G =< a, b; a2 -- b3 = (ababab~)a = 1 >. (c) G is an infinite presentation

solvable group and up to isomorphism has the

G =< a, b; a2

= b3 =

(ababababU)3 = 1 > .

(d) k = 2 and G has a free abelian subgroup of rank 2 and of index 6.

(2) If k = 1 then R(a, b) ~ with1 <_ t < 3, and Gis i somorphic to the alternating group A4. PROOF.Suppose first that k _> 2. If or R(a, b) = abab2 (the two symmetric isomorphic groups). In the first case 1 >=< a, b; a~ = b3 = (ab) 6 = 1 >. This

k = 2 then either R(a, b) = abab words ab2ab2 and ab2ab lead to G =< a, b; 2 =b3= ( abab) 3 = is a Euclidean ordinary triangle

7.3.2

THE TITS ALTERNATIVE - GENERATORS OF FINITE ORDER 187

group which contains a free abelian subgroup of rank 2 of index six. In the secondcase G --< a, b; a2 = ba = (abab~)a = 1 > . This has the symmetric group Sa as an epimorphic image mapping a to a 2-cycle and b to a 3-cycle. Using Reidemeister-Scheier we find that the kernel is free abelian of rank 2 and clearly of index 6 since ]Sa[ = 6. Nowsuppose that k _> 3. The general strategy is the same as in the previous theorem. That is we attempt to find, up to isomorphism, all G such that for an essential representation p : G -* PSL2(C) given by a -~ A, b --* B with tr(A) = 0, tr(B) = 1 then p(G) is elementary, recalling that if G has a non-elementary image it has a free subgroup of rank two. Hence we assume that for each essential representation p : G -* PSL2(C) given by a -~ A, b -~ B with tr(A) = O, tr(B) --- 1 then p(G) is elementary and hence finite or metabelian. To get an essential representation of G in PSL2(C) with image group < A, B > we must choose tr(R(A, B)) -- e, e -- ±1 and we must consider the polynomials gl(x) = f(x) + e, e = ±1 and

which are both in Z[x]. Further ak = 1. tr(AB) is a zero of either g~(x) or of g2(x) and the possible values for tr(AB) are 0,1,-1,~/~,-x/~,v~,-x/~,A,-A,1A and A- 1, where 2 cos(r/5), s =A + 1. Supposefirst that k is odd. Then as in the proof of the previous theorem we must have without loss of generality gl(X) = ((X + 1)(X -- A)(X-- 1 r = (Xa -- 2X -- 1 ) r and g2(x) = ((x - 1)(x + A)(x + r = (x3 - 2x+ 1) r, r e N. This is possible only for r = 1 and then up to isomorphism the resulting polynomials can be realized only for the group G =< a, b; a2 = b3 = (ababab=)a = 1 > . This group was shown to be finite of order 1440. Wenote that M.Conder showed that G is an extension of the cyclic group Z2 by a direct product of the alternating groups At and A5. Nowsuppose k = 21 _> 4. Let y = x2 and we get g~(y) = y~ + dt_~y~-1 + ...

+ d~y + do and

188

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS gu(y) = y~ + ct-ly ~-1 + ... +cly ~- co

with gl(Y) -g~(Y) = ±2. The possible zeros ofgl(y) and gu(y) are 0, 1, 2, 3, A2 2. or (A - 1) If As is a zero of gl(Y) then (A - ~)2 is also a zero of gl (Y) since gl Z[y]. Hence if A2 is a zero of gl(Y) then gl(y)=(y 2 - 3yT1)rh~(y),r where hi(y) E Z[y] and the possible zeros of hi(y) are 0, 1, 2, 3; and 2 and (A - 1) 2 are not zeros of g2(y)- An analogous remark holds for By using the equalities gl(Y) -g2(Y) = ±2, d~_l -- c~-i and d~_: ct_2,(the latter if l > 2) we get, in an analogous manner to that in the previous theorem, that As cannot be a zero of g~ (y) or of g2 (Y). It follows that only 0, 1, 2, 3 are the possible zeros of gl(y) and gu(y). Since g~ (y) - g2 (Y) -- ±2, we have that 0 must be a zero of one of two. Assume, without loss of generality that g~ (0) -- 0. Then 2 must a simple zero of g~(y) since gl(y) g~(y) = ±2. No w Clmust be an odd integer and hence 0 is a simple zero of gl (Y)- 3 cannot be a zero for g~ (y), again because gl(Y) g~(Y) = ±2. Si nce I _>2 we therefore obt ain g~(y) = y(y 3) t-~ and g2(Y) -=- (Y 2)(y - ~-1. This is possible only for l = 2 and hence k -- 4. polynomials gl (X) = 4 -- 3 X~ and

This leads to the

g~(x) = ~ -3x2 + 2 2. replacing y by x Up to isomorphism these polynomials can be realized only for the group G = . Weclaim that this group is infinite solvable. Let x --- ababab2ab2, y = abab2 ab2 ab, u -- ab2 ab2 abab and v = ab2 ababab2. Let H be the subgroup of G generated by x, y, u, v. H is normal in G with G/H ~- Z3 × $3 of order 18. H is center-by-abelian with commutator subgroup H~ of order less than or equal to two lying in the center of G. Hence G is solvable. G is infinite because the Euclidean triangle group T(2,3,6)----<

a,b;a2=b3--(ab)6=l

7.3.2

THE TITS ALTERNATIVE - GENERATORS OF FINITE ORDER 189

is an epimorphic image of G. This completes the proof of Theorem7.3.2.4 for k _> 2. If k = 1 then certainly we may asume that R(a, b) = ab so is isomorphic to the alternating group A4. Themostdifficult cases are (2, n, 2),(3, 3, 2),(3, 4, 2),(3, 5, 2) and (3, The result of J.Howie H 9 (see Theorem7.3.2.9) settles the case (3, 6, For the remainder we have only partial results depending on the length of the relator. Wefirst summarizethe situation for (3, 3, 2). THEOREM 7.3.2.5. sentation

Suppose G is a generalized triangle

group with pre-

G =< a,b;a 3 = ba = R2(a,b) -- 1 whereR(a,b) -- aPlbqlaP:bq2 ...aPUbqk with k _~ 1 and 1 _< Pi -< 2, 1 _< qj _< 2 for j = 1,..., k. Then if k <_ 4, or alternatively the syllable length is 8 or less, then the Tits alternative holds [or G. In particular G has a free subgroup o~e rank 2 or is finite. The cases where G is finite are precisely when up to isomorphism (1) G =< a, b; a3 = b3 = (ab) 2 = 1. In this case G is isomorphic to A4 and hence has order 12. (2) G =< a, b; a3 = ba = (abab2)~ = 1 >. In this case G has order 180. (3) G -=. In this case G has order 288. PROOF.The proof is based on arguing on the syllable length 2k and then using the techniques and strategies of the previous proofs in this section. That is we attempt to find essential representations with elementary images. If the image group can only be non-elementary, then the group G must have a free subgroup of rank 2. To determine, for given syllable length k, the groups which can only have elementary images we make a detailed analysis of the trace polynomials. The case of syllable length 2, that is k -- 1, has already been handled, so we start with k = 2. Since (p, q, m) = (3, 3, 2) up to isomorphism, possible relators R(a, b) are abab, abab2 and aba2b2. The first case is the hyperbolic ordinary triangle group T(3, 3, 4) which has a free subgroup rank 2. In the other two cases we get the groups G1--
a=b3--(abab~)~=1>

and

G2 = . Both of these groups can be shownto be finite of the given orders, either using a subgroup or a computer calculation. For example consider G1. Let N1 be the cyclic subgroup generated by (ab) 5 and let N~ be the

190

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

closure of a in G. An elementray computation shows that (ab) 5 commutes with both a and b and hence N1 is normal in G. Further (ab) ~5 = 1. Then G/N~ =< u, v; u3 = v3 = (uv) ~ = (uvuvg-) 2 = 1 >~- As. Now G1/N2 2< x;x 3 _- 1 > so G1 is not isomorphic to A~. Therefore G~ has order IGll = (3)(60) = 180. G2 can then be handled by Reidemeist er- Schreier method. Now consider syllable length 6, that is k -- 3 so that R(a,b) a~lbtlaS2bt2aSab t3 with 1 _< s~ _< 2, 1 <_ ti < 2,i = 1,2,3. By examing the trace polynomials, as in Theorem 7.3.2.2, we can show that the only group, up to isomorphism, for which there are only elementary representations is 3=b3 G=< =(ababa2b~)2=1>. a,b;a A subgroup analysis now shows that this group contains a non-abelian free subgroup. Let H1 be the normal closure in G of ab. This has index 3 and a presentation H1 =< x, y; (xyxy2) 2 = (xay) 2 = (x-ly2) 2 = 1 > where x = aUba-~,y -- ab. Nowlet H~ be the normal closure in H~ of x. This has index 2 and a presentation H2 =< u,

V;

(u4v3)

2

= (UZV4)

2

= U2V2U3V2U2V

3

=

1 >.

The last relation implies that the words in the first two brackets are conjugates, so the second relation can be deleted. Let K be the normal closure in H2of the elements ur, uv-1. K has a transversal {1, u, u2, uz, ua, u~, u6} in H2 and is generated by zl = u~, z2 = vu-~, z3 = uvu-z, -~, za = u2vu -4, -5], -6, z5 = u3vu z6 = u4vu z7 = u5vu and z8 = u6v. If the relations z~ = z~ = z~ = Zs = 1 are added to K, the resulting factor group, after simplification, has the presentation ~ =< x,,x~,xz;x21

= x~ = x~ = (xlx~) 2 = 1 >~- 92 * Z2

where D2 is the Klein 4-group. Since this is a free product not Z2 * Z2 it follows that K is SQ-universal and therefore has a non-abelian free subgroup. Hence K and also G, has a non-abelian free subgroup. Further since K has finite index in G this argument also shows that G is SQ-universal. The final situation of where k = 4 so syllable length 8. This handled in much that same manner. Up to isomorphism all groups have essential non-elementary representations with two exceptions. The first is the group G1 =< a, b; a3 = b3 = (ab2aZbZaba2b)2 = 1 > .

7.3.2 THE TITS ALTERNATIVE- GENERATORSOF FINITEORDER

191

The normal closure H1 of b in G1 has index 3 and has a presentation H1=< x, y, z; x3 -- y3 _- z3 = Ix, y]2 = Ix, z]2 = [y, z]~ --- 1 > The normal closure K of z, yx in H~ then has index 3 and is generated by al ~ yx, a2 = xy, a3 -~ x-lyx-l~ a4 ~ z~ a5 -~ xzx -1 and a6 = -I. x--lzx Adding the relations a~ = 1 and a~ = a3 to K we get a factor group with a presentation

= (a4a~l) 2 = (alasala-~l) 2 = (alahala~l) 2 = 1 >. In K let H be the normal closure of a4,ah,a6. This has index 2 in K and is generated by xl = a4, x2 ~ ah, x3 -- a6, x4 -- ala4al, x5 -= alahal, and x6 = alasal. Adding to its presentation the relations xl = x2 = x4 = x5 gives the factor group ~ =< Xl,

Z3,

X6;

Xl 3 -~-

X33 = X6 3 = (XlZ3) 2 = (TlX6) 2 = 1 >.

This is the free product of the groups < x~,x3;x3~ = x~ = (xlx3) ~ = 1 > and < xl,x6;x~ -- x~3 : (XlXs) ~ : 1 > with the subgroup < Xl;X~ : 1 > amalgamted. From a result of Lossov [Los] H is SQ-universal and hence it follows that G~ is also SQ-universal. From this it follows that G1 contains a free subgroup of rank 2. The second exception is the group G~ =. In this case the normal closure presentation

H1 of < b > in G~ has index 3 and a

H~ =< x,y,z;x 3 = y3 = z3 = (xyxz2)~ = (yzyx2)~ = (zyzy2)2 = To show that H~ has an essential non-elemenmtary representation in PSL2(C) it suffices to find a map p : H~ -* PSL2(C) with p(x) = p(y) = Y, p(z) su chthat the t race s of X, Y and Z are one, a nd < X, Y > is non-elementary, and the values t~ = tr(XY), t2 = tr(XZ), t3 = tr(YZ) are non-zero solutions of the system of equations

t2-t2t3-$1~-O,

192

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS t3 -- tatl -- t2 =O.

The solutions of this system are the three (different) zeros of the polynomial f(t) = 3 - 3t2 + 3, irreducible over Q. All are real and one of them, which may assume to be tl, is greater than 2. In particulax, it follows that if ( is chosen so that ( + ~1 = tl, then with X’=( 0-1

II)’Y----(,ll

~)’

the group < X, Y ) is non-elementary. Finally, in a straightforward manner, we may construct a matrix

directly to satisfy the conditions tr(Z)

= a+d =

det(Z) = ad - bc = tr(XZ) = c - b + d = t2 and tr(YZ) = a - ~c + -1 = t3 . Hence, H~ and G2 have a f~ee subgroup of rank two. The eases (3, 4, 2) and (3, 5, 2), (also (3, 6, 2) although this is covered [H 9]) can be handled in an analogous manner. For syllable length less than or equal to eight the Tits alternative holds. A finite group can only occur when the syllable length is two, and the group is isomorphic to the symmetric group $4 or the alternating group As. The details are lengthy and can be found in [R 16,20] and [L-R 1,2]. THEOREM 7.3.2.6. sentation

Suppose G is a generalized triangle

group with pre-

G =< a,b;a 3 = bq-- R~(a,b) = 1

with q = 4, 5 or 6 and where R(a, b) = aPlbqlaP2bq2.., aP~bq~ with k >_ 1 and 1 <_ p~ < 2, 1 <_ qj <_ q for j = 1,...,k. Ilk <_ 4 then G satisfies $he Tits alternative. The final situation is (2, q, 2). The next theorem represents several results for this final case due to Rosenberger[R 16,20] and Levin and Rosenberger [L-R 1,2]. The proofs follow along the lines of the preceding parts of this section.

7.3.2

THE TITS ALTERNATIVE - GENERATORS OF FINITE ORDER 193

THEOREM 7.3.2.7. sentation

Suppose G is a generalized triangle group with pre-

G =< a,b;a 9 = ba = Rg(a,b) = 1 > where R(a, b) = abqlabq~.., q~ with k > 1 and 1 < qj < q[orj = 1,. .. , k. If k < 4 then G satisfies the Tits alternative. Further, if 2 < k < 4 then G is finite if and only if G has a presentation (a) G = whic h is f ini te of o rder 120; (b) G --< a, b; ~ =b5= (ababab4)~ = 1> which is f ini te of o rder 1200; (c) G = which is f ini te of o rder 1200; (d) G = whic h is f ini te of o rder 192; (e) G = which is f ini te of o rder 24; (f) G = whic h is f ini te of o rder 48; (g) G = whic h is f ini te of o rder 120; (h) = whic h is f ini te of o rder 576; Before movingon to the case where there are generators of infinite order we mention one further general result from the paper mentioned above. THEOREM 7.3.2.8. sentation

Suppose G is a generalized triangle

group with pre-

G =< a,b;a ~ = bq = R2(a,b) = 1 where R(a, b) = abqlabq~...ab ~ with k = 2kl > 2 and 1 < qj < q for j = 1,...,k. Then, if q > 7, q ~ 8, 10, 16 then G has a free subgroup of rank 2. We now present Howie’s Theorem [H 9] which settles the case where s(G) = 1. The proof uses some similar techniques to the preceding proofs, some of the preceding results and a general result (also proved in [H 9]) concerning free subgroups of the fundamental group of a finite 2-complex. For the details we refer the reader to the paper. Equivalence of R in the theorem is equivalence under the equivalence relation generated by automorphisms of the cyclic groups generated by a and b, inversions, cyclic permutations of R(a, b), and the interchange of a and b in the case where p=q. THEOREM 7.3.2.9. presentation

[H 9] Suppose G is a generalized triangle group with G =< a, b; ap = bq = Rm(a, b) = 1

where 2 < p < q, R(a, b) is a cyclically reduced word in the free product on a and b involving both a and b, which is not a proper power, m > 2 and let s(G) = (l/p) + (l/q) + (l/m). If s(G) = 1 then G contains a free

194

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

subgroup of rank 2, except in the case where the relator R is equivalent to ab in which case G is isomorphic to the solvable Euclidean triangle group T(p, q, r). In closing this section we give the following conjecture. CONJECTURE. The Tits alternative holds for any generalized group where the generators have finite order. 7.3.3

The Tits Alternative

o Generators of Infinite

triangle

Order

Wenow consider where one or both of the generators has infinite order, that is where p _> 2, q = 0 or p = q = 0. Where both generators have infinite order, that is where p = q = 0, the theorem of Ree and Mendelsohn guarantees a non-abelian free subgroup. Recall that for a 2-generator group G a generating pMr {u, v} is a Ree-Mendelsohn pair or RMopair if there exists an integer t such that < u*, v* > is a free subgroup of G of rank 2. Clearly the existence of an RM-pair shows the existence of a non-abelian free subgroup. In this section we interchange the role ofp and q and consider a to have finite order and b to be of infinite order. THEOREM 7.3.3.1. sentation

Suppose G is a generalized triangle

group with pre-

G =< a,b;a v = R’~(a,b) = 1 where 2 < p, 2 <_ m and R(a, b) is cyclicaJly re duced word in the free product on a and b involving both a and b. (1) I.f p >_3 or m >_3 then G contains an P, AYl-pair. (2) Suppose p = m = 2 so that = an suppose R(a,b) = ab’~labn2...ab "k with k >_ 1, hi # O,i = 1, ...,k. Then (a) H k >_ 2 or Ini[ >_ 3 [or some i, then G has a free subgroup of rank 2. In particular ilk is even or k is odd, k >_3 and n l + n2 +...+ nk ~ O, or k = 1 and [nl[ _> 3, then G has an RM-palr. (b)/_fk = 1 and [nl[ _< 2, then G is infinite and solvable. PROOF.We consider first case (2) of the theorem where q -- m -Suppose that G =< a,b;a 2 = R2(a,b) = 1 with R(a, b) = ab’~labTM ..... ab’*~, k >_1, and ni # 0, i = 1, ....k. Wewant to show that (1) Ilk >_ 2 or [ni[ _> 3 for somei, then G has a free subgroup of rank (2) If k = 1 and In1[ _< 2, then G is infinite and solvable. These results will be a consequence of the following technical lemma.

7.3.3 THE TITS ALTERNATIVE - GENERATORS OF INFINITE ORDER195 LEMMA 7.3.3.1. Let G be as above. Suppose one of the following holds: (1) k is even (2) k is odd, k >_3 and nl + n2 + .... + nk ~ 0 (3) k: 1 and]nil >_3 Then G contains a Ree-Mendelsohn pair. PROOF.(of Lemma) Suppose first that k is even. Then R(a, b) gT(a, b) with t E Z and T(a, b) in the commutator subgroup of the free product on a, b. Now choose

and as in the proof of Freiheitssatz choose a w0 so that tr(R(A, B)) 2 cos(r/2) -- 0 so that < A, B > represents Further we can choose w0 so that A and B have no commonfixed points. To see this notice that A has the fixed point zl = i and z2 = -i while B has only the fixed point z(wo) = w0 - 1. If A and B have a commonfixed point, then the group < A, B > is metabelian. Under a suitable conjugation by C ~ PSL2 (C) we obtain

CBC-~ = B~ -- :k

1 "

Because R(a, b) = b~T(a, where T(a, b) is in thecomm utator subg roup in the free product on a and b we must have R(AI,B1)= :k (1

~) for some c ~ C.

But R(A1, B~) has order 2 in < A~, B1 >, so this is a contradiction. Therefore A and B have no commonfixed points. Further ~r(AB) ~ O. Otherwise tr([A,B]) = tr(ABA-1B -1) -~ (tr(AB)) q- 2 = 2 which would imply that A and B must have a commonfixed point. Since tr(AB) ~ and A, B have no common fix ed poi nt, exa ctly the same argument as was used in the proof of Theorem 7.2.1 can be used to show that G has a Ree-Mendelsohn pair. This handles the case when k is even.

196

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Nowconsider the case whenk is odd, k _> 3 and nl Choose two projective matrices A:±(_01

~)andB---±(~

+ n2 ~- ......

~- nk ~

O.

~)withweC

with no commonfixed point. By induction we obtain tr(R(a, b)) = ±(aaw~ ÷ .......

÷ alw) with

ak =--nln2 ...... n~ ~ 0 and al -~ (-1)t+l(nl -~ ......

nk) ~ 0, k --- 2t+ 1.

Since tr(R(a, b)) is a non-constant polynomial in w of degree _> 3 with al ~ 0 we may choose w0, not equal to 0, for B such that tr(R(A, B)) Then R(A, B) has order 2 in < A, B > which then represents G. The argument used in the first case now goes through. Finally suppose that k = 1 and Inll _> 3. Then G has the form < a,b;a 2 = R2(a,b) = 1 > with R(a, b) = ’~1. This ha s as a f actor gro up ~--
~--bnl

=1>

with Inll >_ 3. This is a free product of cyclics and not infinite dihedral and therefore ~ has a Ree-Mendelsohn pair and hence so does G. This completes the proof of the lemma. Wecan now complete the proof of case (2) in the theorem. If k _> then the result follows from the lemmaif either k is even or k is odd with n~ + .... + nk ~ O. Supposethen that k is odd, k _> 3 and n~ + ... + nk = O. Let a~ = b-~ab~, for i ~ Z. Since the exponent sum of b in R(a,b) is zero, it follows that R(a, b) can be written as a word T on (ai}. Let n, M be the minimum and maximumi such that ai appears in T. Then G can be written as an HNNgroup G =
a~M= T~ = 1, b-~a~b = a~+l, i = m, .... M-1 >

This has free part generated by b and associated subgroups K1---< am,....... aM-1;a2m ..... K2~--~ am+l~...... 2aM; a

a2M_l = 1 > a2M---- 1 ~

7.3.3 THE TITS ALTERNATIVE - GENERATORS OF INFINITE ORDER197 Since k _> 3 it is clear that G has a free subgroup of rank 2. Nowconsider the case when k = 1. If Inll _> 3 the result follows from the lemma. If Inll = 1 then G =< a,b;a 2 = (ab) 2 = 1 > . This is infinite dihedral (Z1 * Z2) and solvable. Finally if In~l = 2 then G has the form < a,b;a 2 = (ab2) 2 = 1 >. It follows that ab2a = b-2 and so < b2 > is a normal subgroup of G. Further the factor group G/< b2 >-~< a, b; a 2 = b2 = 1 > which is infinite dihedral and therefore solvable. Then G is solvable since both < bu > and G/< b2 > The above completes the situation when p = 2 and m = 2. We note however that the technique we used of representing the group G in PSLu (C) with tr(B) = does no t gi ve th e desired re sult in thecasewherek is o dd, k >_ 3 and n~ + ... + nk = O. To see this consider the group < a, b; a2 = (ababab-2) ~ = 1 >. This group has a free subgroup of rank 2 but the image group < A, B > with tr(B) = is sol vable und er any repr esentation a -~ A, b -* B of G in PSL:(C). Wenow turn to case (1) of the theorem where either p >_ 3 or m _> Wehandle m _> 3 first and then consider the situation where p _> 3 and m=2. LEMMA 7.3.3.2. Suppose G =< a, b; ap = R’*(a, b) = 1 > with p >_ m >_ 3 and R(a,b) a cyclically reduced word in the ~ree product on and b involving both a and b. Then G contains a Ree-Mendelsohn pair. Pr~oor. (of Lemma)The proof is similar to the proof of the previous lemma. We find a homomorphic image of G in PSL2(C), G =< A, B such that A and B have no commonfixed points. The arguments used in the preceding results then carry through. As in the proof of the previous lemmachoose A--+(

0_1

~)withy:2cos(r/p)and

where wo E C so that tr(R(A, B)) = :t:2 cos(r/m) so that < .4, B > represents G. As in the proof of the previous lemmaw0 can be chosen so that A and B have no colnmon fixed points. To see this notice that A has the fixed points zi = cos(r/p) + i sin(r/p) and z2 = ~- = cos(r/p) - i sin(r/p), has the lone fixed point z(w) = w- 1. Choose w0 such that tr(R(A, B)) 2cos(r/m) as in the previous proofs, w0 is then a zero of a real polynomial, f(x) - 2 cos(r/m) with f(x) non-constant in x. Since this is a real polynomial ~-5 is also a zero.

198

ALGEBRAIC GENERALIZATIONS OF DISCRETEGROUPS

If z(wo) ---- Wo-- 1 # Zl and z(wo) ~ z2, then A and B have no common fixed point. Suppose then that Zl = wo- 1. Wewill then find a Wl such that R(A, B) still has the appropriate order and such that Z(Wl)¢ zl z2. An analogous argument would work if z2 = w0 - 1. To do this choose wl so that tr(R(A, B)) = -2 cos(r/m). Then R(A, B) still has order m in PSL2((~) and so < A, B > still providesa representation of G. Further Wlis a zero of the real polynomial f(x) + 2 cos(r/m) with the same f(x) as before. Then Zl # w~- 1 since zl = Wo- 1 and Wo~ wl. To see that w0 ~ wl suppose they were equal. Then wo would be zero of the polynomial f(x) + 2 cos(r/m) - (f(x) - 2 cos(r/m)) = 4cos(r/m). Howeversince m > 3, 4 cos(r/m) # 0 and so Wl # W0. Suppose then that zz = wl - 1. Then wl - 1 = z2 = Zl = w0 - 1. Thus Wl= w--6 and so w~is a zero of the polynomial

f(z) + 2cos(r/m) - (f(x) - 2 cos(=/m)) = acos(r/m) as before, giving a contradiction. ThereforeZ1 # Wl -- 1 and z2 # I/)1 -- 1 and therefore we can find A and B having no commonfixed points and providing a representation. The proof nowproceeds as in Theorem7.2.1. Weconsider the product AB. Suppose AB is loxodromic (see Chapter 4). Then BABis also loxodromic since tr(BAB) = 2tr(AB) - Now {AB,BAB} is a gener ating pair for < A, B >. Wewill show that these form a Ree-Mendelsohnpair for < A, B > - that is there exists a t e Z such that {(AB)t, (BAB)t} is a basis for a free subgroupof rank 2 in < A, B >. Since these are the images of a, b ~ G, it wouldfollow that a, b constitute a Ree-Mendelsohn pair in G. There exists a positive integer s such that [tr(BAB)8[ > 2 and [tr(AB)S[ > 2. It follows, as in Theorem7.2.1, from the work of Majeed [Maj] that there is a positive integer t such that {(AB)t and t} (BAB) is a basis for a free subgroupof rank 2. Now suppose AB is non-loxodromic. Since p > 2, m > 3 and A and B have no commonfixed point it follows that AB cannot be elliptic of order 2. Then after a suitable conjugation we may assume that < A, B > c PSL2(R). The result then follows from the work of Rosenberger [R 3,13]. This completes the proof of Lemma 7.3.3.2 and the case where m > 3. The final lemmahandles the case wherep > 3 and m = 2. Its proof will complete the proof of Theorem7.3.3.1.

7.3.3 THE TITS ALTERNATIVE - GENERATORS OF INFINITE ORDER199 LEMMA 7.3.3.3. Suppose G =< a,b;a p = R2(a,b) = 1 > wi~h > 3 and R(a, b) a cyclically reduced word in ~he free produc~on a and b involving both a and b. Then G contains a Ree-Mendelsohn pair. P~tOOF. (of Lemma)The proof is similar to the arguments used in the proofs of the previous two lemmas. Weconsider several cases and in each case we find a homomorphic image G =< A, B > in PSL2(C) such that and B have no commonfixed points. The arguments used in the preceding results then carry through. Wefirst handle the case when p is not a power of 2. Case (1) p -- 2~(2k ÷ l) with k _> 1. It is enough to prove the statement for p = 2k ~ 1. Consider a homomorphicimage if necessary. Assume then that p = 2k q- 1 _> 3. Choose A = :k ( 0-1 ~1with r = 2c°s(~r/p)

B=B(w)--5:(~

2w-w~-l) 2- w

and

with w E C.

A has fixed points zl = cos(w/p) q- i sin(r/p) and z2 = ~ = cos(r/p) i sin(~r/p). B has the lone fixed point z(w) = w- 1. Choose w0 such that tr(R(A, B)) -- 2 cos(~r/2) = 0 as in the previous proofs. If A and B a commonfixed point, then < A, B > is metabelian. Conjugating by a suitable C ~ PSL2 (C) we obtain -~ = A~ --- qCAC

0)

q-1 with q -- z~ or q = z2 and

CBC_l:Ul:q_( Then R(A1, B1) + q-

¯

~ 1) 1 "

X fo r so me c e C an d

x e Z with Because R(A1,B~) has order 2 in < AI,B1 > we have either x > 0 or x = 0, c ~ 0. This leads though to a contradiction since q2X ~ +1 if x > 0 because p = 2k + 1 _> 3 and 2c ~ 0 if x = 0. Therefore A and B have no commonfixed points and the previous arguments concerning free subgroups carries through.

200

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

If p is a powerof 2 we must consider several different forms of the relator R(a, b). Further if p = 2v it is enoughto prove the statement for v -- 2 that is p = 4. Then: Case (2). p = 4 and R(a, b) = aTbST(a, with r, s EZ,0 < r < 3,s ¢ 0 and T(a, b) in the commutator subgroup of the free product on a and b. Choose

1

6-w

,wEC.

A has the fixed points Zl = cos(r/4) + i sin(~r/4) and z2 = z-~ cos(Tr/4) is in(r/4). B has th e fi xed po ints za (w) = w - 3 + ~ an z4(w) = w-3- vr~. Nowchoose w0 so that tr(R(a, b)) = th at is so tha t < A, B > represents G in PSL2(C). Suppose A and B = B(wo) have at least one commonfixed point. Then the group < A, B > is metabelian and tr([A, B]) = tr(ABA-1B-1) = 2. Conjugating by a suitable C e PSL2(C~) we obtain CAC_I=A,=+(q

0 with q = Zl or q = z2 and 0 q-1

CBC-1 ~- BI = +

t_ ~

with cl e (~ and t -- 3 + x/~ or t = 3 - v~. Then under this conjugation the relator R(A~, B1) must have the form R(A~,B~)

+ 0 q- rt-s "

Because s ¢ 0, R(AI, B~) cannot have order 2 in < A~, B~ > which is a contradiction. Therefore A and B = B(wo) have no commonfixed points and the method of the previous cases carries through. Case (3). p -- 4 and R(a,b) = aTT(a,b) with r ~ Z, r ¢ 2, 0 < r < 3 and T(a, b) in the commutator subgroup of the free product on a and b. As before choose A=+

-1 x (0 1) with B = B(wo) +

x= 2cos(r/4)

--

V~ and

Wo- 1) (~o 2wo- 2-Wo~

7.3.4

THE FINITE GENERALIZED TRIANGLEGROUPS

201

with Wo~ C chosen so that < A, B > represents G - that is R(A, B) has order 2. NowB has only Z(Wo) = Wo- 1 as a fixed point. If A and B had this as a commonfixed point then < A, B > is metabelian and a suitable conjugation would lead to a contradiction on the relator R(a, b). Therefore A and B would have no commonfixed points and we argue as in the previous cases. Nowthe final case. Case (4) p = 4 and R(a, b) = a2T(a, wit h T(a, b) in the comm utator subgroup of the free product on a and b. Since p = 4 the group ~ =< ~,~;~2 = R2(~,~) = 1 >=< ~,~;~2 T2(~, ~) = 1 > is a homomorphicimage of G with T(~, ~) in the commutator subgroup in the free product on ~ and ~. If T(~, ~) = 1 then ~ = 7,2 * and the statement of the lemma holds for G and therefore also for G. If T(~, ~) # 1 then T(~, ~) is conjugate to a cyclically reduced word TI(~,~) involving both ~ and ~. The result then follows from Lemm.a7.3.3.2. This completes the proof of Lemma7.3.3.3 7.3.3.1.

and the proof of Theorem

Summarziugthe statements in Theorems 7.3.2.1 to 7.3.2.6 7.3.3.1 we get our main result. THEOREM 7.3.2. tion

Let G be a generalized

triangle

and Theorem

group with presenta-

G =< a, b; ap = bq = Rm(a, b) = 1 where p <_q, p >_2 or p = O, q >_2 or q = O, R(a, b) is a cyclically reduced word in the free product on a and b involving both a and b and m >_ 2. Then G satisfies the Tits alternative except possibly whenp >_2, q >_ 2, m = 2, (l/p) + (l/q) _~ and the relator R(a, b) has s ylla ble lengt greater than 8 in the free product on a, b. 7.3.4

The Finite

Generalized

Triangle

Groups

From the preceding section we see that a generalized triangle with presentation

group G

G =< a,b;a p = bq = R’~(a,b) = 1 with 2 _< p _< q is infinite if s(G) -- (l/p) ÷ (l/q) + (l/m) _< 1. For ordinary triangle groups (see section 7.3.1) this is enoughto characterize the finite groups - that is T(p, q, m) is finite if and only if(1/p)~-(1/q)+(1/m) 1. Howeveras we can see from the results in the previous section there are examplesof infinite generalized triangle groups for every possible triple with

202

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

q _> 3. This raises the question of whether the finite generalized triangle groups can be completely classified. The answer is yes and the classification was carried out by Howie, Metafsis and Thomas(see [H-M-T1]) and Levai, Rosenberger and Souvegnier [L-R-S]. The first three people determined an almost complete list which classified all the finite groups except with two presentations undecidable. The second three people determined that one of these possibilities was infinite while the other was finite of large order. This final group has been called the LRS monster. The classsification proceeds in the following manner. Let G =< a,b;a p = bq = Rm(a,b) = 1 with 2 <_ p _< q be a generalized triangle group with s(G) (l /p) + (l/q) + (l/m) > 1. As in the previous section, if G were to be finite, leaves only the followingpossible triples: (2, 2, n), (2, n, 2), (2, 3, 3), (2, (2, 4, 3), (2, 3, 5), (2, 5, 3), (3, 3, 2),(3, 4, 2) and (3, 5, 2). Suppose that R(a,b) = ~’lbql . ..a p~bq~ with k > 1 and 1 __ 2. Again the work in section 7.3.2 shows that for any relator R(a, b) the triples (2, 3, 4), (2, 4, 3), (2, 3, 5) and (2, 5, 3) alwayslead to infinite From Theorem7.3.2.3 the case (2, 3, 3) is infinite unless G, up to isomorphism, has the presentation G = . In this case G is finite of order 1440. This leaves the possibilites of (2, q, 2), (3, 3, 2), (3, 4, 2) and(3, 5, The generalized triangle group G admits an essential representation p : G --~ PSL2(C). Let a --* A, b --* B, so then A has order p, B has order q and R(A, B) has order m. If G were finite, the image group G =< A, B > would be a finite subgroup of PSL~(C) generated by an element of order p and element of order q. As we used in the last section, it is knownthat the only finite subgroups of PSL~ (C) are up to isomorphism the cyclic groups for all n E l~l, the dihedral groups D,~ for all n E N, the alternating groups A4 and A5 and the symmetric group $4. Thus if G were finite, we know the possible structures of the image under the essential representation p. Recall that we can choose A,B ~ PSL~(C) such that tr(A) = 2cos(~r/p) and tr(B) -- 2 cos(w/q). It follows as before that tr(R(A, B)) is a polynomial of degree exactly k in x -- tr(AB) where k is as above, tr(R(A, B)) --- f(x) is

7.3.4

THE FINITE GENERALIZED TRIANGLEGROUPS

203

the trace polynomial for G. Since the possible finite subgroups of PSL2(C) are knownthis gives a finite list of possible values for tr(AB) and hence a finite list of possible values for the roots of the trace polynomial. Define a polynomial

= II(f(x) where the product is taken over all primitive m-th roots a of -1 with nonnegative imaginary part. Thus, if m -- 2, 3, 4, 5 then a(x) is respectively f(x), (/(x))2 _ 1, (f(x)) 2 - 2 or (f(x)) 4 -- 3(f(x)) 2 ÷ 1. Then a(tr(AB)) -- 0 if and only if tr(R(A, B)) --- a ÷ -1 fo r so me primitive root a of -1, which is the case if and only if R(A, B) has order m in PSL2(C). Howie, Metafsis and Thomas [H-M-T 1] then proved the following. LEMMA 7.3.4.1. Let G be a generalized triangle group, f(x) its trace polynomial and a(x) the polynomial as defmed above. If ~r(x) has a multiple root, then G is infinite. This lemma, whose proof we will refer to the paper [H-M-T 1], provides a bound on the degree of the trace polynomial and hence a bound on the length of the relator R(a, b). These reductions leave only a finite numberof possibilities in the (2, q, 2), (3, 3, 2), (3, 4, 2) and (3, 5, 2) tions to check. Building on the previous results of Baumslag,Morganand Shalen [B-M-S], Conder[Co 1] and Fine, Levin and Rosenberger [ F-L-R 1], [R 16,20],[L-R 1,2], Howie, Metafsis and Thomas [H-M-T 1] using a mixture of representation-theoretic, geometric group theoretic and computational group theoretic techniques identifed eleven finite generalized triangle groups (with k _> 2), numbers(1) through (11) in the classification theorem and showed that there were only two other possibilities,(12) and a group G1. Levai, Rosenberger and Souvignier [L-R-S] were able to show that G l was infinite and that the group (12) was finite. The final group (12) determined to be of order 424673280 = 220345 using the computational group theory system GAP. Howie, Metafsis and Thomas have called it the LRS-Monster and have studied some of its subgroups. Wenow list the classification. THEOREM 7.3.4.1. Let G be a finite generalized triangle group and not an ordinary triangle group. Then G is, ~p to isomorphism, one off (I) < a, b; 2 =ba= (abab2)2 = 1>~- A4 xZ2 oforder 24; (2) < a,b; ~ -- b 3 :(ababab2)2 = 1 > oforder 48;

204

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

(3) < a,b;a 2 = b 5 = (abab2) 2 = 1 >~A~ xZ2 o/:order 120. This is ~ -- 1 >; isomorphic to the group < a, b; a~ = b3 --- (abababab~) ~ 3 2 (4) < a, b; =b = ( abab2) = 1>~A5x Z3 of order 180; (5) < a,b; ~ =b4= ( ababab3)~ = 1> of order192; (6) < a, b; ~ =b3= (aba~b~)2 = 1> oforder288; ~ --- 1 > oforder 576; (7) < a, b; 2 -- b 3 --- ( ababab2ab2) 2 3 ~ = 1> of order720; (8) < a, b; =b = (ababab~abab~) 2 5 2 (9) < a, b; =b = (ababab4) = 1:>-- -< a, b ; a 2 =b5 = (abab2ab4)~ = 1 > of order 1200; (10) < a,b; ~ =b~= (ababab~)a = 1> of order1440; (11) < a,b; ~ =ba= ( ababababab2ab2)2 = 1> oforder 2880; (12) < a,b;a 2 = b3 = (abababab2ab2abab~ab~) ~ = 1 > of order 424673280-- 220345. PROOF.Weshow that the list in the statement of the theorem is complete up to equivalence under the equivalence relation generated by tomorphisms of the cyclic groups generated by a and b, inversions, cyclic permutations of R(a, b), and interchange of a and b in the case where p -- q. That is we show that any generalized triangle group presentation leading to a finite group is equivalent to one of the 14 presentations in the list. Nowsuppose that G is a generalized triangle group with presentation G--< a,b;a p -- bq -- Rm(a,b) = 1 where 2 <_ p <_ q,m _> 2 and R(a~b) = aPlbal ...aP~b a~ with k _> 1 and 1 <_pj 1 so we are reduced to considering the triples considered in section 7.3.3.:(2,2, m), (2, q, 2), (2,3,3), (2,3,4), (2,4,3), (2,3,5), (3, 3, 2), (3, 4, 2) and (3, 5, 2). As before the case (2, 2, m) reduces nary triangle groups. If k = 1 then Theorem 7.3.2.2 shows that G is infinite unless G is an ordinary triangle group. Therefore we can restrict attention to where k _> 2. From Theorem7.3.2.3, if G were to be finite we must have m _< 3 so we can restrict to the triples (2, q, 2), (3, 3, 2), (3, 4, 2), (3, 5, 2) and (2, 3, q_>3. The case (2, 3, 3) was covered by Theorem7.3.2.4. Here the only finite group, up to isomorphism, is G =< a, b; a~ = ba = (ababab2)~ = 1 > which has order 1440 and is number (10) on the list. In the case (3, 3, 2) a check of the possible finite groups in PSL2(C) together with the fact that the trace polynomial must have only simple

7.3.4

THE FINITE GENERALIZED TRIANGLEGROUPS

205

roots by Lemma7.3.4.1, if G were to be finite, shows that k _< 4. This was then handled in Theorem7.3.2.5 where it was shown that the only two finite groups are G =< a,b;a 3 = b3 = (abab~) 2 = 1 > which has order 180 and is number (4) on the list and G =< a, b; a 3 = b3 = (aba2b2)~ = 1 > which has order 288 and is number (6) on the list. In the cases of the triples (3, 4, 2) and (3, 5, 2) from Lemma7.3.4.1 also that 2 < k < 4. There situations with 2 _< k _< 4 were covered by [R 16,20] and [L-R 1,2,3] where it was shownthat there were no finite groups. Weare nowreduced to the cases (2, q, 2) with q _> 3. Supposeq _> 6, then G and hence an essential image of G in PSL2(C) will contain an element of order q _> 6. The only finite non-cyclics subgroups of PSL2 (C) having elements of order greater than 5 are the dihedral groups. This implies that if < A, B > provides an essential image of G, then the only possible value for tr(AB) = 0. Hence the trace polynomial has degree 1 and R(a, b) has syllable length 2, that is up to equivalence R(a, b) -~ abk. If gcd(k, q) 1 then this reduces to an ordilaary triangle group while if gcd(k, q) ?~ Theorem 7.3.2.2 shows that G must be infinite. It follows that if G were to be finite we must only nowconsider the triples (2, 3, 2), (2, 4, 2), (2, 5, These are completed by the following three lemmas. LEMMA 7.3.4.2. Suppose G is a generalized triangle group with triple (2, 5, 2) and relator R(a, b). Then G is finite only if, up to the equivalence as defined at the beginning of the proof,, (i) R(a, b) = ab in which case G is an ordinary triangle group o[ orderlO. (2) R(a, b) = ~ in w hich case G hasorder 120. T his i s number(3) on the list. (3) R(a, b) ~ ababab4 in which case G has order 1200. This is number (9) on the list. (4) R(a, b) abab~ab4 in which cas e G a ls o has orde r 1200. This is number (9) on the list. R(a, b) aSabab4 and R( a, b) = aba52ab4 def ine iso morphic groups. PI~OOF. (of Lemma7.3.4.2) Since G has an element of order 5, if were finite, an essential image must be either the dihedral group D5 or the alternating group As. Suppose G --< A, B > is a finite essential image of G. If G - D5 then tr(AB) ~- while if G --- A5 the poss ible valu es for tr(AB) are :1:1 or q_ 1-~_~. (Not q- ~+~-~since < A,B > is finite). It follows that the trace polynomial has degree at most 5 and so R(a, b) has syllable

206

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

length at most 10 - that is k _< 5. If k _< 4 this has already been handled in the the previous section giving the finite groups on the list. If k -- 5 and G is finite, then all possible roots occur in the trace polynomial. A computer search (see [H-M-T1]) then shows that the only possible relators are then R(a, b) = ababaab~ab4abk,1 _< k <_ 4. An analysis of pictures (see Chapter 5) which is beyond the scope of these notes reveals that in these cases G must be infinite. LEMMA 7.3.4.3. Suppose G is a generalized triangle group with triple (2, 4, 2) and relator R(a, b). Then G is finite only if, up to the equivalence as defined at the beginning of the proof, (i) R(a, b) = ab in which case G is an ordinary triangie group order 8. (2) R(a, b) = ababab3 in which case G has order 192. This is number (5) on the list. PI~OOF. (of Lemma7.3.4.3) The only non-cyclic finite subgroups PSL2 (C) containing an element of order 4 are the dihedral group of order 8, D4 and the symmetric group $4. The corresponding values for tr(AB) are then :t:1 or 0. It follows that the trace polynomial has degree at most 3 so that R(a, b) has syllable length at most 6 - that is k _< 3. The results then follow from the results of the previous section. LEMMA 7.3.4.4. Suppose G is a generalized triangle group with triple (2, 3, 2) and relator R(a, b). Then G is finite only if, up to the equivalence as defined at the beginning of the proof, (1) R(a,b) = ab in which case G is an ordinary triangle group of order 6. (2) R(a, b) = 2 in w hic h case G hasorder 24. Th is i s number(1) on the list. In this case G (3) R(a, b) = ababab2 in which case G has order 48. This is number (2) on the list. (4) R(a, b) = abababab2 in which case G has order 120. This is number (3) on the fist. In this case G is isomorphic to A5 x Z2 which is isomorphic to the group with triple (2, 5, 2) ~. and R(a, b) = abab (5) R(a, b) -- ababab2ab~ in which case G has order 576. This is number (7) on the list. (6) R(a, b) = ababab2~ in which caseG hasorder 720. This i s number (8) on the list. (7) R(a, b) -- ababababab2ab2 in which case G has order 2880. This is number (11) on the list. (8) R(a, b) = ababababUab2abab2ab2 in which case G also has order 424, 673,280 -- 22°345. This is number (12) on the list.

7.3.4

THE FINITE

GENERALIZED TRIANGLE

GROUPS

207

PROOF.(of Lemma7.3.4.4) Since G is generated by an element of order 2 and an element of order 3, G is an epimorphic image of the Modular Group M = PSL2 (Z) ~- Z2 * Za. M.Conder[Co 1] determined all finite quotients of Mby a single relator of free product length at most 24. In particular this includes R~(a, b) if R(a, b) has length at most 12. His listing gives the finite generalized triangle groups (1) through (7) in the Lemma and restricts our attention here to length 14 or greater. Possible essential finite images of G in PSL2(C~)are $3, A4, $4 and As. Therefore the possible traces for tr(AB) are 0, ±1, ±x/~ and ~ which are then the possible roots for the trace polynomial JR(X). If G has an essential representation into a cyclic group then G is infinite. Hence if G has essential images both $3 and A4 then the induced representation onto S~b x A~b --- Z~ is also essential and so G is infinite. It follows that we may assume that if 0 is a zero of the trace polynomial, then ±1 are not and vice versa. In particular the trace polynomial fR(x) has degree at most 8 so that k _< 8 and R(a, b) has length at most 16. If k -- 7 then fR(x) is an odd polynomial of degree 7, so by the above its zeros must be 0, :t:v~, 2 ~ " In other words R(a, b) must be mapped to a permutation of order 2 under each of the permutation representations: Pl: a ~ (12), b --~ (123); p~.: a --~ (12), b--* (234); p3: a--* (12)(34),b--* (135); but not under P4 :a -~ (12)(34), b -~ (123). Since G has no cyclic essential images, but maps onto Z2 in such a way that a and R(a, b) map non-trivially, it does not admit Za as a homomorphic image. It follows that the number of b2 letters in R(a, b) is not congruent to 2 modulo 3. By symmetry, together with the fact that R(a, b) is not a proper power we may assume that this number is either 1 or 3. Up to equivalence there are only 4 possibilities; but 2) p2((ab)4(ab2)3), p3((ab)~ab and pa(ab(abab2)3) do not have order 2, so that the only remaining case is R(a, b) = (ab)3ab2ab(ab2)2. If k = 8 then fn(x) must have zeros =t= 1, ± ~/~,2 ~ Hence P2 (R(a, b)), p3(R(a, b)) and p4(R(a, b)) must all have order 2. In particular G must have Za as a homomorphicimage, so the number of b2 letters in R(a, b) must be congruent to 1 modulo 3. By symmetry we may assume that this number is 1 or 4. Since R(a, b) is not a proper power, there are 6 possibilities up to equivalence, but none of (ab)~ab2, a, (ab)4(ab2)

208

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

(ab)aab2ab(ab2) 3, (ab)2ab2)2(abab2) 2 are mapped to elements of order 2 by p2, while pa((ab)2ab2ab(ab2)2abab 2) does not have order 2. This then reduces to the final case 2. R(a, b) = (ab)a(ab2)2ab(ab2) Consider the first possibility so that G has a presentation < a, b; a2 = b3 = ((ab)aab2ab(ab2)2)2 = 1 >. The commutator subgroup G~ of G is generated by x = b, y = aba and has a presentation GI =< x, y; x a = y3 = xy-lx-lyxyx-lyx-ly-lxyxy-1 Let F D Q be an algebraic t s - 3t 3 + 1 --- 0. The map

x--,

X=

0 E

1 -A

= 1 >.

number field containing an element t with

,y--,

Y =

3(Bt1

D)

-D -Ct 0 0

where A = -3t 4 + 8t, B = -4t a + llt, C = 2t 3 - 6,D = -5t 5 + 14t 2 and E = -Tt 5 + 19tu extends to a representation of GI over F. (XY)2 has trace 3 which implies that XY has infinite order. It follows then that in this case G is infinite. This leaves the final possibility: G .-~< a, b; a~ = bs = ((ab)3(abg)~)2ab(ab2)~ = 1 > . A computer computation using GAPdetermined that of the given order.

this was finite

and

These lemmas and their proofs complete the proof of Theorem 7.3.4.1. Wenote that Howie, Metafsis and Thomas have also done some studies on the structure of the finite generalized triangle groups (see [H-M-T2]). 7.4 The Virtual

Torsion-Free

Property

The next linear property we consider is that of being virtually torsionfree, that is having a torsion-free normal subgroup of finite index. The theorem of Fenchel and Fox (Theorem 4.3.6) says that a Fuchsian group is virtually torsion-free. This was extended by Selbcrg (Theorem 4.3.5) any finitely generated linear group over a field of characteristic zero. The Selberg theorem is not true over characteristic a prime. Theorem3.4.4 due to Fischer, Karrass and Solitar proves the corresponding result for onerelator groups. In this section we give certain sufficient conditions on the

7.4 THE VIRTUAL TORSION-FREEPROPERTY

209

relator so that a one-relator product of cyclics with proper power relator is virtually torsion-free. Recall from Chapter 6 that an essentially faithful representation of a group G is a finite dimensional linear representation p over a field of characteristic zero with torsion-free kernel. In Theorem6.2.1 we saw that a finitely generated group G admits an essentially faithful representation if and only if G is virtually torsion-free. Ou~technique in showing that these one-relator products of cyclics are virtually torsion-free is then to showthat they admit essentially faithful representations. Suppose G is a group with the presentation (7.4.1)

G =<

¯ °’ al,...,a~,a

...

1a1 "=RT .......

R~ = 1 >

where e~ = 0 or e~ _> 2 for i = 1,..,n, mj >_ 1 for j = 1,...,k and each R~ is a cyclically reduced word in the ~ product of the cyclic groups < a~ >, ..., < a~ >, of syllable len~h at least two ~nd suppose p is an essemial representation ~om G into a fi~te dimensional linear ~oup so that for each i = 1, ..., n, p(a~) has i~nite order if ei = 0 or exact order e~ if e{ ~ 2 and for each j = 1, ..., k, p(R~) has order m~. If the conjugscy cla~es of torsion elements in G ~e precisely given by the powers of the generators (if of fi~te order) or the ~wers of the other relators, then ~sential representation is essemially faithS. ~om Theorem 6.3.2 - the ~eiheitssatz for on,relator products of cyclics - a on,relator product of cyclics with proof power relator always ad~ts an essential representation. O~ first result is the following. THEOREM 7.4.1. Let G =< a~,...an, ¯ a~ ..... where n,m ~ 2, e~ = 0 or e{ ~ 2 [or i = 1,..n,and U = U(a~,..ae),V = V(a~+~, ...an), are non-tri~l words i~ $hc ~ee prod~c~s o~ a~, ...,a~ and ae+~,...,a~ respectivel~ for some 1 ~ p ~ n - 1. Then G is virtually torsion-[r~. PROOF.The proof depends on the follow~g lemma w~ch character~es the conjugacy closes in G. LEMMA 7.4.1. Le$ G be as i~ Theorem Z 4.1. Then any element order is conj~e to ~ power of UV or of some a~. PROOF.(of lena) Without loss of generality we can a~e that UV involves all the generators and that UVis cyc~cally reduced. Assume first n = 2. Then G has the form G =< a~,a~;a~ ~ = a~ (a~a2) = 1 >. We may assume that s,t ~ 1 and by sy~etry, since (UV) m = 1 an~VU)~ = 1 ~e equiwlent that s ~ t. We m~y also ~e t~t s is a divisor of e~ if e~ ~ 2 and t is a divi~r of e2 if e2 ~ 2. If s = t = 1

210

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

then G is an ordinary triangle group and the fact that G is virtually torsionfree follows from the Fenchel-Fox Theoremin the hyperbolic case, from the fact that its virtually free abelian in the Euclidean case and from being finite in the spherical case (see section 7.3.2). If s _> 2 then s is a nontrivial divisor of el if el _> 2. NowG is the free product of H1 =< al; a~’ = 1 > and H2 --< a2,a;a~ 2 = ae’/8 = (aat2) ’n --= 1 > with < a > and < a~ > amalgamated. Since elements of finite order in a free product with amalgamation are conjugate to elements in the factors the result follows. Nowsuppose n >_ 3. By hypothesis one of U, V has infinite order. Let a = UV and suppose without any loss of generality that p _< n - 2. Then G can be expressed as a free product with amalgamation G =/’/1 *n/’/2 where H~=< a~, ..... %, a; a"~ = a~~ =. .... = a~,~ = 1 > > ~- _e,~=l a H2-----< ap+l, .... an. ~ %+1 ap+ 1 :- ...... n and H =< U-la;

>=< V; >.

The result follows since both factors are now non-trivial free products of cyclic groups. This completes the proof of the lemma. The proof of Theorem 7.4.1 now follows directly. Let G be as in the theorem. Then G admits an essential representation into PSL2(C). Since the conjugacy classes of torsion elements in G are precisely given by the powers of the generators and the powers of the single other relator then this essential representation is essentially faithful. G is then virtually torsionfree from Theorem 6.2.1. Results of Collins and Perraud [C-P] and J.Howie [H 3,4] allow us to extend this to further one-relator products of cyclics with a relator of order at least 4. THEOREM 7.4.2. Let G =< al, ...an; a~1 ...... ae~ " = Rm(al,...an) 1 > with m >_ 4,n > 2 and e~ = 0 or e~ _> 2 for i = 1, ...n and R(a~, ..,an) a cyclically reduced word in the free product on al, .., an which involves all the a~. SupposeR(al, .., an) is not conjugate in the free product on a~, .., a~ -1 for some word U and 2 elements X and Y to a word of the form XUYU ofordersp >_ 2,q > 2 respectively where (l/m) + (l/p) + (l/q) Then G is virtually torsion-free. PROOF. First suppose m > 6. Then the symmetrized closure of R(al, .., an) satisfies a small cancellation condition. If R is not conjugate -1 for some in the free product on ax, ..,a,~ to a word of the form XUYU word U and 2 elements X and Y of orders p > 2, q > 2 respectively where

7.5 FREE PRODUCTWITH AMALGAMATION DECOMPOSITIONS 211 (l/m) -t- (l/p) -b (l/q) > 1, "~ is not conjugate in the fre e pr oduct on al, .., an to its own inverse. From results of Collins and Peraud [C-P] this insures that every element of finite order in G must be conjugate to a power of R or a power of some ai. The proof then proceeds exactly as in theorem 7.4.1. If m -- 4 or m -- 5 the result follows from analagous work on small cancellation diagrams done by J. Howie [H 3,4] which also allows the classification of elements of finite order. As a direct corollary we get. COROLLARY 7.4.1. Let G be a one-relator product of cyclics with relator Rm(al, ...a,,). If m >_ 4 and each a~ has odd order then G is virtually torsion -free. Using identical proofs we can give the following extensions of theorems 7.4.1 and 7.4.2 to one-relator products of groups which admit faithful representations in PSL2 (C). THEOREM 7.4.3. Let G = (A * B)/N(R "~) where A and B are finitely generated groups which admit faithful representations in PSL2 (C) and is a cyclically reduced word in the free product A, B of syllable length >_2. (i) If m >_ 2 and A and B are locally indicable, then G is virtually torsion-free. (2) If m >_ 2 and R = UV where U is a non-trivial element of A and is a non-trivial element orB, then G is virtually torsion-free. (3) If m >_ 4 and R is not conjugate in A ¯ B to a word of the form -~ for some word U and 2 elements X, Y of respective orders p >_ XUYU 2, q >_2 with (l/m) -b (l/p) q- (l/q) then G is virtu ally torsi on-free. From work of Duncan and Howie [D-H 1,2] Theorem 7.4.2 and 7.4.3 can be extended to the ease where m -- 3 and the relator R(ax, ..., an) contains no letter of order 2. Weclose section 7.4 with the following conjecture. The Fischer-KarrassSolitar Theorem answers it in the affirmative for free groups while the results in this section give somefurther supporting evidence. CONJECTURE. Let A and B be finitely generated groups which admit faithful representations in PSL2 (C). Let R be cyclically reduced in A ¯ B of syllable lengfh >_ 2. Then if m >_ 2 the group G = (A * B)/N(R’*) virtually torsion-free. 7.5

Free Product with Amalgamation Decompositions

Wenow consider the property of being a non-trivial free product with amalgamation. Via the Poincare presentation any Fuchsian group with at

212

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

least (algebraic) rank 3 admits such a decomposition and as we saw Corollary 3.4.2 any one-relator group with at least three generators also admits such a decomposition. The reason that amalgamconstructions can be considered as linear properties is through the variety of representation techniques of Helling and Culler and Shalen Werepeat the brief description given in Chapter 3. Suppose G is a finitely generated group with generators gl, .-., g,~- Any representation p : G -~ SL2(C) can then be considered as a point (p(gl), ...,p(gn)) (SL2(C))’~. These re presentation po ints wi ll be subjected to various conditions reflecting the relations in the group. These relations are polynomial relations on the matrix entries and hence the set of all possible representation points for G define an affine algebraic set in C4’~. Call this R2(G). SL2(C) acts on R2(G) by conjugation in the natural way. Thus we can form the categorical quotient X2(G) of Ru(G) under this action. X2 (G) is called the at~ine algebraic set of characters or character space of G. It can be thought of as the parametrization of the inequivalent semi-simple representations of G in SL2(C). Culler and Shalen then proved that (Theorem 3.4.7) if G be a finitely generated group and if the dimension of the character variety of G is positive then G decomposes as either a non-trivial free product with amalgamation or as an HNNgroup. Thus if there are "many" representations into (SLu(C)) then the group must decompose as an amalgam. Using this Culler and Shalen result we can prove that any one-relator product of cyclics with proper power relator and at least 3 generators must be a non-trivial free product with amalgamation. 1 ...... THEOREM7.5.1. Suppose G =< al,....an;a~ a~ ~ = Rm(al, ...an) = 1 >. Hm>_2,n >_ 3 and e~ = 0 or e~ >_2 for all i = 1, ...,n, then G is a non-trivial free product with amalgamation. PROOF.Wemay assume that each ei >_ 2. If not we introduce relations ai A = 1 with f~ >_ 2 for each ai with ei = 0 and write fj = ej if ed > _ 2. The group G* =< al, .... an;all 1 ..... aln ~ = Rm(ax,...an) = 1 > is an epimorphic image of G. It is knownthat if ¢ : G1 -~ G2 is an epimorphism and G2 is a non-trivial free product with amalgamation, then G~ is also via

¢. Hence let e~ _> 2 for all i = 1, ...n. This condition indicates that G is generated by elements of finite order. Wemay assume that R is cyclically reduced in the free product on a~, ..., an and involves all the generators for if a generator is omitted in R(a~, ...an), then G is a non-trivial free product since we are assuming that n _> 3. Since rn _> 2 and n _> 3 there exists an irreducible essential representation of G in PSL2(C), and therefore especially the space of representations of

7.5

FREE PRODUCTWITH AMALGAMATION DECOMPOSITIONS213

G in (PSL2(C)) "~ is non-empty. Here an irreducible representation means that in the image group of G there are at least two elements which have no commonfixed points considered as linear fractional trasnsformations. From the result of Culler and Shalen mentioned above if the dimension of the character space of G in PSL2(C) is positive {as an affine algebraic set}, then G admits a decomposition as either a non-trivial free product with amalgamtion or as an HNNgroup. From the proof of the existence of irreducible essential representations for such one-relator products of cyclics we have that each of n - 1 matrices have two degrees of freedom with the trace and determinant being specified while the final matrix has one degree of freedom - the trace, determinant and relator condition specified. Therefore from the work of Culler and Shalen [Cu-S] (or also from Helling [He] and Rosenberger [R 9] in a different setting) the dimension of the character space is 2(n- 1) + 1- 3 = 2nThis is positive if n _> 3 . Thus if n _> 3, G splits as a non-trivial free product with amalgamation or an HNNgroup. However G is generated by elements of finite order so its abelianization Gab is finite. Therefore G cannot be an HNNgroup and must therefore be a non-trivial free product with amalgamation. The case of 2 generators - the generalized triangle groups is quite different. These group someti~nes admit splittings as non-trivial free products with amalgamation and sometimes don’t. The ordinary triangle groups T(p, q, m) = with p, q, m _> 2ar e n ever n ontrivial free products with amalgamation . However the Fuchsian groups G =< a, b; [a, b] ~ = 1 > have been shown to be free products with amalgamation if n _> 3 and n is not a power of 2. This was done by Rosenberger[R 21] by mapping onto another Fuchsian group with an obvious splitting. Recently, Long, Maclachlan and Reid [L-M-R] proved that such a splitting exists for all n > 1, although they did not provide an explicit splitting. NowDunwoody and Sageev [D-S] have given the following explicit splitting. Let H, K be given by the following presentations: H =< a,x;x ’~ = 1 >, K=< b,w;> and let L=< a,z-laxz > wherez=xax E H. Then the following is a splitting of G as a free product with amalgamation: G--H*LK with the identifications a -- b and z-laxz -- w-lbw. Using the theorem of Bass classifying the finitely generated subgroups of GL2(C.) (Theorem 3.4.6) we can give a decomposition result for two

214

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

generator one-relator groups with torsion if the relator is not in the commutator subgroup. Recall from Corollary 3.4.2 that if there are three or more generators, every one-relator group admits a proper free product with amalgamation decomposition. THEOREM 7.5.2. Let G =< a, b; R’~(a, b) = 1 > with m >_ 2 and R(a, a cyclically reduced word not contained in the commutator subgroup of the 1?tee group on a and b which involves both a and b. Then G is decomposable as a non-trivial t~ee product with amalgamation. PROOF.Bass’s result (theorem 3.4.6) says that if G is a finitely generated subgroup of GL2(C) then one of the following cases must occur. (1) There is an epimorphism f : ~ ~ Z such that f(U) = 0 for all unipotent elements U E ~. (2) G is an amalgamated free product G G2 such that every finiteley generated unipotent subgroup of G is contained in a conjugate of G1 or (3) G is conjugate to a group of upper triangular matrices all of whose diagonal elements are roots of unity. (4) G is conjugate to a subgroup of GL2(A) where A is a ring of algebraic integers. Nowlet G = be as in the statement of the theorem. Since R is not in the commutator subgroup we may write R(a, b) = aPbqc where p ¢ 0 or q ~ 0 and c is in the commutator subgroup of the free group on a and b. Assume p ¢ 0 (an analogous argument works if q ¢ 0). Choose in SL2 (C) the two matrices A=(O_I xl ) with x a real transcendental

B=B(w)=(

wl

2w-w2-l)2-w

number greater

than 2 and

withwEC.

Let A, B be the images of A, B under the natural epimorphism SL2 (C) PSL2(C). Nowchoose w = w0 so that tr(R(A, B)) = 2 cos(~r/m) as in the proof the existence of essential representations. Then < A, B > represents G in PSL~(C) and R(A, B) has order m in PSL2(C) while R(A, B) has order 2m in SLy(C). Further -I ~< A,B >. Considered as linear fractional transformations A, B have no common fixed point. If they did conjugating by a suitable D E SL2(C), we obtain DAD-1 = A1 =

(0 0)

y_~ with

y +

= x > 2

7.5

FREE PRODUCTWITH AMALGAMATION DECOMPOSITIONS 215

Then since R(a, b) = anbqc we get R(A1,B1)

:

(yo

p

z

)

y_~ for some z E C. Since p ~ 0 this implies that R(A1, B1) has infinite order in SL2((~) which is a contradiction since by the construction it must have order 2m. Let G =< A, B >CSL2(C). Since A, B have no commonfixed point, case (3) of Bass’ theorem cannot hold for G. Since x is a real transcendental number and trace is preserved under conjugation, case (4) of Bass’ theorem cannot hold for G. Assumethat there is an epimorhism jr : G -~ Z such that f(U) = 0 for all unipotent elements U in < A, B >. Then f(B) = O. It follows that f(A) -- k ~ si nce f is an epi morphism. Then f(R(A, B)) = ~’) = k p ~ 0 and f(R2~(A,

B)) = 2mkp

which is a contradiction since R2m(A, B) = I. Therefore (1) cannot hold for G. Thus case (2) must hold and G = Kx *r K2 with K1 # K # K2. Now-I E K since -I ~ G and -I ~ Z(G). Therefore < A, B >= K~ *~ with K~ # K # K~. where K1, K, K2 axe the natural epimorhic images of K1, K2, K in PSL2 (C). < A, B > is an epimorphic image of G by construction, so therefore G decomposes as a non-trivial free product with amalgamation. The same argument can be used if one generator has finite order and the relator is suitably restricted. Wehave. THEOREM 7.5.3. Let G =< a, b; as = R’~(a, b) = 1 > with m >_ 2, n _> 2 and R a cyclically reduced word in the free product on a and b which involves both a and b. If R(a, b) = aPbqc with 0 (_ p < n, q ~ 0 and c the commutator subgroup in the free product on a and b, then G admits a splitting as a non-trivial free product with amalgamation. PROOF. Choose A--- (_01 ~) witht----2cos(r/n)

and B=B(w0)= (w l° xw°-w~°-l)x-wo

216

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

with x _> 2 a real transcendental number and w0 chosen so that < A, B > represents G in PSL2(C). The proof then goes through exactly as in the previous theorem. These results leave open the question of when R(a, b) is in the commutator subgroup. CONJECTURE. A two generator one-relator group with torsion poses as a non-trivial free product with amalgamation.

decom-

7.6 Euler Characteristics Suppose F is a finitely sentation of the form: F=<el,..,ep,

hl,

generated Fuchsian group with a Poincare pre-

h~,al,bl,...,ag,

bg, e~ =l,i=l,..,p,R=l>

whereR = el..eph~..ht[a~, b~]...[ag, ba] and p _> 0, t _> 0, g _> 0, p+ t + g > 0, and rn~ _> 2 for i = 1, ...p. Recall then (see Chapter 4) that its Euler Characteristic is given by x(F) = -p(F) where #(F) = 2g - 2 + t ~ -1 1/mi). 2~r#(F) represents the hyperbolic area of a fundamental ~=1( polygon for F. If H is a subgroup of finite index in F, then the RiemannHurwitz formula says that x(H) ---- [G : H]X(G ). Bass [Ba 3], K.Brown[Brn], Chiswell [Ch], C.T.C. Wall [Wa] and others have extended the concept of a rational Euler characteristic to more general finitely presented groups and have shown that in many cases these can be computed very easily from given finite presentations or given amalgam decompositions. Further these general rational Euler characteristics satisfy the Riemann-Hurwitz formula. In particular, Chiswell [Ch] has shown that if G is a one-relator group with torsion so that G =< xl,...,xn~R > with m _> 2, then the rational Euler characteristic of G is given by x(G) = 1 - n (l /m). Wh ile th e ge neral de velopment of Eul er cha acteristics and their uses is beyond the scope of these notes, what we will do in this section is show that for certain one-relator products of cyclics with proper power relators there is a rational Euler characteristic in the sense above. Further this can be computed very easily from the given presentation. As a consequence of the construction we prove that such groups are of a certain finite homological type. Werefer the reader to the book by K.Brown [Brn] for any necessary terminology and a complete development. Wenote however that a group G is of type FL if Z admits a finite free resolution over the group ring ZGand is of typeWFLif G is virtually torsion-free and every torsion-free subgroup of finite index is of type FL. FL and WFLare very strong finiteness conditions (see IBm- Chapter VIII]).

7.6 EULERCHARACTERISTICS THEOREM 7.6.1.

217

Suppose

G = 2,m >_ 4, ei = 0 or ei >_ 2 for i = 1,...,n and S(al,...,an) is a cyclically reduced word not a proper power in the free product on al,..., an which involves a//the generators. Suppose fi~r~her that S(al, ..., a,~) is not conjugate Jn the free product on al ,..., an to a word of the form XY for some elements X, Y ofrespective orders p _> 2, q _> 2 where (1/m)+(1/p)+(1/q) > 1. Then G is of homological type WFL, with vcd(G) <_ 2 and G has rational Euler characteristic x( G) given x(G)

= l-n+

Z~,

1 i=I

whereN = 0 ife~ = 0 andN = (1/ed ifei _> 2. ~rtheriflG ://I < oc then x(H) is defined and X(H) = IG : HIx(G) {Riemann-Hurwitz formula}. PROOF.From results in section 7.4 it follows that under the conditions on the relator, and if m _> 4 , G is virtually torsion-free. This was a consequence of work of J. Howie (see section 7.4) on one-relator products with high-powered relators. This work implies that under the conditions above any finite subgroup of G is cyclic and any element of finite order in G is conjugate to a power of S(al, ..,an) or a power of some ai. Further the orders of generators ai are precisely ei if ei _> 2 and infinite if ei = 0 and the order of S is m. Construct X, the Cayley complex for the given presentation {see [Brn] }. From the work of Howie{[H 3],[H 4]} the relation module splits as a direct sum of cyclic submodules.{see IBm]} Therefore the complex X is acyclic and hence X is a two-dimensional contractible G-complex such that every isotropy group is trivial or finite cyclic and X has only finitely manycells rood G. The construction is straightforward and similar to the case of onerelator groups. {see K. Brown[ 2,Sec.II.5] and also [Hu] }.This implies then that X has an equivariant Euler characteristic xa(X). It follows that the equivariant Euler characteristic Xa (G) = Xa (X) is defined. Since the group G is virtually torsion-free it follows from K. Brownthat G is is of homological type WFLwith vcd(G) _<2 and there is an ordinary Euler characteristic X(G) with X(G) = xa(G) [Brn,Secs. XI.11 and IX.7]. The Riemann- Hurwitz formula then follows also directly from K. Brown IBm, Sec.IX.7]. A generalization of Howie’s work used in the proof of the virtual torsionfree property by Duncan and Howie [D-H 1,2] can be used to show that

218

ALGEBRAIC

GENERALIZATIONS

OF DISCRETE

~ROUPS

G is still virtually torsion-free if m= 3 but no letter {with respect to the free product on {al, .., a~}} appearing in S/a1, .., a~) has order 2. Once G is established to be virtually torsion-free the proof of the theorem goes through in an analagous manner. Hence we have the following corollary. COROLLARY 7.6.1. Let G, S(al,..,an) be as in theorem 7.6.1. Suppose m = 3 and no letter {with respect to the Tree product on al, .., a~} appearing in S/a1, ..,an) has order 2. Then the same conclusions follow. That is, G is o~ homological type WFLwith vcd(G) <_ 2 and G has a rational Euler characteristic as given in theorem 7.6.1. Theorem 7.6.1 and the corollary can be viewed as a direct extension of the result of Chiswell mentioned above on Euler characteristic for onerelator groups with torsion. THEOREM 7.6.2. Let G = where n > 3, m > 2,ei = 0 or ei > 2/or i = 1,...,n and for some p with 1 _< p < n, U is a non-triv/al cyclically reduced word in the Tree product on al, ..., ap whichinvolves a11 these generators and V is a non-trivial cyclically reduced word in the Tree product on ap+l, ..., an which involves all these generators. Then G is o/finite homologiced type with vcd(G) <_ 2 and G has rational Euler characteristic X/G) given by 1 m

i:1

where ~3i = 0 ffei = 0 and Bi = (1/ei) i/e~ > 2. Further fflG : H[ < oc then x/H) is defined and x/H) = IG : HIX(G){ Riemann-Hurwitz formula}, PROOF.From Theorem7.4.1 it follows that G is virtually torsion-free. Since n > 3 at least one of U, V has infinite order and involves at least two generators. Let V have infinite order and involve at least two generators and let a = UV . Then the group G decomposes as a free product with amalgamation G = H1 *A H2 where Hi =< al~...~

¯ el ap~ a 1 :a~ 2 ~ ....

//~ ---< ap+l, ...,a~;ap+~ A =< U-la

.....

a~P : am = 1 > a: ~ ---- 1 > and

>=< V >

Except for the fact that vcd(G) < 2 the result then follows from the the fact that G is virtually torsion-free together with K. Brown[Brn, Sec.

7.6

EULER CHARACTERISTICS

219

IX.7]. To prove vcd(G) _< 2 consider the tree X associated to G with its corresponding fundamental domainand consider an arbitrary torsion-free subgroup H of finite index in G (see [Brn, Chapter II,Appendix]}. The equivariant cohomologygroups of the H-complexX vanish in dimension > 3. Because X is contractible, the equivariant cohomologyis isomorphic to the usual cohomology [Brn, Sec. VII.7]. (See [St] for details}. For the generalized triangle groups we can give similar results under additional conditions. THEOREM 7.6.3. Let G = < a, b; av = bq = (aSbt) m = 1 > where p = 0 orp> 2, q=0orq > 2, s > 1,t > land1< s 2and 1 < t < q,t]q ifq > 2. ThenG is oftinite homologScaltype with vcd(G) < 2 and G has a rational Euler characteristic X(G) g4ven (1) x(G) = -l+f~: +f12+ ~ ifp = 0 or q = 0 or p,q > 2 and ~+~+_~_s t ~ <1. Heref~l =0 ifp =0 or f~ = ~1ifp _>2 andf~ = 0 ifq=O or ~ = -~1 ifq>2 _ 1 ~pS .~qt st 1 (2) X(O)=-:+ +--+:ifp,I q>2and~+:+:>l. In the latter case (ii) it follows that in fact vcd(G)< 1 and therefore G is finite or has a free subgroupof finite index. PROOF. Nowsuppose that G is a generalized triangle group with a presentation G =< al, a~; a~1 = a~2 = (a~a~)’~ = 1 > where a[ # 1 and at~ # 1. Wemayassume that s, t > 1 and that s is a proper divisor of e: if e: > 2 andthat t is a properdivisor of e2 if e2 > 2. Gis virtually torsion-free. If s = t = 1 then Gis an ordinary triangle group and the result is well known.If (i/p) + (l/q) + (l/m) > 1 then Gis of order 2((1/p) + (l/q) + (l/m) - 1)-: and G has the Euler characteristic X(G) = (1/2)((1/p)+(1/q)+(1/m)-l). Suppose that s > 2 or t > 2. From the analysis of the generalizedtriangle groupsgiven in section 7.3 it follows that Gis infinite in any case and weobtain that G has finite homological type from the following free product with amalgamationdecomposition. Suppose s > 2. Then G decomposesas a free product with amalgamation G = H~*A H2 where H: =< a; a~ = 1 >, H2 =< b, c; bq = c, r = (cbt) ~ = r= l l> > = A =< .a; (aS), FromK. Brown[Brn, Sec. IX.9], if t = 1 and p = 0 or q = 0 or 2 < p, q with (s/p) (l /q) + (l /m) < 1,then G is of finite homological type andG has a rational Euler characteristic given by: 1

220

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

+-t--± where 71 : 0 if el : 0 or 3’1 : ~ if el _> 2. That is x(G)_1----~ : - ~1 ~2 as desired. If t : 1 and 2 < p, q and (s/p) (l /q) + (l /m) > 1 th en H2is fin ite and again from K.Brown[Brn,Sec.VIII.11], we get a rational Euler characteristic given by x(G)

= ~(-1+

1 s + 1+ ~)+ p P q

s p

1 2-s 2+’-~P--p

1 1 +~qq+2m

as desired. If t > 2 then consider the free product with amalgamation decomposition of G given by G = K1 *B K2 where K~ =< a,d;a p = d~ = (aSd) m = 1 > ,K2 =< b;b q = 1 > and B =< bt;(bt)~ = 1 >=< d;d~ = 1 >. Again from the analysis in section 7.3 we know that K1 is infinite. From the arguments for t = 1 the results concerning the Euler characteristic holds for K1 . Therefore again from K. Brown [Brn,Sec. IX.7],the result concerning the Euler characteristic follows directly also for G if p = 0 or q = 0 or 2 _< p, q and (s/q) + (t/q) (l /m) _< If 2 < p, q and (s/t)+ (t/q)+ (l/m) > 1, then from [Brn] G has a rational Euler characteristic given by 1 1 t 1 2-s t x(G) = -3 + + + + - q- -q

1 2-s 2-t 2+’-~--p +’~q

1 +2m

as desired. In this latter case it follows in fact that vcd(G) _< 1 and hence that G either finite or virtually free {see [Brn] and references there for details}. The Ricmann-Hurwitz formula and vcd(G) _< 2 follow as in theorem 7.6.2. In closing we pose the following question. Are groups G of the form given in the statement of Theorem 7.6.2 of homological type WFL?

CHAPTER VIII

GROUPS

OF F-TYPE

8.1 Groups of F-Type The last chapter concentrated on linear properties of one-relator products of cyctics where the relator is a proper power. If the relator is not a proper power, almost nothing can be said in general. This is nicely illustrated by the following example of M.Conder [Co 3] following a suggestion of B.H.Nelmmnn. Neumann suggested that the group < x, y; x2 = y3 = R(x, y) = 1 > is iIffinite as long as the relator R(x, y) has length greater than a certain bound. Conder’s example refutes this. Let m be any positive integer congruent to 1 or 3 mod 6. Then the group < x, y; x2 : y3 = (xy),~y = 1 > is trivial. To see this notice that y commutes with xy and hence also with x = (xy)(y-1). Therefore G is abelian and is therefore a factor group of Z2 × Z3 and so has order less than or equal to 6. From 1 = (xy)my -~ xray m+l we have x-m = ym+l. But on the other hand x-’~ = x and y’~+~ = y or y-1. Since x and y have coprime orders it follows that x = y = 1. This situation is analogous to the situation in one-relator group theory. Although the classical Freiheitssatz holds for arbitrary relators, the situation in one-relator groups with torsion is much simpler than that in the general torsion-free case (see Chapter 3). To proceed further with the theory of one-relator groups it is necessary to restrict the form of the relator. An exampleof this is the case of cyclically pinched one-relator groups. Analogously, to continue in our program for one-relator products of cyclics, we must restrict the form of the relator if it is not a proper power. Our motivation here now is both Fuchsian groups, specifically the Poincaxe presentation, and the theory of cyclically pinched one-relator groups. Recall that a finitely generated Fuchsian group has a standard Poincare presentation of the form (8.1.1) F =<el, .., %,hi, .., ht, al, 51, ..., ag, bg; er~i = 1, i ---- 1, ..,p, R = 1 > where R ---- UV with U = el...eph~...ht, V -= [a~,b~]...[ag,bg] and p _> 0, t >_ 0, g >_ O,p+t+g > 0, and mi _> 2 for i = 1,...p. A presentation 221

222

ALGEBRAICGENERALIZAT|ONSOF DISCRETE GROUPS

of the form (8.1) actually presents a Fuchsian group if #(F) > 0 where #(F) = 2g-2+t+Y~(1-1/m d. A group G which has a presentation of the form (8.1.1) is called an F-group and so clearly all Fuchsian groups are groups. If t ¢ 0 then G is a free product of cyclics, so all such free products of cyclics are F-groups. A group theoretical discussion of F-groups is given in the book of Lyndon and Schupp [L-S]. Most of the algebraic results on Puchsian groups are proved for the full class of F-groups without recourse to their geometric nature. Very important in this algebraic analysis is the fact that the sum of the absolute values of the exponents of a generator occurring in the relator is _< 2. These are called quadratic relators (see [L-S]). Using these ideas most of the results on F-groups can be extended to the class of groups with presentations similar to the Poincare presentation but with the relator replaced by its non-orientable analog. That is groups G with presentations of the form (8.1.2)

G =< el,..,ep,

h~,..,ht,a~,...,ag,e~ . m~ = 1,i=l, .., p,R=l

>

where R =UV with U el...ephl...ht, V= a~a2...a 2 2 ~2and p _> 0, t >_ 0, g _> = O,p+t+g> 0, and rni _> 2 for i = 1,...p. To continue further we define a group of F-type to be a group G with a presentation of the form (8.1.3) G ---< a~,. .... , a,~; a~1 ......... a~" = 1, U(al,..., ap)V(ap+l,.., a,~) --where n _> 2, ei = 0 or ei _> 2, 1 _< p _< n - 1, U(al, .., ap) is a cyclically reduced word in the free product on al, ..., ap which is of infinite order and V(ap+l, ..., am) is a cyclically reduced word in the free product on ap+~, ..., aN which is of infinite order. With p understood we write U for U(al, .., ap) and V for V(ap+l,..., Nowif U = a~~ then el must equal zero since we assume that U has infinite order. In this case G reduces to G =< a~, ....,

aN; a~~ ........

a~~ = 1 >

which is just a free product of cyclics and thus an F-group. Therefore if p = 1 (or p = n - 1) we restrict in this chapter groups of F-type those where U = a~ (or V = a~) with Irnl _> 2. Thus a presentation for a group of F-type clearly generalizes the Poincare presentation and the terminology group of F-type was chosen to accent this tie to Fuchsian and F-groups. On the other hand a group of F-type also generalizes, by allowing possible torsion in the generators, a cyclically pinched one-relator group

8.2 FREIHEITSSATZ AND SUBGROUPTHEOREMS

223

and therefore an alternative designation could be a cyclically pinched one-relator product of cyclics. In all cases, a group of F-type, G, decomposesas a non-trivial free product with amalgamation: G = G1 *A G2 where the factors are free products of cyclics ~1 ~---< al, ....

,ap:a

¯ e, ~ I

...

= a~~ = 1 >

~+~1 = ..... G2~-< ap+l, .... ,an:¯ ap+ A =<

U -1

>--<

a~" = 1 > and V >

.

This free product with amalgamation decomposition together with the existence of essential representations, which we will prove in the next section, will be the basic tools for studying groups of F-type. If G is a group of F-type, its factors are the subgroups G1, G2 in the free product with amalgamation decomposition and its relator is UV. The main focus of the study of groups of F-type is the following general question: Given an algebraic property of F-groups how does it extend (if at all) to the more general groups of F-type. Wewill see that the majority of algebraic results on F-groups do carry over (however sometimes with modifications) to groups of F-type. 8.2 Freiheitssatz

and Subgroup Theorems

The basic tools used so far in the study of one-relator products of cyclics were the Freiheitssatz and the existence of essential representations. In this section we show that these can be extended to groups of F-type. Wewill also prove a subgroup theorem for subgroups of small rank analogous to a result on cyclically pinched one-relator groups. THEOREM 8.2.1¯ Let G be a group ofF-type with factors G1,G2 and relator UV. Then G admits a representation p : G ~ PSL2 (C) such that PlG~ and Pla2 are faithful. Further if neither U nor V is a proper power, then G has a faithful representation in PSL2 (C). PROOF.First suppose that UV omits some generator. For instance suppose that UV does not involve al. Then G is a f~ee product Hi * H2 where e~ = 1 > and H1 1 =< al; a H2 =< a2, ...,

an; a~~ ......

a~" = UV = 1 > .

224

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

If H2 admits a representation < a2, ....,

¢ : He -~ PSL2 (C) such that

ap; a~2 .....

a~~" = 1 >-~ Ha ¢-~ PSL2(C)

and G2 ~ H~ ¢-~ PSL2 (C) are faithful then G, has a representation p :G --* PSL2(C) such that Plol and PIG2 are faithful. Therefore we can assume that UV involves all the generators. Choose faithful representations ffl : G1 ~ PSL2 (C) a2 : G2 ~ PSL: (C) such that 0 t~ -1) and~r2(V) where tl and t: are transcendental over Q. This may be done since U and V both have infinite order and if Gi, i = 1 or 2, is non-cyclic, then the dimension of the character space as an affine algebraic set is positive (see the proof of Theorem7.5.1) From work of Shalen (1emma3.2 of [Sh]) there exists an automorphism a of C such that a(tl) = t2. Wedefine a faithful representation pl of G1 by Pl (~5) (g) = (~(crl (i,j) (g)), i = 1, 2 and j = 1, 2, wherep~ (~,j) (g) a~(i5)(g)) is the (i,j) entry of p~(g) (respectively of al(g)). Further p~ = a2. Then P2 is faithful on G~. G~ and G~ generate G and let p be the representation of G induced by p~ and p2. This gives the desired representation of the theorem. We show that if neither U nor V is a proper power then the above construction leads to the existence of a faithful representation of G . In [Sh]P. Shalen proved the following result LEMMA 8.2.1. (Proposition 1.3 of [Sh]) Let G1 *H G2 be a free product with amalgamation and let n > 1 be an integer. Suppose that there exist faithful representations p~ : G~ --* SL,~(C), (i = 1, such that (a) PllH = P21H (b) p~(h) is a diagonal matrix for every h ~ H and (c) Pi(n,1)(g) £ 0fo r ev ery g ~ G~\ Uand p~(~,~)(g) ~ 0 g~G2\H.

fo r ever y

8.2 FRE|HEITSSATZ AND SUBGROUPTHEOREMS Then G1 *H G2 ha~ a faithful

representation

225

in SLy(C).

Wenote first that if the original group is centerless then this faithful representation can be extended to a faithful representation into PSL2(C). Nowconsider the representation of the group of F-type G, constructed as above. Hypotheses (a) and (b) of Lemma8.2.1 are obvious from construction. We now check hypothesis (c). Suppose Pl(2,1)(g) = some g E G1 \ H. Then Pl(g) is an upper triangular projective matrix in PSL2(C). Hence, the subgroup K =< g, U-1 > of G1 is solvable. Since G1is a free product of cyclics this implies that K is either cyclic or infinite dihedral. K cannot be cyclic since we assumed that U is not a proper power in G1 and g ~< U-1 >. Therefore K must be infinite dihedral. Then necessarily we have that g and gU-1 have order 2. From this we obtain that

pl(U_l)

~_

0 0 t~ ~)

2

andpl(gU-1)=(~

2 xt~l 1) --Wt~

with w(t2-t~ ~) = 0. This gives a contradiction because t2 ¢ t~ 1. Therefore Pl(~,l)(g) ~ 0 for all g E G1 \H. In a similar manner, since V is assumed not be a proper powerin G2 it follows that P20,2) (g) ~ 0 for all g e G~ It follows than from Lemma8.2.1 and the remark after it that G has a faithful representation in PSLg (C). Wenote that if both U and V are proper powers, then there is no faithful representation in PSL2(C). If U = Uff,a >_ 2, and V = V~,/~ _> 2, and p : G -~ PSL2(C) is a representation where p(U) and p(V) have infinite order then p(U~) and p(V1) must commute. However U1 and V1 do not commutein G. {Non-elliptic elements of PSL2 (C) commuteif and only they have the same fixed points (see Chapter 4). Therefore p(U~) commutes with p(V1~) implies that p(U1) commuteswith p(V~).} There are several immediate consequences of Theorem 8.2.1 which we describe in the next section. Wenote here that using exactly the same argument the following generalization of Theorem8.2.1 can be obtained. TI-IEOREM8.2.2. (1) Let H1 and H2 be groups and U1, U2 elements of infinite order in H~, H2, respectively. Supposethat each H~, i -- 1, 2, admits a faith&l representation p~ in PSL2(C)such that tr(p~(U~)) is transcendental. Let H = (Hi * H2) / N ( Ui U2 ) be the one-relator product of Hi, H2 relator UiU2. Then H admits a representation p : H -~ PSL2(C) such that PlH~and PlH~are faithful.

226

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

(2) Suppose H is as in (1) but assume further that each solvable subgroup of H~,i = 1, 2, is either cyclic, finite or infinite dihedral andUi is not a proper power in Hi. Then H ad~nits a faithful representation in PSL2 (C). Manyof the consequences which follow for groups of F-type from Theorem 8.2.1 will also hold for groups H of the form of Theorem 8.2.2. We will point these out also in the next section. The theory of one-relator products of cyclics has been built so far on the foundations of essential representations and the Freiheitssatz. The next two results give the Freiheitssatz for groups of F-type. THEOREM 8.2.3. (Freiheitssatz for Groups ofF-type) Let bea gro up of F-type. If the relator UV involves all the generators, then any subset of (n - 2) of the given generators generates a free product of cyclics of the obvious orders. PROOF.Let G be a group of F-type with decomposition (4) and let G1, G2 and p be as in the decomposition. Suppose{x l, ...., x~-2 } is a subset of {al, ...,a,~}. Suppose first that p - 1 of the x~ are elements of GI and n - p - 1 of the x~ are elements of G2. Assumewithout loss of generality that x~ -- a~, i -- 2,. .... , n- 1. Since UVinvolves all the generators, it follows that < a2, ..., ap > contains no element which is conjugate to a nontrivial power of U. Similarly < av+l, ..., a,~-i > contains no element which is conjugate to a non-trivial power of V. Recall from Theorem 2.6.2 that if G = Hi *A H2 and {Zl,...,Zm} is a finite system of elements in G, then there is a Nielsen transformation from {Zx, ...,z,~} to a system {yl, ...,y,~} for which one of the following cases hold: (i) y~ --- 1 for somei ¯ {1, ..., m}. > can be written as w = ~l~=~v~, ~ = (ii) Each w E< Yl,.-.,Ym 4-1, ei ---- ei+l if ~i ---- ~+~withL(yg~)<_L(w)for i -- 1, ..., q;. ~ ei a (iii) There is a product a =l~q ~ ~i=~ y~,, ~ 1, with y~ ¯ A(i -- 1, .., q) and in one of the factors H~ there is an element x ~ A with x-~ax ¯ A;. (iv) There is a g ¯ G such that for some i ¯ {1,...,m} we have y~ ~ gAg-1, but for a suitable natural number k we have yk~ ¯ gAg-~. (v) Of the yi there are p _> 1 contained in a subgroup of G conjugate to H~ or H2 and a certain product of them is conjugate to a non-trivial element of A. Further the Nielsen transformation can be chosen so that {Yl, ...,Y,~} is shorter (with respect to the length and a suitable order) than (z~..., z,~} the lengths of the elements of {Zl, ..., zm}are preserved. From the above result it follows that < a~, ...,a~-i > is a free product of cyclics of the obvious orders. Next consider the case where p of the

8.2 FREIHEITSSATZ AND SUBGROUPTHEOREMS

227

xi axe elements of G1 and n - p - 2 of the xi axe elements of G2 ¯ The remailfing case where p - 2 of the xi are in G1 and n - p of the xi are in G2 is handled identically. Assumethen that xi = ai, i = 1, ..., n - 2. Then G1 =< al, ...,ap > and we have V = gWh where g,h ¯< ap+l, ...,an-2 > and the norlnal form of W(with respect to the free product of cyclics G2 ) begins with a non-trivial power of an-i or aN and also ends with a nontrivial powerof a,~_l or a,~ since UVinvolves all the generators. The system {V, ap+h.... , a,~_~} is Nielsen equivalent to {W,ap+~, .... , a,~_2}. A product u~vi .... urv~ with r _~ 1, 1 ¢ u~ ¯ G1 and 1 ~ vi ¯< ap+l,...,an-2 > for i = 1,...,r can be trivial only if a product VSlw~ .... VS~w~with k >_ 1, 1 ~ w~. ¯< %+1, ...,a~_~ > and s~ ¯ Z\ {0} for j = 1,...,k is trivial. But this latter product is trivial only if a product W~Iz~...W~’~Zmwith m_> 1, 1 ~ z~ ¯< a~+l,...,a,~-2 > and t~ ¯ Z\ (0} forj = 1,..,m is trivial. But this last product can’t be trivial. Therefore u~v~ .... u~vr is non-trivial if r > 1 and 1 ¢ u~ ¯ G~ and 1 ¢ vi ¯< ap+i,...,a,~_: > for i = 1,...,r. This implies that < a~, ..., an-2 > is a free product of cyclics of the obvious orders. Wenote that nevertheless example consider

the algebraic

rank of G may be n - 2. For

G -=. G can be generated by x = ala2 and y = a3al since [x, y] = (a~a2a3) 2 = a4 ~ 1. a3a2a

Wealso note that in general a subset of (n - 1) of the given generators need not be a free product of cyclics. For example in < al~a2~a3~a4~a

¯ ~ a~ ~ = a~~ = a~4~ = ala2aaaa = 1 > 1 ~

with e~ _> 2 the subgroup generated by a~, a2, a3 has the presentation < al, a2, a3; a~1 = a~~ = a~3 = (aia2a3) ~ = 1 > which is not a free product of cyclics. Howeverif we allow that the relator UVinvolves all the generators and that U and V are proper powers in the respective free products of generators which they involve, then any n - 1 of the given generators generates a free product of cyclics of the obvious orders. Specifically: THEOREM 8.2.4. Let G be a group of F-type with relator UV such that UV involves all ~he generators. If both U and V are proper powers (U = Uff , V = V~k with m > 2, k > 2 in the respective free products on

228

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

the generators which they involve), then any subset of (n 1)of the given generators generates a free product of cyclics of the obvious orders. In a orientable surface group of genus g _> 2 any subgroup with three or fewer generators is a free group. An algebraic proof of this for two generators was given by G. Baumslag [G.B. 2] and for the case of three generators by G.Rosenberger [R 8] (see Chapter 3). Our next result says that under certain restrictions any two generator subgroup of a group of F-type must be a flee product of cyclics. THEOREM 8.2.5. Let G be a group of F-type. Suppose further that neither U nor V is a proper power or is conjugate to a word of the form XY for ele~nents X, Y of order 2. Then any two generator subgroup of G is a free product of cyclics. PROOF.Without loss of generality we may assume that UV involves all the generators. Recall as in the proof of Theorem8.2.3 that from Theorem 2.6.2 it follows ifG -- HI*AH2and {xl, ..., x,~} is a finite system of elements in G, then there is a Nielsen transformation from {xl, ..., x~} to a system {yl, ..., y~) for which one of the following cases hold: (i) Yi = 1 for somei e {1, ..., m}. q (ii) Each w E< yl,...,y,~ > can be written as w = Hi=I Y~,ei = =t=l, e~ = e~+l if ui = u~+l with L(y~,) <_L(w) for i = 1, ..., q. r~’q el a¢ 1 with y~, e A, (i = 1, .., q) and (iii) There is a product a = ¯IL=I Y~,, in one of the factors H~ there is an element x ¢ A with x-lax ~ A. (iv) There is a g ~ G such that for some i ~ {1,...,m} we have Yi gAg-1, but for a suitable natural number k we have -1. y~ ~ gAg (v) Of the yi there are p _> 1 contained in a subgroup of G conjugate H~ or H2 and a certain product of them is conjugate to a non-trivial element of A. Since the group of F-type G decomposes as a free product with amalgamation with amalgamated subgroup A =< U >=< V >, the result will follow from Theorem 2.6.2, described above, if we can show that there are no non-trivial powers U~, Vt~ and elements X1 E GI\ < U >, X~ ~ G2\ < V > such that X~U*~X~~ = U~: for some s2 ~ 0 or X2Vt~X~1 = Vt~ for some t~ ¢ 0. Suppose there exists a non-trivial power U~ of U and an element X ~ G~\ < U > such that XU*X-~ t= U for some t 2~ 0. The argument for V works identically.. We consider G~ a discrete subgroup of PSLu (R) which has no parabolic elements. Then is hyperbolic. Let z~, z2(z~ ~ z2) be the fixed points of U. Then either X fixes both Zl and z2 or X interchanges zl and z2. If X fixes both z~ and z2 then X and U commute and s = t. Since G1 is discrete this implies that X = Uq~, U = U[ for some q, r ~ Z \ {0} and

8.2 FREIHEITSSATZ AND SUBGROUPTHEOREMS

229

U1 E G1. In this case Irl _> 2 since X ~< U >. But this implies that U is a proper power, contrary to our assumptions. If X interchanges zl and z2 then X fixes the midpoint of the axis of U. This implies that s = -t and X has order 2. In particular XUX-1 -1 =U which implies that (XU)~ = 1 since X2 = 1. Then < X, U > is infinite dihedral. Let Y = XU. Y has order 2 and U = XY contrary to our assumptions. Therefore there is no X ~ GI\ < U > and non-trivial power -~ = Ut for some t 7~ 0. The identical argument works U~ such that XU*X for V and therefore by Theorem 2.6.2 the theorem holds. Wenote that without the additional hold. For example suppose (7=< al,a2,a3,

hypotheses the theorem does not

aa;a~ 1 -=a~ 2 =a~3 =a~4 =(ala2)S(aaaa)S=l

Then < (alan) ~, (aaaa) ~ > is not a free product of cyclics. A less trivial example is given by G =< al,a2,a3, a4;a21 ---- a~ = ag --= a34 = ala2a3a4= 1 >. Since [a~a~,a3al] = a4 it follows that G =< ala2,ala3 > and thus < a~a2,a~a3> is not a free product of cyclics. Finally we close this section with the statement and proof of a result of A. Hempel [Hem] which gives a strengthened version of the two-generator subgroup theorem for groups of F-type (Theorem 8.2.5). THEOREM 8.2.6. Let G be a group ofF-type and let H be a non-cyclic two generator subgroup of G. Then H is conjugate in G to a subgroup < x, y > satisfying one of the following conditions. (i) < x,y > is a free product of cyclic groups (2) t i s i n =< V > fo r so me natural nu mber t and y- ~x~y is in < a~,...,ap > ory-~xty is in < ap+~,...,a,~ >. PROOF.G is a non-trivial free product with amalgamation G = GI*HG~ with G~ =< a~,...,ap >, G~ =< ap+l,...,o~ > and H =< U >=< V >. Any subgroup entirely contained in a conjugate of either G~ or of G~ is clearly a free product of cyclics. Now suppose that U = U(a~,..,a~) = U], with q >_ 1 and U1 not a proper power in G1 and that V = V(ap+~, ..,a~) = V[, with r _> 1 and not a proper power in G2. Choose in each Gi (i = 1,2) a system Li of left coset representatives of H in Gi normalized by taking 1 to represent H and taking the inverses L~-1 of the left coset representatives as a system of right coset representatives.

230

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Introduce a length L and an order in G as described in section 2.6. If g is given in the symmetric normal form g = sl...s,,kr,,...r~ then L(g) = if k E H and L(g) = 2m+ 1 if k it H. Suppose that K =< a, b > is a non-cyclic two-generator subgroup of G. Wemay assume that {a, b} is minimal with respect to L and the order. From Theorem 2.6.2 we may assume that < a, b > is either a free product of cyclics or after a suitable conjugation a is in either G1 or G2 with at E H for some natural number t. Without loss of generality we may then assume that a -- U1, and let b = sl...s,~kr,~...r~ be given in symmetric normal form. If L(b) _< 1 there is nothing to prove so suppose that L(b) > 1 and then m >_ 1. From Nielsen reduction in free products with amalgamation or from the proof of Theorem8.2.5 it follows that ifg is in G~ and gU~g-1 is in H then gU~g-~ = U~t and if g is in G2 with gU~g-~ ~ H, then t = dq for some integer d and gU~g-~ = U~t . Further if g is in G~ and gU~g-~ t= U~ for some non-zero integer t, then we must have already gUlg-~ = U~1 and ifg is in G2 and gud~qg-~ = U~dq for some nqn-zero integer d then we must have already gU~g-1 = U1~q. Again from Nielsen reduction in free products with amalgamation it follows that if L(bUd~qb)< 2L(b) - 1 for some non-zero integer d, then r~ = si -~ and r~Ud~qr{1 = U~dq. -~, More generally if ri = s~ ¯ ~dq --1 ~dq rit~ 1 ri = U1 for i = 1, ...,~, 0 _< j < m and L(bud~qb) < 2L(b) - 2j for some non-zero integer d, then r~+l = sj-~l and +dq. rj+lUdqr-~_~ = U -~. Analogous statements hold for b-~Udlqb Suppose b = Sl...s,~kr .... rl as above and suppose further that there are at least two different letters g~,g2 of the s~,.,s,,,r,~,..,rl if k ~ H and s~, .., s,~, k, r,~, .., r~ if k it H such thatgiU~gi t -1 is not inH, i=l,2, ~ d: bt: withd~,d2 for all non-zero integers t. Then in any product bt~ ud~bU 1 non-zero integers and bt~ 7£ 1 7£ bt~ the letters gl, g~ remain uncancelled in b, (consolidations to elements of length _> 1 are allowed). The analagous statement holds for products of the type bt~ud~ibt~Ud~bta with dl, d2 nonzero integers and bt* ?~ I for i = 1, 2, 3. Therefore in this case using standard cancellation arguments K ----< a, b >= is a free product of cyclics. Nowsuppose that there is at most one letter g of the Sl,., s,~,r,~, .., rl if k ~ H and Sl,..,s,~,k,r~,..,rl if k it H such that gU~lg-~ is not in H for all non-zero integers t. Then after a suitable conjugation applied to the subgroup < U1q, b > we get a conjugate < U~, b~ > of < U1q, b > with ~ in G1 or in G2 with e = :t:1. This gives the desired result. b~-~U~b Workof Rosenberger [R 8] can be used to give a complete classification in certain groups of F-type of subgroups of rank less than or equal to 4 satisfying a quadratic condition. Specifically we have. THEOREM 8.2.7.

Lef G be a group ofF-type

such ~hat neither

U nor

8.3 LINEAR PROPERTIESGF GROUPSOF F-TYPE

231

V is a proper power or is conjugate to a word of the form XY with X, Y elements of order 2. Let ul, ..,un, 1 ~_ n ~_ 4, be elements of G. Suppose that the system {ul, .., un} is not Nielsen equivalent to a system containing an element which is conjugate in G to an element of the amalgamated subgroup < U >--< V >. Suppose W(ul, ...,u,) = 1 where W(xx, ...,x,~) is a quadratic word in the tkee group on Xx, .., xn and let H be the subgroup generated by ul, .., un. Then H is either trivial, a bee product of cyclics or a surface group (oriented or non-oriented). 8.3 Linear Properties

of Groups of F-Type

The existence of an essential representation for a finitely presented group G indicates that the group is "almost" a finitely generated linear group. This was the idea behind the investigations in the last chapter on torsion one-relator products of cyclics. Here we consider the basic linearity properties for groups of F-type. As we will see, most of the algebraic properties of F-groups carry through to this extended class. In this section we consider the linearity properties studied in the last chapter. These are (1) The virtual torsion-free property; (2) The Tits alternative; (3) SQ-universality; (4) The existence and computation of rational Euler characteristics. As already described, the fact that a group o£ F-type decomposesas a free product with amalgamation is immediate from the definition. As a direct consequence of this decomposition we get our first results which mirror exactly the situation for F-groups. Throughout this section we suppose that G is a group of F-type with presentation (8.1.3) G =< al,. .... , an; a~1 ......... a~~ = 1, U(a~,..., ap)V(av+~,.., an) = where n _> 2,ei = 0 or ei > 2, 1 _< p < n - 1, U(al,..,ap) is a cyclically reduced word in the free product on al,...,ap which is of infinite order and V(ap+~, ..., a,~) is a cyclically reduced word in the free product on %+~,...,an which is of infinite order. Ifp = 1 (or p = n - 1) we restrict groups of F-type to those where U = a T (or V = a~) with lm[ >_ 2. Welet el G1 ~ %+1-G2~<:ap+l, .... , an; ap+ --a~ ~-1> and 1 - .... -A =< U-~ >=< V >, so that G = G~ *A G2. G1, G~ axe the factors of G and UV is the relator.

232

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

THEOREM 8.3.1. Let G be a group ofF-type with presentation (8.1.3). Then (1) If e~ >_2 then a~ has order exactly e~. (2) Any element of finite order in G is conjugate to a power of some a~. (3) Any finite subgroup is cyclic and conjugate to a subgroup o[ some ~ai~. (4) Any abelian subgroup is either cyclic or ~ee abelian of rank PROOF.This th~rem follows directly from the ff~ product with amalgamation decomposition and the fact that the factors are ff~ products of cyclics. ~ompa~ (2) it follows that any essential representation m~t be essentially faithf~. The e~stence of an essential representation then guarantees, as we saw in Chapter 6, that any group of F-type is virtually torsion-fr~. Fu~her we saw in Theorem 8.2.1 that if neither U nor V is a proper power, then there is actually a fai~hf~ representation into PSL2(C) and hence G can be co~idered as a linear group. Therefore under these conditions G is residually finite and also Hopfian from a th~rem of Malcev [Mall. Wewill see in the next section, that relative to the prope~ies of residual fi~teness and Hopfian, we can remove the restrictions on U and V. THEOREM 8.3.2. Le~ G be a group oT F-type. Then (1) G is vir$ually torsion-[ree. (2) H nei$her U nor V is a proper power, $hen G is residua~y finite and thus Hopfian. The bu~ of the last chapter was devoted to proving that almost all torsion one-relator products of cyclics satis~ the Tits alternative. The situation for groups of F-type is even simpler. A group of F-type will either contain a ff~ subgroup of ra~ 2 or is itself solvable. This latter situation can o~y occur for t~ ~oups. THEOREM 8.3.3. Le$ G be a group ofF-type. Then either G has a ~ee sub~oup of rank 2 or G is solvable and isomorphic to groups wi~h one o[ the following presentations: O) H1 =< a,b;a2b ~ = 1 >; (ii) H2=< a, b, c; a2 = b2 = abc~ = 1 >; Oil) Ha =~ a, b, c, d; 2 =b~~ c 2 = d2 = ab~= 1 > . PROOF. Suppose n > 4. Then a flee product of cyclics injects into G and therefore G has a flee subgroup of ra~ 2. n = 4 and not all ei = 2. Then a flee product of two cyclic Z2 * Z2 injects into G and therefore G has a ff~ subgroup of

of ra~ ~ 3 Next suppose groups not ra~ 2.

8.3 LINEAR PROPERTIESOF GROUPSOF F-TYPE

233

If n = 4 and all ei = 2, then necessarily, since U and V have infinite order, we must have p = 2. Then G has a presentation < a,b,c,d;a 2 = b2 = c 2 = d2 = (ab)S(cd) t = 1 > with s > 1, t > 1. G then has as a factor group the free product ~ =< a, b, c, d; a2 = b2 = c2 = d2 = (ab) s = (cd) t = 1 > . This is a non-trivial free product ~ = * < c, d; c 2 = d2 = (cd) t = 1 >. If (s, t) # (1, 1) then G has a free subgroup of rank 2 and therefore also contains a free subgroup of rank 2. If (s, t) = (1, 1) then G has the presentation < a,b,c,d;a 2 = b2 = c2 = d~ = abcd = 1 > which is solvable. If n = 3, then at least one ei = 0, since U and V are assumed to have infinite order. Supposewithout loss of generality that et = 0 so that al has infinite order. If p = 2 then a free product of two cyclic groups not Z2 * Z2 injects into G and therefore G has a free subgroup of rank 2. Nowlet p = 1. Then U = a +s with s >_ 2. Supposefirst that (c2, ca) # (2, 2). Then a t product of two cyclic groups not Z2 * Z2 injects into G and therefore G has a free subgroup of rank 2. Next suppose that (e~, e3) = (2, 2). Then G a presentation < a,b,c;b 2=c2=aS(bc)t=

l

where s > 2 and t >_ 1. G then has as a factor group the free product ~ =< a, b, c; as = b2 = c2 = (bc) t = 1 >, ~=< a;a ~ = 1 > * < b,c;b ~ = c 2 = (bc) t = 1 > . If s > 2 or t > 2 then a free product of cyclics of rank > 2 and not Z2 * Z2 injects into G and thus G has a free subgroup of rank 2. It follows that G does also. This leaves the case where s = 2 and t = 1. G then has the presentation < a, b, c; b2 = c 2 = a2bc = 1 > which is solvable.

234

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Finally suppose n = 2. Then since U and V have infinite order both generators must have infinite order and G must have a presentation < a,b;aSb ~ = 1 > with s _> 2, t _> 2. G then has as a factor group the free product of cyclics ~ =< a,b;a s = bt -- 1 >. If (s, t) # (2, 2) then G has a free subgroup of rank 2 and therefore does also. This leaves the situation where (s, t) = (2, 2). G then has presentation < a, b; a2b2 = 1 > which is solvable. Recall from our previous discussions the close ties between the existence of non-abelian free subgroups and SQ-universality. The next result says that for group a of F-type G either it is SQ-universal or isomorphic to one of the three exceptions given in Theorem8.3.3. THEOREM 8.3.4. Let G be a group ofF-type. IT G is not solvable then G has a subgroup of finite index which maps onto a [-ree subgroup of rank 2. In particular a group of F-type G is either SQ-universal or solvable. PROOF.Suppose G is not solvable¯ First assume that n _> 5 or n -- 4 and at least one ei ~ 2. Wemay assume without loss of generality that each ei _> 2 passing to an epimorphic image if necessary. The first part of the proof mirrors the proofs of Theorem6.2.3 and of Theorem7.3.2.1. Let p : G -~ PSL2(C) be an essential representation. It follows from the work of Selberg that there exists a normal torsion-free subgroup N of finite index in p(G). Let ~r be the canonical epimorphism from p(G) onto p(G)/N. The composition ¢ of the maps in the sequence G ~ p(G)

--~ p(G)/N

gives a representation of G onto a finite group. Further ¢(a~) has order in p(G)/N since p is essential and N is torsion-free. Nowconsider the free product of finite cyclic groups X :(al,...,an,

1 ~- . ...

~

.

There is a canonical epimorphism e : X --~ G. Consider the sequence X _A_+ (7 ~ p(G)IN.

8.3 LINEAR PROPERTIESOF GROUPSOF F-TYPE

235

Let Y = ker(¢e). Then Y is a normal subgroup of finite index in and YVI < a~ >= {1} for i = 1,...,n. Then by the Kurosh theorem Y is a free group of finite rank r. The finitely generated free product of cyclic groups X may be considered as a Fuchsian group with finite hyperbolic area. Therefore from the Riemann-Hurwitz formula we have

j,(x) = ,(Y) where #(X) = n- 1 - ((1/el)

.. ......

+

#(Y) = - 1. Therefore r : 1 - j((1/el) + .... + (1/en) - n + 1). The group G is obtained from X by adjoining the additional relation UV = 1 and thus G = X/K where K is the normal closure of UV in X. Since K C Y the factor group Y/K may be regarded as a subgroup of finite index in G. Using work of Baumslag, Morgan and Shalen (corollary 3 of [B-M-S]) Y/K can be defined on r generators with j relations. Thus the deficiency of this presentation is given by d = r-j = 1 - j((1/el)

+ ....

+ (1/e,~) - n+

= 1 + j(n - 2 - (1/e~) - ....

- (1/e,)).

If n_> 5or n=4andat least oneei ~ 2thend_> 2. It follows, then from the theorem of B.Baumslag and Pride, that G contains a subgroup of finite index which maps onto a free group of rank 2. Next suppose n = 4 and all ei = 2. Then necessarily p = 2 and U = (ala2) 8, Y = (aaa4)t with Is[ _> 1, It[ _> 1. Further since G is non-solvable we must have Is] _> 2 or It[ _> 2. Without loss of generality we assume that s _> 2 and t _> 1. Then G has as a factor group the free product. ~=< al,a2,aa;a

~ 8 T = a2 = (a~a2) = 1, a~ = 1 >.

G has as a normal subgroup of index 2 a group isomorphic to the free product of three cyclic groups 8=y2=z2=1>. <x,y,z;x Therefore G and hence G also has a subgroup of finite index mapping onto a free group of rank 2. Now suppose n = 3 and at least two ei ¢ 2. Then without loss of generality we may assume that p --= 1, al has infinite order and U = a~

236

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

with s >_ 2. Suppose first that a2 also has infinite order and V = a~t with t _> 2. Then G has as a factor group the free product of cyclics G=
a~=aa

=1>.

G and therefore G also has a subgroup of finite index mapping onto a free group of ramk 2. Nowsuppose that V ---- a~, e = ~:1. Then G is isomorphic to the free product Z * Z¢3 which has a subgroup of finite index mapping onto a free group of rank 2. Therefore we may now suppose that V involves both a2 and a3. Choose a number m _> 7 such that m is relatively prime to s. Then G has as a factor group G " "~ a = a ~:=a~ 2 =< a, a2, a3~

s=aV _- 1> .

Eliminating a = V-~ we get that G =< a2, a3;a~~ = a~3 -~ V(a2, aa) "~ = 1 >=< x,y;x p -- yq = V(x,y) m = 1 > with mk 7 and_ (p, q) -- (eu, e3) ¢ (2, 2). Thus ~ is a generalized triangle group with s(G) < 1. It then follows from Theorem7.3.2.1 that ~ and thus G has a subgroup of finite index mapping onto a free group of rank 2. Next suppose n -- 3 and that two ei -- 2. Then we may assume without loss of generality that a~ has i~ffinite order, U = a~ with s _> 2, (e~, ea) --(2, 2) and V ~- (a2a3)t with t _> 1. If t _> 2 then G has as a factor group the free product t -. s= 1, a~2 = a~ G ~ (a2a3) 1 > 1 =~ al,a2~ a3~ a which has a normal subgroup of index two iso~norphic to the free product of cyclics < x,y,z;x t = yS = zs = 1 > . Hence G and therefore also G has a subgroup of finite index mapping onto a free group of ra~k 2 if t _> 2. If t --- 1 then G has the presentation G ---< al, a2, a3; sa22 = a~ = a1a2a3 ~ 1 ~( al~ a2; a22 = I, ~as la2) ~2

=1>.

Since G is not solvable we have s _> 3. Then G has as a factor group the free product of cyclics ~ al, a2~a1 ---- a = 1 > .

8.3 LINEAR PROPERTIESOF GROUPSOF F-TYPE

237

Since s > 3 this has a subgroup of finite index mapping onto a free group of rank 2 and therefore G also has a subgroup of finite index mappingonto a free group of rank 2. Finally suppose n = 2. Since U and V are assumed to have infinite order and G is not solvable it follows that G must have a presentation < x,y;xSy ~ = 1 > with s k 2 and t _> 3. Then G has as a factor group the free product of cyclics G =< x,y;x ~ = yt = 1 > . Since (s, t) ~ (2, 2), G and hence G also has a subgroup of finite mapping onto a free group of rank 2. This completes the proof.

index

One of the most powerful techniques in the study of Fuchsian groups is the Riemann-Hurwitzformula relating the Euler characteristic of the whole group to that of a subgroup of finite index. As we remarked in section 7.6 Bass [Ba 3], Brown [Brn], Chiswell [Ch], Wall [Wa], and others have extended the concept of a rational Euler characteristic to more general finitely presented groups and have shown that in many cases these can be computed very easily from given finite presentations or given amalgam decompositions. Further these general rational Euler characteristics satisfy the Riemann-Hurwitz formula. In section 7.6 we showed how to compute the rational Euler characteristic for certain torsion one-relator products of cyclics. Weclose this section using results of C.T.C. Wall [Wa] and K.Brown[Brn] to extend this Euler characteristic to groups of F-type and give a Riemann-Hurwitz type formula. THEOREM 8.3.5. Let G be a group ofF-type. Euler characteristic x(G) given

Then G has a rational

where ai = -1 ire~ = 0 and ai = -1 + (l/e/) fie/k If IV: HI< oo then x(H) is defined and x(H) = IG HIx(G) Hurwitz formula). In addition G is of finite homological type and vcd(G) <_

(Riemann-

PROOF.C.T.C. Wall in [Wa] defined a rational Euler characteristic for a class of groups which includes the trivial group, the infinite cyclic group Z and all finite groups and which is closed under free products with finitely manyfactors and subgroups of finite index. If G is a group, we denote this

238

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Euler characteristic by x(G). From Wall we have that x(H) = 1/IH I for a finite group and x(H) = 0 if H is infinite cyclic. Further if H is a free product, H = A * B, then x(H) = x(A) + x(B) - 1. It follows that if H is a free group of finite rank r then x(H) = 1 - r. Further Wall shows that if H1 is a subgroup of finite index in H then x(H1) -- IH HIlX(H). Nowlet G be a group of F-type. From theorem 8.3.2 it follows that G is virtually torsion-free. Using this fact we can apply the techniques of Brown [Brn] to get an Euler characteristic for G by the formula x(a) X(G1) -b X(G2) -

x(A

where G1, G2 are the factors of G from the above decomposition and A =< U >--< V > is the amalgamated subgroup. Since A is infinite cyclic, we have x(A) = and th erefore x( G) = x( GI) + x(G2). Fu rther G~ and are free products of cyclic groups, so we can apply the computation rules of Wall above to obtain n

=2 + i~l

ai = -1 if e~ = 0 and ai = -1 + (1/e~) if ei _> 2. Wecan also write this as

x(G)

= 2- n+

where f~s = 0 ifei = 0 and f~i = 1/ei ifei k 2. This second form is somewhat closer to the form usually written for F-groups. The RiemannHurwitz formula follows directly from Wall. Wenote further that the Euler characteristic of a group of F-type can be zero. In fact x(G) = if n =2. Since a group of F-type G is virtually torsion-free, it follows from K. Brown, as in section 7.6, that G is of finite homological type with vcd(G) _<2. 8.4 Additional

Results

Finitely generated Fuchsian groups are linear, so it follows that they are residually finite and Hopfian. G.Scott proved that they are subgroup separable while Fine and Rosenberger showed that they are conjugacy separable (see Chapter 4). In this section we extend these four properties to groups of F-type. Suppose G is a group of F-type with relator UV. If neither U nor V is a proper power, then from theorem 8.2.1 G has a faithful representation in PSL2 (C) and thus can be considered to be a linear group. It follows

8.4 ADDITIONAL RESULTS

239

then that under these conditions G is both residually finite and Hopfian. From work of F. Tang [Tan 2] and R. Allenby and F. Tang [A1-T 3] we can relax the restrictions on U and V and conclude that these results are true in general, that is, it is not necessary to have non-proper powers. THEOREM 8.4.1. A group of F-type is residually

finite

and Hopfian.

PROOF.If H is a group and h E H then we say that H is < h >-potent if for any m _> 2 there exists a finite quotient of H where the image of h has order exactly m. This is clearly a strong form of residual finiteness. From F. Tang [Tan 2] and Allenby and Tang [A1-T 3] we have the following result. Let G = A *H B where H =< h > is infinite cyclic. If both A and B are < h > - potent and < h >-separable then G is residually finite. Groups of F-type have exactly this form where A and B are free products of cyclics. The theorem then follows since in any free product of cyclics K, K is < g >-separable and < g >-potent for any element g of infinite order in K. The < g >-separability follows from a result of P. Stebe [St] while the < g >-potency can be deduced in the following manner. Let K be a free product of cyclics with presentation Suppose g is an element of infinite order in K and m _> 2 a natural number. If g is a power of a generator of infinite order, the < g >-potency follows directly, so suppose that g has free product length _> 2. Let K map onto K=. K is then a torsion one-relator product of cyclics, so there exists an essential representation of K onto K in PSL2 (C). Then from Selberg there exists a torsion-free normal subgroup N of finite index in K. The sequence K---~K---*K---~K/N then gives a homomorphismof K onto a finite group where g has order m. Therefore K is < g >- potent. It follows then from the Theoremof Allenby and Tang that a group of F-type is residually finite and from the result of Malcev that it is Hopfian. Recall that a group G is subgroup separable or LERF(Locally extended residually finite) if given any subgroup H of G and any element g not in there exists a subgroup H* of finite index in G such that H is a subgroup of H* and g is not in H*. Againthis is a very strong form of residual finiteness. Brunner, Burns and Solitar [Br-B-S] proved that cyclically pinched onerelator groups are subgroup separable (Theorem3.5.10) while G. Scott [Sco] proved the same for finitely generated Fuchsian groups (Theorem 4.3.11). From two results, one of Allenby and Tang [A1-T 2,3] and one of Niblo [Ni 2], it can be deduced that groups of F-type are subgroup separable.

240

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

THEOREM 8.4.2.

A group ofF-type

is subgroup separable.

PROOF.Allenby and Tang’s method [A1-T 2,3] is purely group theoretic and proceeds by showing that the amalgamated free product of two virtually free groups amalgamating a cyclic group is subgroup separable. Since free products of cyclics are virtually free, this result can be applied both to groups of F-type, and also to the case of a group G with a presentation a=< ax,. .... , an; a~~ .........

a~~ = (UV) m = 1 >

with 1 # U = U(al,...,ap), 1 ~ V = V(av+l~..,an)~ 1 _< p _~ n- 1 and m _> 2. This case will be considered in the next result, due to Allenby on conjugacy separability. Niblo [Ni 2] using some topological arguments shows that amalgamated free products of Fuchsian groups, amalgamating a cyclic subgroup are subgroup separable. Since free products of cyclics can be faithfully represented as Fuchsian group his result applies to groups of F-type. Wemention that M.Aab[Aa] has established the subgroup separability of a more general class of fundamental groups of graphs of groups. Recall that a group G is conjugacy separable if given any two dements of g, h of G, either g is conjugate to h, {g ,-, h} or there exists a finite quotient G of G where the images of g and h are non-conjugate. In Chapter 4 it was proved that Fuchsian groups are conjugacy separable. Allenby [A] has extended this to groups of F-type. THEOREM 8.4.3.

A group ofF-type

is conjugacy separable.

Allenby actually proved the following more general result from which the second part is just an addendumto his result¯ Wereproduce Allenby’s proof which is quite lengthy.. THEOREM 8¯4.4. (8.4.1)

Suppose

G=
= a~~ = 1, (UV)m : 1 >

where n >_ 2, ei = 0 or ei >_ 2, 1 <_ p <_ n- 1, U = U(al,..,ap) a non-trivial cyclically reduced word in the free product on al, ..., ap and V = V(ap+l, ...,a~) is a non-trivial cyclically reduced word in the free product on ap+l, ..., an and m >_ 2. Then G is conjugacy separable. Farther if neither U nor V has finite order in the respective free product on the generators which they involve, then G is conjugacy separable if m>_ 1. PROOF.The second part of the statement of the theorem, that is when neither U nor V has finite order clearly implies the conjugacy separability of groups of F-type.

8.4 ADDITIONAL RESULTS If V has infinite amalgamtion of

241

order and m >_ 2 then G is the free product with

1A, a;a al,..,a s =< ~

...

B =< 1ap+l, ...,

a~p = am = 1 > and

~P+I an; ap+

...

a~~ = 1 >

with the cyclic subgroup < h >=< U-la >=< V > a~nalgamated. If V has finite order and U has infinite order swap the roles of U and V. If both U and V have finite order, then both are conjugates of some generator and we may assume that G has the form G =< a~, .....

, a,~; a~~ .........

a~~ = 1, (a~a~)"~ = 1 >

withi 7& j. Assume further that rn_> 2. ThenG = H*RwhereHis ej ¯. i e a~ = a~ = the free product of cyclics on the ak, k ~ i,j and R =< a~, a~, (aiaj) = 1 Let k = e~/gcd(e~,c~), ~ = ce/gcd(e~,c~), 1 = ej/gcd(ej,~), f3/gcd(ej, ~). Then R is obtained from Ro =< x, y; xk = yt = (x~yS) "~ = 1 > by two successive free products with amalgamation. From a result of Dyer [Dy] these will be conjugacy separable if R0 is. However Ro ~-< x,y;x k = yt = (xy),~ = 1 > which, since m >_ 2, is an ordinary triangle group. From Theorem 4.3.10 Ro is conjugacy separable. Therefore R is conjugacy separable. A free product of cyclics is conjugacy separable and so in this case G = F * R is conj~acy separable. Therefore for the remainder of the proof we will assume that V has infinite order. Let first m _> 2. Wework with the above decomposition and use the fact that G now is actually a group of F-type. The proof with V of inifnite order and m > 2 then follows from the following series of lemmas. Following Allenby we use the notation A, h etc. to denote homomorphic images. LEMMA 8.4.1. Suppose G has form (8.4.1) ht ~ h-t -~. if and only ifh ~ h

and h = V as above. Then

PrtOOF. (Lemma8.4.1) Let A,B be the factors of G. Then if c = h~ ,-~c -t h -~ d then from the free product with amalgamation decomposition there exists a sequence h~,..., h~ of elements of < h > such that ((1))

ht ~Ahi~"~Bhi~ "~A .... . B h~ = h-t = d.

A and B are free products of cyclics and h has infinite order, so inside A,B we can only have ht ,~ ht or ht ~ h-t (see the arguments in 8.2.5 and 8.2.6). Thus each ik is a t or a -t. Further if say ht = b-lh±tb, then h is conjugate to h±~ using the same element b. Hence (1) implies that h ~ -1.

242

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

LEMMA 8.4.2. Let A be a free product of t~nitely many cyclic groups and h E A of infinite order. Then there exists a finite homomorphicimage ~ of A in which -~ has prescribed order t >_ 2. PROOF.(of Lemma8.4.2) (See also the proof of Theorem 8.4.1.) Without loss of generality we can assume that all the generators of A are involved in h. Consider the one-relator product of cyclics H = A/N(h~). Then there exists an essential representation of G in PSL2(C) and hence there is a homomorphic image ~ in PSL2 (C) where ~ has order t. ~ is finitely generated subgroup of PSL2 (C) and hence residually finite so there exists a finite homomorphicimage of A and therefore also of A where h has order t. LEMMA 8.4.3. Let A and h be as in Lemma 8.4.2. and let F be a normal free subgroup of finite index in A contained in the kernel of the natural map from A onto the direct product of its (finitely many) cyclic factors. Let h~ -1. ~ F, w >_ 1 and suppose h is not conjugate in A to h Then ~here exists a finite homomorphicimage ~ of A in which -~ 7~ 1 and notation {g}~ to denote the set of conjugates of g in G. PROOF.(of Lemma8.4.3) Let = {o ~1 ~- 1, o~2 , ... , at} be c oset representatives for A mod D where D = FC ~ with C the centralizer of h in A. Then every eleinent of A is of the form yfa where y ~ C,f ~ F, a ~ S. Let {~}~ be the set of conjugates of hTM in A. Hence {hw}A -~ {a~f-~h~fa~}k#~ U {f-~h~f} where f runs over all the elements of F. Now{f-~h~f : f e F}Q < hw >-- {hW}. (This is a singleton set since h~ ~ h~’ is ruled out by hypothesis if w~ = -w and by order considertaions if [w’[ ~ Iw[.) Similarly {a;~f-lh~fak}C~ < h~ >C {h~}. But, if k ¢ 1, a~f-~h~fa~ -- h~ implies that fak ~ C C D which implies that a~ ~ D, which is impossible by the choice of S. Hence {a~f-lh~fak}n < h~ >= ~ if k ~ 1. Consequently {f-~h~f} ~ {akh~a-~ ~} = O (if k ~ 1) - an empty intersection holding in a free group F. From results of Dyer [Dy] there exists a finite nilpotent homomorphicimage FIX of F in which {~--1]-~]~-}~ < ~-~ >= ~ for all ak ~ 1. Since X has finite index in A, we can assume without loss of generality that X is normal in A. To modify X further so that also {y--l~vy}~ ( ~-w )} = {~--w} in A/X, take Y = F~(F) the sth term of the lower central series for F where s is such that h~ ~ F~_~(F) \ F~(F). Let E be a characteristic subgroup of F ~ that F~(F) C E C F~_~(F) and EV~ < h >=< >where v is the orde r of ~ in A/X. Then ~ has order v in the residually finite group AlE and therefore A has a normal subgroup Z such that [A/Z[ < oc and in which

8.4 ADDITIONAL RESULTS

243

hZ has exact order vw. Replace X by X V~ Z = U. Then A = A/U satisfies the conclusions of the lemma. LEMMA 8.4.4. Let < g >, < k > be cyclic subgroups of a free group F and let s ¯ F\ < g >~ k >. Then there exists a finite homomorphic image P of F in which -~ ¯ -~\ < y > < ~ >. PROOF.(of Lemma8.4.4) Suppose [g, k] ~ 1 in F and suppose s-7 y~fl’ expresses the image of an element of < ~,~ >C F/F~(F) = F~.. Suppose[g,k] ¯ F~_I(F)/F~(F).. If for i _> t all the a.i are equal, then for somej _> t we have ~3j # f)t since a free group is residually nilpotent, that is V~IF~(F) =< 1 >. But this implies that ~ has order {f)j - ~t] in Ft which is a contradiction. Thus a~ ~ at for some r > t and then ~-~-~ = ~ in Ft for some f~ _> 0. f) = 0 leads to a similar contradiction to the one just noted. If/~ >_ 1 then ~,~]~ = [~,~] = 1 in F~ which contradicts ~,~] ~ 1 in Ft. Thus there is a finitely generated torsion-free nilpotent homomorphic image ~ of F in which ~ ~< y >< ~ >. From results of Stebe [St] then there is a finite homomorphicimage of F in which ~ @<~ >< ~ >. If [g, k] = 1 then < g >< k >=< g, k >=< gl > with g = g~, k = g~ for some g~ ¯ F and s, t ¯ Z. The desired result again follows from results of Stebe [St]. LEMMA 8.4.5. Let A and h be as in Lemma8.4.2. and let k,g ¯ A be such that k q~ HH~ where H =< h > and H~ = g-lHg. Then there exists ~. a finite homomorphicimage -~ of A in which k ~ HIf PROOF.Let F be the kernel of the natural map from A to the direct product of its finite cyclic factors and let (h~) a denote g-lhf~g etc. If k q~ HF factor out the normal sub’group HF. Otherwise suppose that k ¯ HF. Write H = Ho U hHo U ... t~ h~-iHo where r is the least power of h to lie in F, ho = h~ and H0 ---< h0 >. Then ~ _ HH~ {h~g-lh~g: a,f~ ¯ Z}={h ho(hoh) ~ ~ ~ ~ :O
244

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

i - k = =k(m - j) by order considerations. If i = k then m = j. Otherwise from 8.2.5 and 8.2.6 we deduce that v-lhv = h±1. Note that in case (iii) we have < v-lhv > N < h >-- 1. LEMMA 8.4.7. Let A and h be as in Lemma 8.4.3, and let F be the kernel of the natural mapfrom A onto the direct product of its finite cyclic factors. Let hTM be the least power of h in F. Suppose for all r such that 1 < r < s that uT, vT E A and that ur = h*rvTh~ where i~,j~ are unique. Then there exists a finite homomorphicimage A of A in which h has order wp with p a prime and if for 1 < r < s, -fk~,-~lr is a solution oft ---- x~y then kr =- ir and IT =- j~ modulop. PROOF.(of Lemma 8.4.7) As before let H --< h > and ~ - - g-iHg ~ implies that for Note first that the uniqueness of the solution ur = h*~v~M ~ =< 1 >. This in turn implies that wr = [v~lh~vr, h~] 7~ 1 each r, HNH or else v~lhwvr, hTM generate an infinite cyclic subgroup of F which is a contradiction. Note that each w, lies in F which is free of rank > 1. If [wl, w2] -- 1 set w~ = [w2, z] for suitable z so that [Wl,W~] ~ 1. If [wl, wu] ~ 1 set w~ --- w2. Repeat this with wa, w4 .... in turn and consider q = [Wl, w~, ..., w~] ~ 1 in F. Nowchoose a finite homomorphicimage of F of exponent p where p is arbitrarily large and coprime to ]A/F[. Now pass to a homomorphic image A of A in which ~ has order p. None of the~7~ is trivial in ~since~ ~ 1. But this means that < ~’ > ~ < ~’ >~-7=< 1 >. Therefore suppose that, in ~, which is an extension of a finite group of exponent p by a group of coprime order ]A/F[, we kr -it,

ar = jT-mr. Assuming p does not divide p~ we have that p does not

These together imply that completing this lemma.

~r-1 --wh W ~<

> which is a contradiction

Wenow return to the main proof of Theorem 8.4.4. Wesuppose G has form (1) with m_> 2, V has infinite order and set a = UV. As in the proof of Lemma8.4.1, let A, B be the factors of G and < h >--< V >=< U-la >. Let c, d be elements of G of minimumlength in their conjugacy classes and such that c is not conjugate to d in G. Wehave to show that there is a finite quotient G of G where the images of c and d are not conjugate. If c -- I or d -- 1, then this follows from the residual finiteness of G considered as a group of F-type (see Theorem 8.4.1). Nowsuppose c ~ 1 and d ¢ Let Ix[ denote the length of x written as a normal form in ’the free product with amalgamation (see Chapter 2). Then Ix[ -- 0 if and only if x E< h and Ix[ = 1 if and only ifx ~ A\ < h > or x ~ B\ < h Case (1): Icl -- Idl -- 0. Then c hi ,d = h~wit h c ~ a d. In particular

8.4 ADDITIONAL RESULTS

245

i ~ j. If [i I ¢ IJl from Lemma8.4.2 we can find a homomorphicimage of G which is a free product with amalgamation G of two finite groups with j. So we may assume that I ~] = [illjl so that l~i[ ¢ I~Jl and hence ~i ~U~ i,d= -i i -~i c =h h and h ~G h In p~ticular h ~A h-~ and h~ -i. ~s h A being a ~ product of cyclics contai~ a fi~tely generated ~ group FA as a normal subgroup of finite index and contained in the kernel of the natural map ~omA onto the direct product of its fi~te cyclic factors. Simil~ly B contains such an FB. Let < h~ >= FA ~ F~. By Lemma8.4.2 there exists normal sub~oups X~, Y~ of fi~te index in A and B such that X~ < h >: Y~ < h >:< h~ >. Set A1 : FA ~X1,B~ : FB ~Y1. Then A~, B~ are normal of fiaite index in A, B respectively. Nowlet ~1 = 1, a~, ..., ~, fll = 1, ~2, ..., ~ be coset representatives in A, B respectively, chosen as in Lemma8.4.3 modulo sub~oups DA ALVA, DB = B1CB where CA,C~ are the centralizers in A, B of < h~ >. Use Lemma8.4.3 to find normal ~ subgroups X~, ~ of finite index such that (in A/X2) {~-

S h S~}N

y-

>= @ and in B/Y2

<

< >=

where f ~ A~,~ ~ B~ and 1 < k ~ r, 1 < t ~ s. Using Lemma8.4.2 we c~n modify X2, Y2 to assume without loss of generality that X2N< h >= Y2N h >=< hz~ >. As in the proof of Lemma8.4.3 choose normal sub~oups EA, E~ of fiMte index in A and B su~ that h~*~ > and such that

{VW }n <

EAN < h >= E~N < h

>=

in A/EA and B/EB respectively. Finally set A3 : EA n X2, B3 : EB ~ Y2. Nowlet G : A/A3 *<~> B/Ba the amalgamated ~ product of two finite ~oups. If ~i ~ then h and there exists a sequence of conjugate powers. But by choice of Aa, B~ the only conj~ate of~i~ in A/A3 and B/B3 is ~ itself. Hence ~i ~ ~-i in G. G is conjugacy separable so there exists a fi~te homomorphicimage of G and hence of G where c ~ d completing case (1). Case (2): (i) ~c] = 0,~d~ = 1. Without loss of generality let d (the case d ~ A is handled simil~ly). Thus d e B~H. Let c = i.

246

ALGEBRAIC

GENERALIZATIONS

OF DISCRETE

GROUPS

Since d is assumed to be of minimal length in its conjugacy class we have {d}BA < h >= 9. Suppose w is such that d~,h~ E FB withFB as in case (1). Then {d~}gA < ~ >= Oor els e b-l d~b = h t fol lows. App lying the proof of the first part of Lemma8.4.3 analogously we get a finite homomorphic image B B/M of B such that {~ }~ < >= ~). Case (2): (ii) Icl = 1, Idl = 0. This case may be dealt with as in (2): (i) in a symmetric manner. Case (2): (iii) All other cases where Icl ~ Idl are dealt with by passing to a free product with amalgamation G with finite, factors A and B where I~l = Icl ~ Idl = I~1. This is easily achieved by using the IIc property of the free products A and B (see [St]) to keep images of elements of A\ < and B\ < h > out of < ~ >. Case (3): Nowassume that c ~a d with Icl = tdl _> 2. Let c = ulu2...ur and d = vlv2...vr. Consider the system of equations (I(i))

ui+~ = x~v~x~

¯Ui4-r ~ xr--l~lVrXO

A solution of I(i) is a set of elements ho, h~, ..., h~_~ of < h > which satisfy the system simultaneously. By a result of Dyer [Dy] c ,-~ d iff for some value of i, 0 <_ i < r, the equations I(i) have a solution. Thus since c ~v d we know that for no i does I(i) have a solution. Weshow that there is a homomorphic image G = A *<~> B with A and B finite in which for no i does the corresponding syste~n of equations have a sohition. Since G is conjugacy separable the theorem follows. Nowif for each i ( for which all pairs u~+~and v~ lie in the same factor or B) there exists some t, possibly depending on i, such that u~+t ~ HvtH then by Lemma8.4.5 there is a finite homomorphic image of A (or B) which ~ ~t HKiH. If X is the intersection of all the corresponding kernels in A and Y is obtained similarly in B, then X and Y are easily modified to produce a free product with amalgamation G of finite groups in which ~d. Consequently we may assume that for at least one value of i the equations I(i) have no solution in G ~nd yet each individual u~+j = x~-~_~vjxj, 1 <_ i _< r is solvable in H so that a solution of ~(i) in some generalized free product G = A *<~> B of finite groups cannot immediately be ruled out. Weclaim that, even here, a generalized free product G = A,<~> B of finite

8.4 ADDITIONAL RESULTS

247

groups which is a homomorphicimage of G can be found in which ~ ~U ~. First suppose for each k, 1 _< k _< r, that v-~lhv~¢ = h or h-1. Thus any of the equalities ui+j = h’~vjh ~ may be rewritten as u~+~ = ~-~ h’~+~v~b or h~+’~v~b~+5. . Usingsuch adjustmentswe can find So, s~, ..., sT E Z such that (1)

al ui+l ~- h-a°Vlh ~2 ui+2 = h-~lv2h

Ui+T: ’~ h--a~--ivrh where sT - s0 ¢ 0 since c ~v d. We may further modify this solution by replacing h-~° by h-~°+~ : h~° and hence h~ by h~+~ if -~ v~hv~ = h or by h~-~ if v~hv~ = h. Then replace h-~ -~±~. by the appropriate h Continuing in this way we see that: (i) If in (1) the numbervk for which v~hvk = h-~ is even, then sT will be changed to fit = sT - 5 whereas (ii) in (1) the number of vk for which v-~hva = h-~ is odd, then sT will be changed to fit = sT+6. In case (i) fiT-flo =sT-s0 ¢0. In case (ii) fir - fl0 = sT - s0 - 26. Since 6 can be chosen arbitrarily and since c ~ d we deduce that in (1) sT - s0 must be an odd integer and that rio, fl~ can be chosen so that fit - fl0 is any odd integer. In case (i) consider the maximum, J, of the differences ]sT- s01 as in (1) as i varies over all the integers from 0 to r- 1. If we take normal subgroups MA, MBof finite index in A and B respectively such that, in addition to satisfying all the other sufficient conditions imposed, we also have MA~ A = MA~3 B =< h~ > where ¢ > J, then for no i can the equations 7(i) solved in (A/Ma) *<~> (B/MB) with the proper identifications. A similar result holds for case (ii) if we assume that ¢ is even which possible from the construction of essential representations. Thus ~-Zo # 1 since fl~ - fl0 is odd and ~ has even order. Thus we may assume that for at least one, and hence for all, i that the equations ui+~ = x~,~l_lvdY~5,1 < j < r, have solutions xi,~_~,yi,d in < h > and yet no solution in I(1) and that for each i at least one -1 v the equations, say ui+k(i) = xi,k(i)_ 1 k(i)Yi,k(i) is solvable with unique x, y. For each i select one such equation, say the k(i) th, as above. Fixing i, consider in turn the k(i) 1st, k( i) + 2nd, et c. eq uations of thesyst em, arranging if possible that Yi,k(i) = xi,~(i) etc. Since c ~ d this matching -1 must eventually fail, say at equation ui+i(i) = xij(i)_ 1~)f(i)Yi,I(i). Wenow choose p as in Lemma8.4.7 so that p is larger than all ]si -flil (as i runs

248

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

over all the integers from 0 to r - 1) and where ~i i s t he u nique s olution for xi,l(i)-i in the above equation and h~i is the forced value taken Yi,f(i)-I in the preceding equation. This choice of p leads, as in Lemma 8.4.7 to an amalgamated free product G with finite group factors which is a homomorphicimage of G and in which ~ ~U ~ as required. This completes the proof for m_> 2 by the result of Dyer [Dy]. Nowlet m = 1. Then U and V both have infinite order and G = A*H B with A -=
~p ¯ +I p-l-1 a

~ "’"

a~,p = 1 >

-~ a~" -- 1 > and

H==. The result then ibllows analogously. 8.5 Decision

Problems in One-Relator

Products

of Cyclics

In Chapter 2 we introduced the basic decision problems in combinatorial group theory; the word problem, the conjugacy problem and the isomorphism problem. Two additional decision problems are the generalized word problem and the power conjugacy problem. Let G =< X; R > and let H be a subgroup of G. The generalized word problem, abbreviated GWP,for G with respect to H is to determine if it can be decided algorithmicatly in finitely many steps whether an arbitrary element of G, written in terms of the generators X is in H. Wesay that the group G has a solvable GWPif G has a solvable GWPwith respect to the subgroups generated by recursive subsets of X. If G =< X; R > then the power conjugacy problem, abbreviated PCP, is given two elements in G, written in terms of the generators X to determine algoritmically, if a power of one is conjugate to a power of the other. The situation ibr one-relator groups is as ibllows: (1) Magnus,based directly on the classical Freiheitssatz, proved that onerelator groups have solvable word problem and fi~rther from the analysis it can be shown that they have solvable GWPas described above. (2) B.B. Newman[Ne], using the spelling theorem, proved that onerelator groups with torsion have solvable conjugacy problem (Theorem 3.2.6). Lipschutz [Li], using standard small cancellation theory, showed that cyclically pinched one-relator groups have solvable conjugacy problem (Theorem3.5.5). Juhasz [J 2] has described that all one-relator groups have solvable conjugacy problem, however a complete proof is not, yet available.

8.5 DECISION PROBLEMSIN ONE-RELATORPRODUCTS His methods -use an extension of small cancellation W(6) diagrams (see sections 3.8 and 5.3).

249

theory, the theory of

(3) Various special cases of the isomorphism problem have been settled. Pride [P 4] proved that the isomorphism problem for two-generator, one-relator groups is solvable (Theorem 3.2.7)¯ Rosenberger [R 21], using Nielsen reduction methods, showed that the isomorphism problem for cyclically pinched one-relator groups is solvable. Using essentially the same techniques, together with the Whitehead algorithm, Fine, Rosenberger and Stille [F-R-S 3], solved a restricted version of the isomorphismproblem for a class of para-free one-relator groups introduced by G. Baumslag. This was discussed in Chapter 3 -Thcorcm 3¯6¯5¯ More generally it has been shown that a one-relator group with torsion is word-hyperbolic (see Chapter 3). Sela [Se] has proved that there is positive solution of the isomorphism problem for torsion-free hyperbolic groups. In this section we extend these results to groups of F-type. The first result due to F.Tang [Tan 2] is that groups of F-type have solvable GWP. THEOREM 8.5.2. Let G --< al,. .... ,an; a~1 ......... a~~ = 1, (UV)m = 1 > where n >_ 2, e~ = 0 or e~ >_ 2, 1 <_ p <_ n - 1, U = U(al,..,ap) a cyclically reduced word in the free product on al, ..., ap o[ infinite order and V = V(ap+l, ..., a,~) is a cyclically reduced word in the ~ree product on av+l, ..., an o[ infinite order and m k 1. Then G has solvable GWP. In particular a group o[ F-type has solvable GWP. H m k 2 it is only necessary for one o[ either U or V to have infinite order [or G to have solvable GWP. PROOF.Suppose first G = G1 *A G2 where

that m = 1, then U and V have infinite

Gl=
ap+l, .....

,ap;a~ ~ .........

¯ ep+~ ,amap+l = .......

order and

@v =1 > = ae.n = 1 > and

A =< U >=< V-1 >. Let Wbe a word in al,...,a=. Then W = x~y~x2y2...xky~ where xi are words in a~, ..., av and yi are words in a~+~, ..., a=. Since G~ and G2 are ~ products of cyclics, there is an algorithm to deter~ne whether x~ ~< U > and y~ ~< V > for i = 1, ...,k. Th~ there exists an algorithm to express Was a reduced word in G~ *A G2 ~OIIl which it follows that G has solvable GWP.

250

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Now suppose that m > 1 and that either U or V has i~dinite order. Suppose without loss of gnerality that U has infinite order. Let a = UV so that U = aV-1. Then G = G~ *A G2 where G1--4 al, ..... , ap; a~1 ......... G~ --< ap+l , ..... , a,~, a;ap+l ~P+I - ¯- ......

a~p = 1 >

-- a~~ -- a "~ -- 1 > and

A =< U >=< aV-1 >. Then as in the case where m = 1, G has solvable GWP. Wemention also that in [Tan 2] Tang proved a spelling theorem for torsion one-relator products of finite direct products of free, groups which extended the spelling theorem of B.B. Newman. Tang also studied the concept of Magnus relators. These are defined in the following manner. If Wis a word in the generators X, we use the notation {W}to stand for the elements of X involved in W. Suppose Go =< X; R > where X is a set of generators and R a set of relators in X and suppose S is a cyclically rcduccd word in thc frcc group on X. Thcn S is a Magnus relator for G =< X;R,S > if for every subset J of X with J ~ {S} ~ {S} the subgroup < .1 > of G is isomorphic to the subgroup < ,1 > of Go under the canonical homomorphismof Go to G. The classical Freiheitssatz says that the relator is a Magnusrelator for a one-relator group. The Freiheitssatz Ibr one-relator products of cyclics G--
,a,~;a~~= .......

"~=1> -a,~ =R

where m _> 2 says that the relator Rm is a Magnus relator. Tang proved the following interesting ~reiheitssatz for one-relator products of one-relator groups. THEOREM 8.5.3. groups and R is a {R} C~ {R~} ~ @,i product, G = (A *

HA =< X; R~ > and B =< Y; R2 > are one-relator cyclically reduced word on X [A Y such tha~ {R~} ~ = 1,2 then R is a Magmasrelator for the on,relator B)/N(R).

We now turn to the conjugacy problem and the power conjugacy problem. Because of the structure as a non-trivial free product with amalgamation both have positive solutions. THEOREM 8.5.4. Groups of_bU~ype have bo~h solvable conjug’acy prob1era and solvable power conjug’acy problem. PROOF.Groups of F-type are non-trivial free products with amalgamation whose factors are free product of cyclics. The conjugacy problem and

8.5 DECISION PROBLEMS IN ONE-I~ELATOR PI~ODUCTS

251

PCPare solvable in free products of cyclics in much the same way as they are in free groups.. For example suppose G = A * B with A =< a; ap = 1 > and B =< b; bq = 1 >. If W = aPlbal...aP~b q~ is a cyclically reduced word in A ¯ B then W1"-, W if and only if W1is a cyclic permutation of W modulo the powers. The following result in Magnus, Karrass and Solitar [M-K-S Theorem 4.6] rednces the conjugacy problem to that of the factors and amalgamated subgroup. Theorem8.5.4 then follows from this result. LEMMA 8.5.1. Let G = A*H B. Then every element of G is conjugate to a cyclically reduced element of G. Moreover, suppose that g is a cyclically reduced element of G. Then: (i) If g is conjugate to an element h ~ H, then g is in some factor and there is a sequence h,h~, ...,h~,g where h~ ~_ H and consecutive terms in f, he sequenceare conjugatein a fact, or. (ii) If g is conjugate to an element g~ in some factor, but not in a conjugate of H, then g and g~ are in the same [actor and are conjugate in that factor. (iii) If g is conjugate to an element pi...p~ where r >_ 2 and p~,p~+~as well as pi,p~ are in distinct factors, then g can be obtained by cyclically permutingpi,...,p~ and then conjugatingby an element ot" H. Another solution to the conjugacy problem was provided by techniques of Juhasz and Rosenberger [J-R] whose method of proof led to the fact that groups of F-type satis .fy certain hyperbolicity properties. Juhasz and Roscnbcrgcr’s proof is b~cd on a dctailcd analysis of thc structurc of thc involved Van Kampendiagrams. Juhasz and Rosenberger’s proof has the following interesting consequence. THEOREM 8.5.5. Let G be a group ofF-type. Assume ~hat at least one of U or V is neither a proper power nor a product of two elements of order 2 in the free product on the generators they involve. Then G is hyperbolic. In fact [J-R] contains deeper results on the combinatorial curvature of groups of F-type and related one-relator products. The above theorem can be considered as a special case of a muchmore general results (see.. section 2.7). Bestvina and Feign [Be-F] have shown that an amalgam of two hyperbolic groups over a cyclic subgroup is still hyperbolic, while Kharlamapovich and Myasnikov [Kh-M 3] have a more general result that the amalgam of two hyperbolic groups is again hyperbolic whenever one of the amalgamated subgroups is quasiconvex and malnormal in its respective factor. Thcsc rcsults can bc wcrc summarizcd in Thcorcm 2.7.6 which wc restate here.

252

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

THEOREM 8.5.6. (Theorem 2.7.6) IT H1,H2 are hyperbolic and H1 A H~ = H is a qu~iconvex s~ibgroup, malnormal in either H1 or H2, then ~,he amalgamatedproduct HI *H H2 is hyperbolic. In par~,ic~lax iT W~, W2 are elements oT infinite order in HI and H2 respectively and neither is a proper power then HI *wl=w2H2 is hyperbolic. Since tinitely generated subgroups of free groups axe quasiconvex it follows that i~ A and B are free groups and A ~ B = H is a t~nitely generated subgroup malnormMin either A or B then the amalgamated product A *H B is hyperbolic.

CHAPTER IX RELATED

GENERALIZATIONS

9.1 Related Generalizations

OF DISCRETE

of Discrete

GROUPS

Groups

Wenowconsider two new general classes of groups which are also natural extensions, via presentations, of discrete groups. Further these classes can be studied using the same techniques as in handling one-relator products of cyclics. First we consider a class of groups which arise in the study of nonEuclidean crystallographic groups (NECgroups) {See Chapter 4}. Wecall the members of this class groups of SN-type for Special NEe-type. A precise formulation will be given in the next section but, concisely, they can be described as iterated amalgamsof one-relator products of cyclics. The name arises since they have presentations which are similar in form to those of NECgroups. Weshow that a natural Freheitssatz exists for this class and that these groups admit essential representations into PSL2(C). As a consequence of these representations we obtain results on linearity properties for groups of SN-type analogous to those on one-relator products of cyclics. In particular we show that any group in this class is either SQuniversal or infinite solvable. The class of groups of SN-type contains as a subclass a collection of special NECgroups studied by Zieschang and Kaufman [Z-K]. Our techniques allow us to generalize a result on the ranks of these groups. Second we introduce and study a related class of groups, called generalized tetrahedron groups. The terminology was introduced by Vinberg, who studied them independently IV1. These groups generalize the class of ordinary tetrahedron groups discussed by Coxeter (see [Co - M]). The ordinary tetrahedron groups arise in 3-dimensional geometry in an analogous manner to the way the ordinary triangle groups were constructed in 2-dimensional geometry. The ge~mralized tetrahedron groups will be formally defined in section 9.3 but concisely they can be described as triangular products of generalized triangle groups. As for the class of Special NEC groups, we will prove that each generalized tetrahedron group admits an essential representation into PSL2 (C), and from this deduce many linearity properties. In addition we will use the concept of a Gersten-Stallings angle to prove a version of the Freiheitssatz. In particular we give sufficient 253

254

ALGEBRAIC

GENERALIZATIONS

OF DISCRETE

GROUPS

conditions for the generalized triangle group factors to inject into the group (see section 9.3.3.) 9.2 Groups

of Special

NEC Type

Recall that a non-Euclidean crystallographic group or NEC group, G is any discontinuous group of isometrics of the hyperbolic plane (see section 4.4). The class of NECgroups contains the Fuchsian subgroups as a subclass. Recall further that the geometric analysis of the structure of NECgroups follows a similar pattern to that of the Fuchsian groups. Using this geometric analysis it can be proved that a finitely generated NEC group G has a presentation of the following form (Theorem 4.4.1) (a) Generators: X ~2 J where X and J arc disjoint sets and J itself the disjoint union of ordered sets Jj = (Xl,~l, ..., xj,~), nj _~ (b) Relators: R0 U R1 t.J R2, the union of three sets of relators

2 . z~,h J} (i) R0= { xj,h,

(ii) R~is the union of sets R~,j = (tj,~ , ...,tj,n~ ) wherem~,h are positive integers and tj,h -= Xj,haj,hXj,h+laj--,~ (indexing h cyclically modulo rtj) with aj,h words containing only generators from X. (iii) R2 is a set of wordsof the form km~ where mk are positive integers and sk are words containing only generators from X Finally in the set of all aj,h together with all s~ each generator from X occurs exactly twice. This presentation of G can be deduced from the structure of a surface B. In the presentation, if the surface is orientable we can think of the X generators as the Fuchsian group part and the J generators as reflections. If the surface is orientable and there are no reflections this reduces to the structure obtained via the Poincare presentation. Wenow consider groups which have presentations similar in form to certain of those above. In particular we define a group of SN-type or group of Special NEC-type to be a group having a presentation of the form (9.2.1) G=<

a~,...,a~,b~,...,bk,%

............

Rk

=

1

>

wheren_> 1, k_> 2, e~ -- 0 or ej _> 2 for i = l, .., n, fi=0orfi_>2fori= 1, .., k, m~>_ 2 for i --- 1, .., k andfor eachi = 1, ..., k, Ri = Ri(al, .., a,~, bi) is a cyclically reduced word in the free product on a~, .., a~, bi, which involves b~ and at least one of the a~s and which is not a proper power. In this case we call the Ri the relators. Using our previons terminology we recall that a representation p of a group of SN-type is essential if p(ai) has infinite order if ei = 0 or exact

9.2.1 ESSENTIALREPRESENTATIONSFOR GROUPS OF SN-TYPE 255 order e~ if e~ >_ 2 for i = 1,..,n,p(b~) has infinite order if f~ = 0 or exact order f~ if f~ > 2 for i = 1, .., k and p(R~) has exact order m~for i = 1, .., k. In the next section we show that each group of SN-type admits an essential representation in PSL2(C) and satisfies a FHS. From this follow many of the linearity properties of one-relator products of cyclics. 9.2.1

Essential

Representations

for Groups of SN-Type

Wefirst establish the FHSand the existence of essential representations for groups of SNtype. THEOREM 9.2.1. Le~ G be a group of SN-type, so that G has a presentation of the form 9.2.1

wheren> 1, k > 2,ej =Oorej >_2fori= l,..,n, fi=Oor fi > 2fori= 1,..,k, mi > 2 for i -= 1,..,k and for each i -= 1,...,k , R~= R~(al, ..,a,~,b~) is a cyclically reduced word in the free product on al, .., am, b~ which involves bi, no other b~, and at 1cast one of ~he a~s and which is not a proper power. Then (1) < al, ...,an >c-~< a];a~’ > *...* < a,,;a~c > (2) For any subset {bi~, ..., b~, } we have < al,..,an,bi~,...,bi, ~ < ai,.., a=,b~,.., bit, a~

a~ ~ ~.q b

(3) If each Ri involves a// a~, ...,a,, {a~,..., a~, } of {al, ..., a,~}

= >G b( ’~ q= RT

R.m~* -= 1 >

then for any proper subset

< a~,..,a~t,b~ >~= *...* < a~t;a~’~,* > * < b~;b~~ > (4) There exists an essential representation of G in PSL2 (C) which is faithful on < al, ...,a,~ >~ PROOF.Wesuppose first form G=< al,..,a,~,b~,b2;a~

that

~ ..

k = 2. Therefore the group G has the

a~~ =b(’ =b2/~ = R7~ =/L~ ~ = 1 >

wheree~, f~ are as before, m~, m2> 2 and R~= R~(a~,.., a,~, b~), i = 1,

256

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Let a 1 =~(

al,..

,

an,

bl;

a~ 1 ..

a~e~ = b{1 = R~TM = 1 > and

an, b2; a~~ .. a~~ = b2/~ = R~2 = 1 > .

G2 =< al, ..,

G1 and G2 are one-relator products of cyclics. Since both ~’?.1 and m~are at least 2 we can apply the FHSfor one-relator products of cyclics (Theorem 6.3.2) to say that < al, ...,an >~=< al, ...,an >a2=< a~, ..,an;a~ ~ .... a~~ = 1 >. It follows that identifying {a~,...,an} in G~ with {a~,...,an} in G2 gives a subgroup isomorphism and therefore G is the free product of G~ with G~ amalgamated over < a~, ..., a~ >. Prom the structure theorems on free products with amalgamation, G~, G2 and < a~, ..., a~ > inject into G. If {ai~,.., ai~} is any proper subset of {al, ..., an} then {ail,.. , ai~, bi} generate a subgroup of Gi. If Ri involves all al, ..., an then again from the FHSfor one-relator products of cyclics this must be the free product. This establishes (1),(2) and (3) for the case If k = 3 then G has the form at, .., an, bl, b2; ael~ .... an ~" = b{~ = b2h = ba/~= R~= R~~ = R~a =: l’- Using the result for k = 2 and the FHSfor one-relator products of cylcics G must now be the amalgamated free product of al =< al, ..,

an, bl, b2; a~~ .. a~’~ = b{~ = b~~ = R~1 = R~’~ = 1 > and an, b3; a~~ .. a~’~ = ba~ = R~~ = 1 >.

C~2 =< al,..,

amalgamated over < al,..., an >. Parts (1),(2) and (3) of the theorem follow as before. For general k the first three parts now follow easily by induction. Wenow establish the existence of an essential representation. Again first suppose that k = 2. Let G, G~ and G~ be as before with k = 2. From Theorem6.3.1 there exists an essential representation Pl of which is faithful on < al,...,an >. Let A~,..,A~, B~ be the images of a~, .., an, b~ under p~. Let B2 = ~=

(t,

* 2cos(~r/f~) -

B2 = :t:

2 - t

) iff2>2and -

9.2.1

ESSENTIALREPRESENTATIONS FOR GROUPSOF SN-TYPE 257

with t a variable to be determined. Since tr(B2) = 2cos(Tr/f2) if f2 _> 2 we then have B2/2 = 1 for every choice of t. Weshow there exists a choice of t so that the image of R2 has order m2giving the essential representation. SupposeR2 = Wlb~1 w2 ..... b~~ whereWl, w2,. .... are non-trivial words in ~ where Wi are the corresponding imel, ...,an. Let R-~2 = W1B~IW2...B~2 ages of wl, w2, ..., wn under the representation pl. Since Pl is faithful on < al, ...,an > it follows that W1,W2,... are non-identity projective matrices. Therefore the same arguments as in the proof of Theorem 6.3.1 and Theorem6.3.2 can be used to assert that there exists a choice {perhaps by conjugating the original A~, ..., Anif necessary} so that (d~ d3 d2) W1B2klW2...B 2k~= -td4 where dl + da is a non-constant polynomial in t. From the fundamental theorem of algebra there is a choice tl so that (dl + da)(t~) = 2cos(tim2) With this t~ in B2 we then have tr(R--~) 2cos(tim2) and he nce ~2 2 = 1 Let p be the representation which takes a~,...,an,bl,b2 to A1, ..., An, B1, B2(t~) {or whichever conjugates of A1, ..., A,~, B1 were necessary so that the trace polynoinial was non-constant} respectively. This representation p is then faithful on < a~, ..., an > and from the trace conditions p(b~){= B1},p(b2){= B2},p(R~) and p(R~) have exactly the correct orders - that is p(bi) has infinite order if f~ = 0 or order fi if fi _> 2 and p(Ri) has exact order m~ for i = 1, 2. Therefore the representation p is essential. Once the essential representation for the case k = 2 is established the same argument can be iterated for the general case. This completes the theorem. Theorem 6.3.1 proved that if A and B are groups which admit faithful representations in PSL~(C) and R is a non-trivial word of syllable length at least 2 in A* B then the one-relator product G = (A * B)/N(R’*) with m _> 2 admits a representation p into PSL2(C) such that PlA is faithful, PlB is faithful and p(R) has order m. Using this result and an analagous proof to that of Theorem9.2.1 we obtain: THEOREM 9.2.2. Let Go =< A; re/A > and, /:or Bi; rel B~ >. Suppose that G =< A, B~, ..,

Ba; re/A, re/B1, ..,

i=l,..,k,

re/B~, R~~ .....

let G~ =< R~~ = 1 >

wherek > 2 and, for each i = 1,..,k, R~ = R~(A, Bi) is a non-~rivial cyclically reduced word in the free product A * B~ of syllable length at least two and

258

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

each m~ >_ 2. Assume further that each G~ [or i = 0,1,..,k representation into PSL2(C). Then

admits a faithful

(1) Each G~ [or i = 0,1,...,k injects in G (2) For each subset {il, .., it} o[1,2, .... ,k we have >G=< i~ A,B~, Bi,; relA,..,

< A, Bi~ ..,B~,

tel

RT~ = 1 >

Bi,,RT

(3) There exists a representation p of G into PSL2(C) such that is faithful for i = 0,1..,k and p(R~) has order m~for i = 1,2,..,k. Wenote that the conclusions of Theorem9.2.1 are also valid for certain modifications of the presentation type (2.2). For example the same proof can be used for groups with presentations of the form G =< a~,

..,

a,~+~;

~~ R~ ..... a~n~+~ .. = ~+~

Ra = 1 >

where n _> 1, k _> 2, e~- = 0orej _> 2 for i = 1,...,n+k, mi _> 2 for i = 1,2,..,k and for each i = 1, ..,k, R~= R~(a~,...,a~+,~) is a non-trivial, cyclically reduced word in the flee product on ai, ..., ai+,~ which involves all ai ~ ... ~ ai+,~. In general it is not clear what happens relative to essential representations for general multi-relator groups. Below we present an example of a group G very similar in form to a group of SN-type but which has no essential representation in PSL~(C). Let G =< x,y,z;x 4 = y3 = z ~ = (xy[x,y])~ = (x3z(x[x,y])2z-~)5 1 >. Suppose X, Y, Z are the images of x, y, z under a supposedly essential representation in PSL~(C). Then without loss of generality let tr(X) a = v/~, since x4 = 1, and tr(Y) = b = 1 since y3 =1. Suppose tr(XY) then from a direct computation we have trX[X, Y] = a(a~+ b2+ 2- abe-3) and trZY[X, Y] = c(a~ +b2 +c~ -abc-3). Since (XY[X, y])2 = 1 we have, if the representation is essential, that tr(XY[X, Y]) = 0. Therefore either c = 0 or a~ + b2 + c ~ - abc- 3 = 0. Howeverif c = 0 since a = v~ and b = 1 we have that a2 + b~ + c ~ -abc - 3 = 0 also. Therefore tr(X[X, Y]) = and so (X[X, y])2 = 1. However then from the second relation it would follow that X~s = 1 which coupled with X4 = 1 gives X = 1 and hence the representation is not essential. 9.2.2

Linearity

Results

for Special

NECGroups

In this section we show that groups of SN-type satisfy muchthe same linearity properties as one-relator products of cyclics. Specifically we examine the Tits Alternative, SQ-universality and virtual torsion-freeness.

9.2.2

LINEARITYRESULTSFOR SPECIAL NEC GROUPS

THEOREM 9.2.3. Let G be a group of SN-type. universal or G is infinite solvable.

Then either

259

G is SQ-

PROOF.For a group of SN-type having form (9.2.1) let N = n ÷ k. first prove the result for k - 2 so that N = n ÷ 2. The result for general k then follows in a straightforward manner. Thus we suppose that our group G has the form G=< ax,..,an,

bl,b2; a~1 ....

a~’

:bl

fl

:b~

f2

:

R m -~-

S t:

1

>

where n >_ 1, ej = 0 or ej _> 2 for i = 1,...,n, fi = 0 or fi _> 2 for i = 1, 2, m_> 2, t _> 2 and R = R(al, .., am bl) is a cyclically reduced word in the free product on al, .., a,~, bl which involes b~ and at least one of the ai and S = S(al, .., am, b2) is a cyclically reduced word in the free product on al, .., an, b2 whichinvolves b2 and at least one of the ai. Since, if a quotient of a group is SQ-universal, the group is itself SQuniversal we can, without loss of generality, assume that ej ~> 2 for i = 1, ...,n and fi _> 2 for i = 1, 2, passing to a quotient if necessary. For N _> 5 or N = 4 and (el,e2,f~,f~,m,t) (2 ,2,2,2,2) th e pr oof follows the outline of the proof of Theorem 6.2.3. From Theorem 6.3.2, G adlnits an essential representation p into PSL~(C) which is faithful on the free product on al, .., a,~. Therefore from Selberg’s theorem on fi~fitely generated subgroups of linear groups there exists a normal subgroup H of finite index in p(G) such that p(ai) has order ei moduloH, p(bi) has order fi modulo H, p(R) has order m modulo H and p(S) has order t modulo H. Thus the composition of maps {where 77 is the canonical map} G -~ p(a) -~ p(G)/H gives an essential representation ¢ of G onto a finite group. ~ .. a~~ =bll ~ =b~h= 1 > be the free Let X=< a~,...,a,~,bt,bu;a~ product on {a~, .., a,~, bl, b2} ¯ There is a canonical epimorphismf/: X -~ G. Wetherefore have the sequence X --~ G ~ p(G)/H Let Y = ker(¢o/3). Then Y is a normal subgroup of finite index in X and Y is torsion-free. Since X is a free product of cyclics and Y is torsion-free, Y is a free group of finite rank r. Suppose [X : YI = J. Regard X as a Fuchsian group with finite hyperbolic area p(X). {Every fintely generated free product of cyclics which is not i~ffinite dihedral can be faithfully represented as such a Fuchsian group}. From the Riemann- Hurwitz formula : j#(X)

= #(Y)

260

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

where #(Y) = r and #(X)

= N- 1 - (1 +.. el

+ __ en

),

we obtain 1 1 r=l-j(~+..+--+en~7~2+

1

1 -N+I)

G is obtained from X by adjoining the relations R’* and St and so G = X/K where K is the normal closure of R"~ and St. Since K is contained in Y the quotient Y/K can be considered as a subgroup of finite index in G. Applying the Reidemeister- Schreier process Y/K can be defined on r generators subject to (j/m) + (j/t) relations. The deficiency d of this presentation for Y/K is then d=r

J j m t

_l_j(1

1 1 1 ~+"+--+~en -f~ -f~2

1 1 m t+ +--+--N+I)

If N_>5, or N = 4 and (el,e2, fl,f.2, m,t) ~ ( 2,2,2,2,2,2), t hen d> 2. From the work of B. Baumslag and S. Pride [B.B. - P 2] Y/K and hence G has a subgroup of finite index mapping onto a free group of rank 2. Therefore Y/K is SQ-universal and since this has finite index in G, G is also SQ-universal. Therefore if N _> 5, or N = 4 and (el, e2, f], f2, m, t) (2, 2, 2, 2, 2, 2) then G is SQ-universal. Next suppose that N = 4 and (e~,e2,f~,fu,m,t) = (2,2,2,2,2,2), now G has the presentation (7=< al,a2,

bl,b2;a~

=a~2 =b~ =b~ =R2---o¢2=

1>

where R = R(a~, a2, b~) and S = S(al, a2, b2). We may assume that R and S are not proper powers since otherwise it reverts back to the previous arguments where one of the exponents is not 2. Without loss of generality we must consider the following four cases: (1) R al b~ and S = a~b~ (2) R al b] and S = a2b~ (3) R = alb~ and S involves both al and (4) Both R and S involve a~ and In case (1) if R = a~bl and S = a]b~ then a

=<

al, a2, b,, b2; a~ = a2~ = b~ = b~ = (a~bl) 2 = (a~b~)~ = 1 >.

9.2.2

LINEARITYRESULTSFOR SPECIAL NEC GROUPS

261

Setting al = 1, G can be mapped onto < a2,b~,b2;a~ = b~ = b~2 = 1 >= Z2 * Z2 * Z2 which is SQ-universal and therefore G is. {Recall that any non-trivial free product except the i~dinite dihedral group Z2 * Z2, is SQuniversal}. In case (2), where R al bl and S -- a2b2 the n by set ting bl -- 1, G can be mapped onto < al, a2, b2; a~ = a~ = b~ = (a2b2) ~ = 1 >= Z2 * D: {where D~ is the Klein 4-group }. Since this is a non-trivial free product and not infinite dihedral, G is SQ-universal. In case (3) where R = alb~ and S involves both al and a2 set al = 1, to obtain as an image of G, < a~,51,52;a~ = 55 = b2~ = (a2b~) ~w = 1 >. This is a non-trivial free product Z~ * T where T has the presentation < a2, b2; a~ = b~ = (a2b~)2~’ -- 1 >. For no value of w is this cyclic of order 2 and therefore Zu * T is SQ-universal, and hence G is. Finally suppose case (4) where both R and S involve a~ and a: . Let be the subgroup of G generated by a = a~a.2, b = alb~ and c = a~b2. Then H has index 2 in G and there are three, possibilities for R and S relative to H. (1) Both R and S are elements of (2) R is an element of H but S is not {or vice versa}. (3) Both R and S are not elements of If both R and S are elements of H then R = T(a, b) where T(a, b) is a freely reduced word in the free group on a, b and S = U(a, c) where U(a, c) is a freely reduced word in the free group on a, c. Then H has a presentation H =< a,b,c;T2(a,b)

= T2(a -~,b-~)

= U~(a,c) = U~(a-1,c-~)

=1 >.

Using the arguments in [L-S pg. 293] we may assume that one of the generators a, b has exponent sum zero in T(a, b). If b has exponent sum zero, let a = 1 to obtain the quotient < b, c; c 2w = 1 >= Z*Z2~if w >_ 1 or Z*Zif w = 0. Either is SQ-universal and therefore G is. Ira has exponent sum zero let b = 1 to obtain the quotient < a, c; UU(a,c) a theorem of M. Edjvet [E 3] on groups which have balanced presentations, this quotient has a subgroup of finite index mapping onto a free group of rank 2, and is therefore SQ-universal. Thus G is SQ-universal. This completes the situation where both R and S are elements of H. Nowsuppose that R is an element of H but S is not. Suppose R -- T(a, b) as above then S = a~U(a, c) where U(a, c) is a freely reduced word in the free group on {a, c}. The subgroup H now has the presentation H----< a,b,c;T2(a,b)

= T2(a-~,b-~) ---- U(a,c)U(a-~,c-~) = 1 > .

Let a = 1 to obtain the quotient < b, c; b2w ~- 1 > = Z * Z,~, which as above is SQ-universal and therefore G is.

262

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Finally suppose both R and S ~re not in H. Then R = alT(a, c), S alU(a, c) and H has the presentation H = . Letting a = 1 we obtain the quotient < b, c; >-- Z*Zwhich is SQ-universal, and therefore G is. This completes the situation when N = 4 and all exponents are 2. Wemust now consider the case when N = 3. Thus G has the presentation G =< al,bl,b2;a~ ~l=b~1=b~22=Rm=St = 1> . From the proof given above for N >_ 4 a group with a presentation as above is SQ-universal if the deficiency d of the analogously defined subgroup Y/K is 2 or greater, or equivalently, if (1/e~) + (1/f~) + (1/f2) + (l/m) (l/t) < 2. By a lengthy argument on the possible values of (el, f~, f2, m, the SQ-universality or not of groups with presentations of form (2.4) can established directly. Howeverthis can be simplified using a result of Lossov [Los]. He proved that if G ---- A *H B with IHI < ~, IA : HI _> 2 and IB : HI _> 3 then G is SQ-universal. In (2.4) we have G = G1 *H G2 where ~1 ~___b]fI = .~D~ G1 --< al,bl, . a~ = 1 >,G2 =< 1a~,b~;a~ =bf22 = st = 1 > ~ and H =< a~;a~ = 1 > . Since we can assume that el ~ 2, H is finite cyclic and Lossov’s theorem applies unless both G1 and Gu are finite with IG~ : HI = 2 and IG~ : HI = 2. This would imply that both G~ and G~ are finite dihedral and G would have the presentation G 1 =< al, b~, b2; a ~ = b~ = b~ = (a~bl) =~

(alb2)

2-- 1

>.

This has the structure of a free product of two isomorphic finite dihedral groups amalgamated over their cyclic subgroup of index 2. This group is infinite solvable. Therefore if N = 3 either G is SQ-universal or G is infinite solvable. This completes the proof for k = 2. If k > 2 using the same argument as before, employing an essential representation of G onto a finite group we obtain an analogously defined subgroup Y/K of finite index in G having a presentation with deficiency d given by 1 1 +..+--+ J(e, e~

1 -~1

+..+

1 1 +--+..+--~k

1

(n+k)+l)

ml

1 1 1 =j [ ( n +k ) - l ] - j ( ~ +. . +-- +e,~ ~ ~- " -~

1 ~ ~---1 ~- " -~ -- ) 1 mk

9.2.2

LINEARITYRESULTSFOR SPECIAL NEC GROUPS

263

As in the case when k = 2 the deficiency is 2 or greater mfless n = 1 or n = 2 together with all exponents 2. Thus G is SQ-universal except possibly in these latter two cases. If n = 1 , Lossov’s theorem applies to show that G is SQ-universal if k > 2. If n = 2 and all exponents are 2 then G has the presentation G=< al,a2,

bl,b2,...,bk;a~

=a~ =b21 .....

b~ = R~ .....

R~ = 1 >.

Setting al = a2 = i we get the quotient < bl, b2, ..., bk; b2~ .... = b2~ = 1 >. This is a free product of cyclic groups of order 2 and since k > 2 it has more than two factors and is thus SQ-universal. Therefore G is SQ-universal. This completes the proof. Notice that, if a group is SQ-universal, it must contain a free subgroup of ra~k 2. From this we obtain the Tits Alternative. COROLLARY 9.2.1. Tits Alternative.

Let G be a group of SN-type then G satisfies

the

COROLLARY 9.2.2. Let G be a group of SN-type then G is infinite solvable if and only if G has a presentation < al, bl, b2; a~1 -- b2~ -- b22 = (albl) 2 = (alb2) 2 --- 1 > . Otherwise G is SQ-universal. If G is a group of SN-type then G admits an essential representation. If we can classify the elements of finite order we can apply Theorem6.2.1 to get that G is virtually torsion-free.. THEOREM 9.2.4. Let G be a group of SN-type. Assume each relator is not a proper power and satisfies one of the following conditions:

R~

(1) R~ -- U~V~where U~ -- U~(a~,..,a~) and V~ = V~(a~+~,..,a,,b~), k _< n are non-trivial words in the free product on ai, .., a~ and a~+l,.., a,~ bi respectively. (2) R~ is not conjugate in the free product on a~,..,a,,b~ to a word of the form XY where X, Y are elements of orders p >_ 2, q >_ 2 respectively with (1/m~) + (l/p) + (I/q) and m~ >_ 4. (3) R~ is arbitrary but e~ = 0 for i = 1,...,n, f~= 0 for i = 1,...,k. Then G is virtually torsion-free. PROOF.Let H be a one-relator

product of cyclics with presentation

H =< al,..,a~,b~,a~. ~ ....

a~~ = b~

=1 >

withn_> 1, ei =0or e~_> 2 for i = l, .., n, fj =0or fj _> 2 forj = 1 or 2. If m _> 2 and any one of the three conditions in the statement of the theorem is satisfied then any element of finite order in H is conjugate to a

264

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

power of one of the generators ai or a power of the relator R (see Chapter 7). Nowlet G be a group of SN-type. Wemay assume that k = 2. From the proof of Theorem 9.2.1, G is an amalgamated fxee product of one-relator products of cyclics. Hence, from the torsion theorem for such amalgams any element of finite order in G is conjugate to an element of finite order in one of the factors. Combiningthis with tim statement above on one-relator products of cyclics we have that if each relator Ri satisfies one of the stated conditions then any element of finite order in G is conjugate to a power of a generator al, ...am bl, b2 or a power of a relator R1, R2. Hence an essential representation must be essentially faithful and applying Theorem6.2¯ 1 we get that G is virtually torsion-free. COROLLARY 9.2.3. Let G be a group of SN-type satisfying the conditions o[ theorem 9.2.4 then the conjugacy classes of torsion elements are precisely given by the powers of the generators and the powers of the relators. 9.2.3

Rank Conditions

for Certain

NEC Groups

Zieschang and Kaufman[Z-K] have recently considered the rank { minimal number of necessary generators } of groups with presentations of the form These groups are contained in the class of groups of SN-type and are NEC groups with reflections. Zieschang and Kaufmanproved that if 2 < h < k, then G has rank 3 if h > 2 or h = 2 and k is even. Otherwise G has rank 2. Wenow give some extensions of this. THEOREM9.2.5. Let G (aiaa) k = 1 > with ei = 0 or ei >_ 2 for i = 1,2,3, 2 ~_ h ~_ k. Then 2 <_ rankG <_ 3 and rank G = 2 if and only if G has the presentation < a,b,c;a p = b2 = c 2 = (ab) 2 = (ac) k’ = 1 > withp= 0 orp >_ 2 and k >_3, k odd. { In the case where G has rank 2, G, as above, is generated by x = a and y = cb and G is an epiinorplfic image of the Fuchsian group F k 81,82, 83, 84; 812 = 822 ~--- 832 : 84 = 81S28384 = 1 > with k _> 3 and k odd.} PROOF.Let G be as above. If el = e~ = e3 = 2 then the result follows from Zieschang-Kaufman. Assume that e; where al ¯ ~ = a~~ = (a~a2) h 1 >,G~ =< a~,a3;a~ ~ ~ = a~ ~ (a~aa) = i > and H =< a~; a~~ = 1 > . Clearly 2 _< rank G _< 3. Assume G =< z, y > so that rank G = 2. Weuse Nielsen reduction on {z, B}. In Chapter 2 we stated the following result which is crucial in the proof here.

9.2.3

RANKCONDITIONSFOR CERTAINNEC GROUPS

THEOREM 9.2.6. Let G = H1 *A H2. H {x 1,..., x,~} is a finite elements in G then there is a Nielsen transformation from {Z1, ..., system {Yl,-.., Ym}for which one of ~he following c~es hold:

(i) (ii)

265 system of Zm} tO a

= 1 .. ., Eac~ w ~< y~,...,y~

> ca~ be wri$~e~ as w = ~=~y~,~ =

"~ a ¢ 1, withy~ ~ A(i 1,..,q) and Thereisaproduc~a =~a~=~y,~, in one of the factors Hi ~here is an elemen~ x ~ A wi~h x-~ax e A; (iv) There is a g e G such Chat for some i e {1, ...,m} we have y~ ~ gAg-~, bu~ for n suitable natural number k we have y~ e gAg-~; (v) Of $he y~ ~here are p ~ 1 contained in a subgroup of G conjugate ~o H1 or H~ and a certain produc~ of them is conjugate ~o a non-~rivial elemen~ of A. The Nielsen trnnsformn$ion can be chosen so that {y~, ...,y~} is shorter (wi~h respec$ ~o ~he len~h and a suitable order) Shan {Xl, ..., x,~} or the len~hs of the elements of {x~, ..., x~ } are preserved. ~r~herif {x ~, ..., x~ } is a generating system of G $hen in case (v) we find p ~ 2 for in ~his case conjugations determine a Nielsen $ransformn~ion. g we are interesSed in $he combinn$orial description of < x~, ..., x~ > in $erms of genera$ors and relations we find again $haf p ~ 2 in ease (v), possibly a~er suieable conjugations. (iii)

Case (i) of Th~rem 9.2.6 c~nnot occur for {x,y} in G since G is noncyclic while case (ii) of Theorem 9.2.6 cannot occur for {x,y} in G for otherwise G would be a ff~ product of cyclics. Each G~ is an ordin~y triangle group, so there is no d ~ G~ ~ H with 1 ¢ dt ~ H for some t e N. Therefore case (iv) cannot occ~. If case (v) occ~s we may assume that are both in G~ or both in G2, which is a contradiction since G1 ¢ H ¢ G2. Therefore case (iii) of Theorem9.2.6 occurs. Without loss of generality assume that 1 ¢ x = a~ for some positive integer t and that there is an elemem v in G~ ~ H with v-~a~v = a~ ~ 1 for some u,w ¢ 0 and a[ ~< x >=< a~ > . Regard G~ as a subgroup of PSL2(C). Then a direct computation shows that u = -w if e~ = 0 and u ~ -w rood el if e~ ~ 2 and v2 = 1. Therefore v-~a~v = a~~ so (va~) 2 = 1. Hence e~ = h = 2 and twe may assume that x = al became G =< x, y >=< a ~,y >C< al,y > ¯ Assu~ne that either e~ = 0 and k even or both e~ and k are positive even. Adjoining the relations a~ = (a~a3) ~ = al = 1 we obtain the quotient < ae, au; a~ = a~ = 1 > which is not cyclic. It follows that ~t least one of ez or k is odd and therefore there is no relation v-~a~v = a~ ¢ 1 for some integers s, t and some v e G2 ~ H. Analogously as above if e~ ~ 2 then the gcd of e2 and k is 1. Nowwe come back to our generating system {x, y} with x = a~.

266

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Using the above arguments and Nielsen reduction we may assume that y = hla2h2a2...hma2 with m _> 1 and the hi coset representatives of H in G2. Because there is no relation v-latl v = a~ we must have m = 1 for otherwise x and y cannot generate G. Hence y = hla2 for some hi E G2\H. Further x and y can generate G only if x and yxy -~ or, equivalently, al and h 1 a ~- 1 h ~- 1 generate G2.Fromthe classification of generatingpairs of triangle groups {G2is also an ordinary triangle group} we get that one of e3 or k is 2. Therefore without loss of generality we have that e3 = 2 and k >_ 3 and hence G has a presentation < a,b,c;a p = b2 = c 2 = (ab) 2 = (ac) k = 1 > with p = 0 or p _> 2 and k >_ 3, k odd. This completes the proof when el ~2. Note that, if G has a presentation as above, let u = ab so that G also has the presentation < u,b,c;u 2 = b~ = c 2 = (ub) ~ = (ubc) t~ = 1 > . Let x = a = ub and y -- cu -- cba-1. Then [x,y] = (ubc) ~ and therefore ubc, c, u, b are in < x, y > since k is odd. Therefore {x, y} generate G. From the second presentation it is clear that G is an image of the Fuchsian group . 2 = s~ = s~ = sk~ = sls2sas4 = 1 > Fwithpresentation < sl, 2,sa, s4~s 8 1 with k odd and k _> 3. Nowsuppose that el ---- 2 and e2 _< h and ea _< k. Let {x,y} be a generating pair for G. Again we use Theorem 9.2.6 and case (iii) of that Theorem must hold. Therefore we may assume that x -- al is in A and without loss of generality there is a v E G1 \H with valv -~ -- al. Regarding G1 as a subgroup of PSL2 (C), and using direct calculations we get that e2 = 2 and h = 2r _> 2 and (ala2) ~ commutes with al. As for the case when el ~ 2 we use the classification of generating pairs for triangle groups to obtain that at least one of e3 or k is odd and the gcd of e3 and k is 1. Hence we get that e3 = 2 and k odd since, if e3 _> 3, then G2 cannot be generated by two elements of order 2 (see the arguments for the case el ~ 2. From this we must have h = 2 for, if h >_ 4 then G has rank 3 from the Zieschang-Kaufman result. This completes the proof. Using modifications of the same type of cancellation arguments as in the proof above we can give the following generalization which we just state. THEOREM 9.2.6. Le~ G =< al, a2, a3, ..., am bl, b2; a~1 = a~~ ...... ~ a~ = b{’ = bY2~ = (ala2 .... anbl)h = (ala2 .... anb2) k = 1 > with e~ = 0 or e d >_2fori--1,2,...n, fi=Oorfi>2fori=l,2andh_>2, k_>2. Then n+l_< rankG <_n+2. 9.3 The Generalized

Tetrahedron

Groups

Consider a tetrahcdron in either Euclidean, spherical or hyperbolic 3space. The reflections in the faces generate a group of isometries and we consider the subgroup consisting of the orientation preserving isometries.

9.3 THE GENERALIZEDTETRAHEDRON GROUPS

267

This subgroup is called an ordinary tetrahedron group and is the obvious analog in 3-dimensional geometry of an ordinary triangle group. In Coxeter it is shown that such a group has a presentation of the form (9.3.1)

< 9~,y,

z;

xk

= yl

= zrn

-~-

(xy-1)r

:

(yz-1)p

~ (zx-1)q

1

>

More generally we call a group an ordinary tetrahedron group if it has a presentation of form (9.3.1). Coxeter{see [Co-M] } gave necessary and sufficient conditions for an ordinary tetrahedron group to be finite. Vinberg observed that one can state these conditions as follows: A group G with a presentation of the form (9.3.1) is finite if and only the matrix 1 - cos(n/k) cos(~r/k) .

-cos(r/l) - cos(~r/r)

- cos(r/m) - cos(~r/q)

1 - ~os(r/O- cos(~-/~) - cos(~/p) cos(r/m) - cos(r/q) - cos(~r/p) has positive determinant (see [Cox 1],[Cox 2],[Cox 3]). In the analysis of the Tits Alternative for generalized triangle groups and for special NECgroups subgroups with presentations of the following form related to (9.3.1) often arose. < al,a2,aa,

a~_ = a2 = aa = R’~(al,a2)

= R~(al, aa) = R~(a2,a3)

Vinberg IV] studied these groups independently and called them generalized tetrahedron groups and we adopt this terminology. Specifically a generalized tetrahedron group is a group with a presentation of the form

(9.3.2) < al, a2, an; a~1 = a~2 = aa = R~(al, a2) = R~(a,, a3) -= Rq3(a2, a3) = 1 where ei = 0 or e~ _> 2 for i -= 1,2,3;2 < m,p,q ; R~(al,a2) is a cyclically reduced word in the free product on a~, a2 which involves both a~ and R2(a~, an) is a cyclically reduced word in the free product on a~, 33 which involves both a~ and aa and Ra(a2, a3) is a cyclically reduced word in the free product on 32, aa which involves both a~ and aa ¯ Further each Ri, i = 1, 2, 3 is not a proper power in the free product on the generators it involves. As before a representation p : G -+ Linear Group is essential if p(ai) has infinite order if ei = 0 or exact order e~ if e~ >_ 2 for i -= 1, 2, 3, has order m, p(R2) has order p, and p(R3) has order q. In particular the

268

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

existence of any essential representation implies that the group G is nontrivial. Tsaranov ITs 1],ITs 2] considered the special generalized tetrahedron groups where each Ri(u, v) is a word of the form u’~v b and gave a complete classification of the finite groups in this case, extending the results of Coxeter. Recall that a group G is a polygonal product (see Chapter 2) if it can be described in the following manner. There is a polygon P. Each vertex v corresponds to a group G.. Each edge y corresponds to a group Gy and adjacent vertices are amalgamated via relations along the edges. The group G is then the group formed by the free product of the G. modulo the amalgamating edge relations. The groups G. are called the vertex groups. As mentioned in Chapter 2, Karrass, Pietrowski and Solitar have developed a subgroup theory of polygonal products which is parallel to the theory for free products with amalgamation. If the polygon has four or more sides then the group G decomposes as a free product with amalgamation and it follows that each of the vertex groups inject into the group G. In the language introduced in Chapter 3 we say that the vertex groups are FHSfactors whenever the polygon has four or more sides. If the polygon is a triangle it is called a triangular product. Amongall products {see Chapter 2} the triangular products are the least well-behaved and the question arises as to when the vertex groups inject. For a generalized tetrahedron group G with presentation (9.3.2) let G1 ----< al, a2; a~1 = a~2 -= R~(al, a2) = 1 ¯ el .= a~ .= R~(a~, a3) = 1 G2 1 ~< al~a3~ a G~ --< a2, a3; a~2 = a~a = R~(a2, a3) -- 1 > Each of these groups is a generalized triangle group and the generalized tetrahedron group G is then a triangular product of these groups with edge amalgamations over the cyclic subgroups < a~ >, < a2 >, < a3 >. As mentioned earlier it is not clear when the vertex groups in a triangular product inject into the group. In section 9.3.5 we give conditions for this to occur relative to the generalized tetrahedron case. Wecall G~, G2, G3 the generalized triangle group factors of the generalized tetrahedron group G. As in the analysis of one-relator products of cyclics and groups of SN-type we show first that generalized tetrahedron groups admit essential representations into PSL2(C). In particular this shows that these groups are never trivial. Using the existence of essential representations we give sufficient

9.3.1

ESSENTIALREPRESENTATIONS

269

conditions for both the SQ-universality of generalized tetrahedron groups and the existence of non-abelian free subgroups. Using techniques of Gertsen and Stallings we also give conditions for the generalized triangle group factors to inject. 9.3.1

Essential

Representations

Wefirst prove the existence of essential that these groups are non-trivial.

representations

and thus show

THEOREM 9.3.1. Let G be a generalized tetrahedron group with presentation as in (9.3.2). Then G admits an essential representation into PSL2(C). PROOF.Let G1,G2,G3 be as given above relative to G. Then G1 --< al,a2, at = a2 = R~n(al,a2) = 1 > is a generalized triangle group. Choose an essential representation of G1 in PSL2(C) which is possible from Theorem6.3.2 a.nd let At, A2 be the respective images of at, a2 ¯ If A,B are projective matrices in PSL2(C) and RI(A,B) q~ with all Pi¢ 0 if A has imqnite order and 1 _< Pi < p if A AplBql ..... AP~B has order p and all qi ¢ 0 if B has infinite order and 1 < qi < q if B has order q, then tr(R1 (A, B)) is a polynomial of degree k in tr(AB) with coefi%ients from. Z[tr(A), tr(B)]. {see Chapter 7}. Nowsuppose tr([gl, A2]) ?g 2. In [R 9] Rosenberger proved the following lemmawhich is crucial for the remainder of the proof of Theorem9.3.1. LEMMA9.3.1.

[R 9] Let A,B,C,X

be four elements in SLy(C) at with AB = C. For A we write A = (a~) a3 a4and we use simT and r = ilar notation for B,C, and X. Let x = (xt,x2,x3,x4) (tr(X), tr(AX), tr(BX), T and 1 0 0 al a3 a2 a4 bt ba ba b4 " c1 c3 c2 c4 Then Mx = r and det M= tr([A,

B] - 2.

Then from this lemma we can construct a projective matrix A3 in PSL2(C)such that tr(A3) = 2 if e3 = 0 or tr(Aa) -- 2cos(~r/ea) if ea _> 2, tr(R2(A1, A3)) -- -2cos(trip) and tr(Ra(A2,Aa)) -- -2cos(~r/q). Since a projective matrix B in PSL2(C)has order t if its trace is +2cos(~r/t) the representation defined by at -+ At, a2 -~ A2, a3 -+ A3 gives the desired essential representation. The construction of Aa using lemma9.3.1 used

270

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

the fact that tr(R2(A1, A3)) and tr(R3(A2, Aa)), analogously as above, are non-constant polynomials in tr(A1Aa) and tr(A2A3) respectively and then employed the Fundamental Theorem of Algebra. Nowsuppose al --~ A1, a2 --+ A2 defines an essential representation of G1 with tr([A1, A2]) = 2. Then lemma 9.3.1 cannot be used directly but essentially the same argument goes through as follows. Since tr([A1, A2]) 2 we may assume without loss of generality that < A1, A2 > is an i~nite metabelian but not abelian ~oup. A~er a s~table conjugation we may ass~e that ~ and A2

0

~rther if el = 0 we may assmne that assume that ~ = 0. Hence: A~=~

0

~-~

~ 0

~

~ ~ ~1,~ = 0. Ife~ ~ 2 we may

andA2=

~ 0

~

with 5 ~ 0 since < A~, Au > is non-abelian. We may ass~e further that ~ ~ ~1, for if fl = !1 then e2 = 0 and we may find an essential representation p~ : G1 ~ PSL:(C),aI ~ + ~-~ = tr(A~),tr(A~) ¢ ~2. Therefore assume ~ ¢ ~1. Let A3=~( xix3

x~

with x~,x2,x3,x~ considered as variables and A3 to be considered as an element of PSL:(C). Now tr(R2(A1,Aa)) = -2cos(r/p) and tr( Ra ( A~ , A3 ) ) = -2cos(u/q) are non-co~tant polyno~als in tr( ( A ~ A3 ) and tr(A2A3) respectively. Choose zeros of these two polynomials and let tr(A3) = 2 if c3 = 0 or tr(A3) = 2cos(u/ea)if e3 ~ 2. Assumefu~her that (trA3, tr(A~A3), tr(A2A3)) # (0, 0,0). Nowconsider the system of linear equations: x~ + x~ = tr(A3) (9.3.3)

O~Xl + o~-Ix4 = tr(A1A3) = ~X1 -~- (~X3~- ~--lx4 = tr(A2A3) =

Since det

I 0 ~ 0 ~-~

=~(~-~-~)#0,

9.3.1

ESSENTIALREPRESENTATIONS

there exists a unique solution -i rl(2 -X 1

(2__

(2--1

(Xl,

r2 ~-

(2--1

X3, X4)

with

_(2~-1_~_ ~(2--1 rlH 6~-- ~---1) r2+6-~r3 and

’x3

(2rl X4 --

271

(2

r2

__ (2_1

(2 __ (2--1"

Further this fixes tr(A~A2A3) = (2flxi+ (26X3 + (2--1fl--lx4 automatically. Nowconsider xlx4 - x2x3 = 1. If x3 ~ 0 choose an x~ to satisfy this. This choice of x2 together with (xt,x3, x4) will give an A3 in PSL2(C) completing the desired essential representation. Suppose then that x3 = 0. Then if (tr(Au),tr(AiA3)) = (0,0) must also have (tr(A3),tr(A2Aa)) = (0,0). If q ~ 2 then we may replace tr(R~(A2, A3)) = -2cos(~r/q) by tr(Ra(A~, A~)) = +2cos(~r/q) and then we obtain an A3 = ±

x~

(

X3

X4

E PSLu(C)

as desired because the polynomials tr(R3(A2,A3))2cos(~r/q) and tr(Ra(A2, A~)) + 2cos(~r/q) have no commonzero. Therefore assume that q = 2. If tr(R3(A2, A3)) has two different zeros for tr(A2A3) then we may argue analogously to the above by replacing one zero by another one. Therefore if (tr(A3), tr(AiA3), tr(A2A3)) (0, 0, 0) we areleft with the c ase that q = 2 and tr(Ra(A2, A3)) has exactly one zero (possibly with multiplicity). In an analogous manner we may argue on p and tr(R~(A~, A3)) and hence we may assume that p = 2. Therefore if < At, A2 > is metabelian and (tr(A3), tr(AiA3), tr(A~A3)) # (0, 0, 0) we obtain an essential representation for G except possibly when p = q = 2 and tr(Ra(A~, A3)) has exactly one zero and tr(R2(A~, A3)) has exactly one zero. The above arguments were based on starting with an essential representation of G~. Wemay also start with an essential representation o£ G2 and argue as above if (tr(A2),tr(A1A2),tr(A2A3)) ¢ (0, 0,0) and necessary start with an essential representation of Ga and argue as above if (tr(A~),tr(A1A2),tr(AiA3)) (0,0,0). Th erefore if we ass ume tha (tr(Aa),tr(AtAa),tr(AuA3)) # (0,0, 0), (tr(A2),tr(AiA2),tr(A2A3)) (0, 0, 0) and (tr(A1), tr(AiA2), tr(A1A3)) # (0, 0, 0) we obtain essential representations for G except possibly in the following situation: m = p = q = 2, tr(R3(A2, An)) has exactly one zero,tr(R2(A1, A3)) has exactly one zero, tr(Rt (A1, A2)) has exactly one zero and pi(Gi) is metabelian for each essential representation p~ : Gi --~ PSL2(C), i = 1, 2, 3. Therefore assume that we have the above situation. Since the images of the triangle group factors

272

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

are all metabelian this completely determines the traces of AIA2, A~A3and A2A3. For example if all ei >_ 2 we must have tr(A~A~) = 2cos(r/el ~r/e2), tr(AiA3) = 2COS(TO/el:1: r/e3) and tr(A2A3) = 2cos(~r/e2 zr/ e3). If we now start with A~=+

(: 0)

a- 1 and

As==t=

fl 1

with a + a-~ -- 2cos(Tr/el), ~ ~ 0and/3 + 13-~ = 2cos(~r/e2) asabove, the n in the system of equations (9.3.3) we automatically get that x~x4 ~- 1. This gives A3 in PSL2 (C) again completing the desired essential representation. Finally suppose that one of the triples (tr(Aa), tr(A~Aa), tr(A2Aa)) (tr( A2), tr(A~ A~), tr(A~.A3) and (tr(A~), tr(AiAe), tr(A~Aa)) is (0,0,0). Without loss of generality then suppose that tr(A3) tr(A1A3) = tr(A2Aa) = 0. If at least one of e3,p,q is not equal to 2 then we may choose one of tr(A3) , tr(A~A3) , Tr(A~Aa) non-zero. If for example e3 = 2, p _~ 3 and tr(A~A3) = th en werep lace tr( R2(A1, A3)) = -2cos(~r/p) by tr(R2(A~, A3)) +2cos(~r/p) and th en tr (A~A3) = 0 is not a solution to the equation tr(Ru(A~, A3)) = +2cos(~r/p). Therefore suppose that also e3 --- p = q -- 2. In this case because, of the trace conditions, G has an epimorphism onto the group

which preserves the orders. Recall the trace conditions give us the orders. Let

If e~ ~ 2 using the same argmnents as before we determine first an A~, A~ with tr([A~, A~]) ~ 2. This is possible since G~ is a non-abelian dihedral group. This then gives an essential representation of G~. Wecan then find an A~ to give an essential representation of G* and thus of G. In a symmetric manner we can in G* start with {a~,a3} or {a~,a~} and will obtain essential representations with the possible exceptions when el = e~ = m = 2 and tr(A~A3) = O, tr(A~A~) = O, tr(A~A~)

9.3.1

ESSENTIALREPRESENTATIONS

273

In this final case, again because of the trace and order conditions, G has an epimorphism onto a~ =a~=ag~- (ala2)2-~-

G** =< al:a2,

(ala3)2 ~-(a2a3)2-~

which preserves the orders. This group has the essential given by al -~

10)

,a2-~A2=±

Al=+(_01

and thus G has also.

0

i

representation

,aa-~Aa=±

This completes the proof.

Wemention that Vinberg has given an independent proof of this result IV] when the generalized triangle group factors have only non-metabelian images. Note that included in the proof is a proof of the following corollary which we will need in the next section. COROLLARY 9.3.1. Let p~ : G~ ~ PSL2(C),a~ --~ A~,a2 ~ A2 an essential representation. Suppose that < A1, A2 > is non-abelian and suppose further that one of the following holds: (1) tr([A1, A2]) # (2) tr(A~) # ±2, tr(A2) # ±2 and (p, q) # (2, Then p~ can be extended to an essential representation p : G --~ PSL2 (C) PrtOOF. The proof is a direct consequence of the proof of Theorem9.3.1. If tr([A1,A~]) # 2 then lemma 9.3.1 can be applied directly as in the beginning of the proof of Theorem 9.3.1. For condition 2 above, none of the exceptional cases handled in the proof of theorem 9.3.1 arise so therefore starting with an essential representation of G1 either lemma9.3.1 can be used directly or equations (9.3.3) can be solved in a suitable manner. Weremark that in constructing essential representations p~ : G~ --~ PSL~(C) for the generalized triangle group factors if pi(Gi) abelian we may replace pi by pi* with pi* (i) G non-abelian. Further if tr(Ai) = +2 and not the Klein bottle group, then ei = 0 and we may replace pi by p~* with tr(p~*(a~)) ~ ±2 (see Chapter 7). An iteration of the above procedure yields the following corollary. COROLLARY 9.3.2.

=< al,...,a,~;a~

~

""

Let n > 3 and an

/~1 (al,a2)

Pt 2m:(a2,aa) .. ..

Rm~"(an,al) = 1

where ei = 0 or ei > 2 for i = 1,...,n; mi > 3 for i = 1,..,n and for i = 1, .., n, Ri(ai, ai+l) is a cyclically reduced word in the free product on

274

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

ai,ai+l which involves both ai and ai+l ( with an+l = aa. Then G admits a representation p : G -* PSL2(C) such that p(ai) has infinite order ei = 0 or order ei if ei >_2 and p(Ri) has order mi for i = 1, ..., n. ~rther by a straightforward extension of the proof we get the following generalization. : THEOREM9.3.2. Let G =< al, ..., an~. a~~1 .... a~~ = R~’ (al, a~) .... R~_~ (a~,a~) = R~ (a~,aa) ..... (a2 ,an) = 1 >, where 3 n~ 3, ei ~0 orei ~ 2 [ori= 1,..,n, mj ~ 3[orj = 1,..,2n-3 and k = 1, .., n - 1, Rk is a cyclically reduced word ~ ~he ~ee product o~ a~, a~ i~o1~i~g bo~h a~ and ak and Tot s = 0,.., n- 3, ~+~ is a cyc~cally reduced word i~ She free product o~ a~, a~+3 involving bo$h a~ a~d a~+3. Then G admits a represen$ation p : G ~ PSL~(C) s~ch ~hat p(ai) h~ in~n1~e order iTe~ = 0 or order ei iTe~ ~ 2 and p(Rj) h~ order m~ [or = 1, ... ,2n - 3 .

9.3.2 Some Linearity Properties for Generalized Tetrahedron Groups Groups which admit essential representations in manycases satisfy additional linearity properties - for example the Tits Alternative and the presence of non-abelian free subgroups, SQ-universality and being virtually torsion-free. Wenow consider these linearity properties for generalized tetrahedron groups. First we discuss the Tits alternative and SQuniversality. THEOREM 9.3.3. Let G be a generalized tetrahedron group with pre. el =a = ~ ~3 = R~(al, a2) = R~(al, a3) sentation (2.1) - < al, a2, a3, I 2 a3 R~(a~,a3) = 1 > . If e~ >_ 2 let/3~ = 1/e~ while if e~ = 0 (so that a~ has infinite order) let fl~ = O. If ~3~ + ~2 + fl3 + (l/m) + (l/p) + (l/q) then G contains a subgroup of finite index which maps onto a free group of rank 2. In particular G is SQ-universal. PROOF.Recall that ira group G maps onto a non-abelian free group it is SQ-universal - that is every countable group is embeddable into a quotient of G. Further G is SQ-universal if i~ contains a subgroup of finite index which is SQ-universal. Further since if a quotient of a group is SQ-universal the group is itself SQ-universal we can without loss of generality assume that ei _> 2 for j -- 1, 2, 3, passing to a quotient if pecessary. The proof follows the basic outline of the proof of Theorem6.2.2. From Theorem 9.3.1, G admits an essential representation p into PSL~(C)

9.3.2

LINEARITYFOR GENERALIZED TETRAHEDRON GROUPS 275

Therefore from Selberg’s theorem on finitely generated subgroups of linear groups there exists a normal subgroup H of finite index in p(G) such that p(a~) has order e~ modulo H for i = 1,2,3 and p(R1) has order m modulo H, p(R2) has order p modulo H, p(R3) has order q modulo H . Thus the composition of maps ( where 77 is the canonical map)

a Z o(a) gives an essential representation ¢ of G onto a finite group. Let X =< ¯ el e2 e3 al, a2, aa, al = a2 : a.~ = 1 > be the free product on al, a2, a3. There is a canonical epimorphism f~ from X onto G. Wetherefore have the sequence X 2 G 2 p(G)/H Let Y = ker(¢ o ~). Then Y is a normal subgroup of finite index in X and Y is torsion-free. Since X is a free product of cyclics and Y is torsionfree Y is a free group of finite rank r. Suppose IX : YI = J" Regard X as a Fuchsian group with finite hyperbolic area #(X) (every finitely generated free product of cyclics can be faithfully represented as such a Fuchsian group unless it is an infinite dihedral group). From the Riema~m-Hurwitz formula : j#(X) = #(Y) where

,(Y)= r and

1 1 +--+--)

~(X)=2-(--1 el

e2

e3

Equating these we obtain; r=l

j(

1

1 1 +--+---2)

el

e2

e3

G is obtained from X by adjoining the relations R~"~,R P 2 and R~ and p so G = X/K where K is the normal closure of R~, R~ and R~. Since K is contained in Y the quotient Y/K can be considered as a subgroup of finite index in G. Applying the Reidemeister- Schreier process or repeated applications of Corollary 3 in [B-M-S] Y/K can be defined on r generators subject to (j/m) + (j/p) + (j/q) relations. The deficiency d of this presentation for Y/K is then d=r -j

m

J

J-1 p q

.... j(

1 1 1 1 1 1 +--+--+--+-+--2) el e2 e3 m p q

276

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

By assumption (1/el) + (l/e2) + (l/e3) + (l/m) + (l/p) + (l/q) so d > 2. From the work of B. Baumslag and S. Pride [B.B.- P 2] Y/K and hence G has a subgroup of finite index mapping onto a free group of rank 2. Therefore Y/K is SQ-universal and since this has finite index in G, G is also SQ-universal. Weremark that G is infinite if, in the notation of Theorern 9.3.3, /~1 + ~]2 + ~3 + (l/m) + (l/p) + (l/q) If a group G is SQ-universal then it must contain a non-abelian free subgroup. This would then be the case for generalized tctrahedron groups under the conditions of the theorem above. Vinberg has also obtained a proof of this result. However non-abdian free subgroups for generalized tetrahedron groups can be obtained under other conditions as well. THEOREM 9.3.4. sentation

Let G be a generalized

tetrahedron

group with pre-

~ ~ ~ 1 al~a2, a3,¯ a~ = a~ = a~ = R~(al,a2) = R~(a~,a3) = R~(a:,a3) -- 1 > ~ with k _> 1 and for Suppose that e I ~_ e 2 and R,(a~,a2) = I 0:1 au ~1 . ..% aO~k 2 ~ i=l,...,k, ai 7 0,~3i # 0 and 1 <_ai < e~ if e~ >_2, 1 <_/3~< e2 ire2 >_2. ~rther suppose that either (1) k ~_ 2and one of the following holds: (i) e~ >_ 4 and m >_ (ii)

e~ >_3andre>_4

(iii)

el >_ 3 and > 3

(iv) e~ = 0 and m >_ or

(2) at least one of p, q is not equal to 2, k >_ 1 and one of the following holds : (i)2 _< e~ _< e~ and (1/e~) + (1/eu) + (l/m) (ii)el = Then G contains a non-abelian free group. Symme~ricMstatements made replacing RI by 1~2 or 1~3 .

can be

PROOF.The general techniques of the proof follow the methods of Chapter 7, in partciular the proofs in section 7.3. The proof of Theorem9.3.4 depends on the following three lemmas.

9.3.2

LINEARITY FOR GENERALIZEDTETRAHEDRON GROUPS

277

LEMMA 9.3.2. Let G be as in the statement of the theorem and G1 be as defined in section 2. If G1 admits an essential representation into PSL2 (C) such that the image group is non-elementary then G contains non-abelian free subgroup. LEMMA 9.3.3. Let G be as in the statement of the theorem. Ilk ~_ 2 and one of the following holds: (i) e2 ~_ 4 and m >_ (ii)

e: >_3andre>_4

(iii) el >_3 and m ~_ (iv) el = 0 and m >_ Then G contains a non-abelian free subgroup. LEMMA 9.3.4. Let G be as in the statement of the theorem. If k ~ 1, at least one of p ,q is not equal to 2 and one of the following holds: (i)2 _< el _< e2 and (1/e~) + (1/e2) ÷ (l/m) (ii)el -~ Then G contains a non-abelian Tree group. (Wesplit lemmas9.3.2 and 9.3.3 since the proofs are somewhatdifferent) PROOFS.( of lemmas) (1) ( Lemma9.3.2) From the statement after proof of theorem 1, given an essential representation pl of G1 into PSL2(C) with p~(G~) non-elementary there exists an essential representation p of G extending Pl. If pl(G1) is non-elementary then it must contain a nonabelian free subgroup and hence p(G) does also. Since G maps onto p(G), G must also contain a non-abelian free subgroup. The proofs of lemmas9.3.2 and 9.3.3 are technical but follow esentially by showing that under the stated conditions either G1 admits an essential representation with non-elementary image or if not G has a factor group which contains a non-abelian flee subgroup. This was the method used to analyze free subgroups of generalized triangle groups in section 7.3. (2) ( Lemma9.3.3 ) As before let G~ be the generalized triangle group G~--< a~, a2; a~1 = a~~ ---- r~ RI (al,

a 2)

1>.

Call an essential representation into PSL2(C) suitable if the image group is non-elementary. From Lemma9.3.2, if G~ admits a suitable representation then G will contain a non-abelian free subgroup. From the

278

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

analysis of free subgroups of generalized triangle groups (section 7.3) if the length of the relator is _> 4 and the conditions of Lemma9.3.3 hold then G1 will admit a suitable representation except in the following five cases. (i) (~1 = 2, ¢2 6,m --- - 3, k - -- - 2 a ndG1 h as a presentation G1 -=<x, y; x 2 = y6 _-_ (xyxy3)3 _~ 1 (ii) el -= 2, e2 = m=- 4, k = 2 and G1 has a presentation GI --< x, y; x ~ -= y4 = (xyxy2)4 = 1 (iii)

el = e2 = m = 3, k =- 2 and G~ has a presentation G~ =< x, y; x3 = y3 = (xyxy~)3 _= 1

(iv) el --= e: = m-- 4, k --= 2 and G1 has a presentation G1 :< x, y;

X4 -~-

y4 _= (xy2xy2)4 = 1

(V) el =- 2, e2 -= m = k =- 4 and G1 has a presentation G~--< x, y; x~ -- y4 -- (xyxyUxy2xy~)’~ -= 1 > . In each of the five cases G~ does contain a free subgroup of rank 2. Further we can without loss of generality assume that x --= a~ and y -- an. Suppose first that the generalized tetrahedron group G has as a factor group 2 G* --
2 =(a2a3)

-- 1>.

G* has the subgroup H* of index two where H* has the presentation ~1 __a~~ = RT"(~ H* I :~ al,a2;a ,a~) = R~r~(a-1 ~ ~ ~ , a-l~ ~ ~ =1 >. In each of the five cases above the relation R~(a-~~, a~-~) -- 1 follows from m a hereH*= therelations a~~=a~2= R1 (1,a~) =1 ~nd therefore Thus G has G~ as a sub~oup and therefore G has a ~ subgroup of ra~ two. Suppose then that G does not have G* as a factor group and co~ider first case (i). Here G1 has a representation p~ onto the fi~te dihedral ~oup D~ =< A1, A2; A~ = A~ = (A1A2)2 = I > and a second representation p12 onto the (2,3,6) triangle ~oup T1 =< A;, A~; A~ = A~ = (AIA~)3 = 1 >. T~ is i~nite and metabelian. Let G3 =< a2, a3; a~ = a~~ = R~(a~, an) 1 >. If there is ~n essential representation p3 of G3 with respective images

9.3.2

LINEARITYFOR GENERALIZED TETRAHEDRON GROUPS

279

A2, Aa such that tr([A2, Aa]) -- 2 then we may choose the image < A2, Aa to be infinite and metabelian with tr(Aa) ~ +2 if e3 -- 0. Because m = the representation P3 of (~3 may be extended to an essential representation of G with non-elementary image (see the proof of theorem 1~ and therefore G will contain a free subgroup of rank 2. Suppose then that there exists no essential representation of Ga with tr([A2, A3]) = 2. If there is a suitable representation of G3 then G has a free subgroup of rank 2. If there is no suitable representation Pa of (~3 then Ga has the factor group G;" --< a2, a3;a~ ~- a~ = (a2a3)2 ~- 1 > and hence G has the factor group G** --< al, a2, a3; a~ = a26 = a~ = (ala2ala~) a = (alan) p = (a2aa) 2) = 1. If p _> 3 then we extend p12 and we can obtain a free subgroup of G** of rank 2 and hence of G. If p--- 2 then G* is a factor group of G. This handles case (i). Case (ii) is handled analagously. G~ has a representation p~ onto the finite dihedral group D4 =< A1, A2; A2~ = A~ = (A1A2)2 = 1 > and a second representation p~ onto the (2,4,4) triangle group T~ =< A1, A2;A~ = A~ = (A1A~)~ = 1 >. If we argue as above we must in addition consider the possibility that the finite symmetric group $4 is an image of Ga. Hence, G has a free subgroup of rank 2 or, possibly after changing the generators of G3, G has the factor group G**= and a second essential representation P12 onto the (3,3,3) triangle group Tt =< A~,A2;A~ = A~ = (AIA2) 3 -- 1 >. Since (p,q) ¢ (2, 2) we may extend pt2 directly (tr(Aa), tr(AiA3), tr(A2A3) (0, 0, 0 ) t o a representation p : G -- ~PSL2(C) such that p(G) is non-elementary. If (tr(A3), tr(A~A3), tr(A2A3) = (0, 0, 0) then G* is a factor group of G. This handles case (iii).

280

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

In case (iv) introduce the relation (ala2~ala’~) 2 = 1 and in case (v) the 2 2 2 2 = 1. In both these cases the corresponding relation (ala2ala2ala2ala~) factor group of G1 has a suitable representation into PSL~(/2) and therefore G contains a non-abelian subgroup completing the proof of Lemma9.3.3. (3){ Lemma9.3.4} Suppose that the conditions of Lemma9.3.3 hold that at least one of p, q is not equal to 2 and one of the following holds: (i)2 _< e~ _< e~ and (1/e~) + (l/e2) + (l/m)

(ii)el = Without loss of generality let q ¢ 2. Suppose (71 is as before and suppose that A1, A~ , with < A1, A2 > non-abelian, are the respective images of al, a2 under an essential representation of G1. For both cases (i) and (ii) above we may find a projective matrix Aa such that the triple (tr(A3), tr(A~Aa), tr(AuA3)) ¢ (0, 0, 0) as in the proof of Theorem9.3.1. Consider first case (i) above and assume further that k = 1. If rn _> 7 m = 5 then G1 has a suitable representation and as before since q _> 3 this extends to an essential representation of G. Let 2 _< rn _< 4 or rn = 6 and let pl : (71 -~ PSL2(C) be an essential representation, pl can be chosen be suitable except in the following four exceptional cases: (a) rn = 6, e~ = 2, e2 = 6 and Rl = ala~ (b) rn = 6, et = 3, e2 = 6 and Rt a~a~ (c) m = 6, el = 6, e~ = 6 and R~ a~a232 orR1 = ala 232 (d) rn = 4, e~ = 4, e~ = 4 and R1 at a~ orR1 = a~a2 In cases (a),(b) and (c) introduce the relation R~to obtain a factor G~ which has a suitable representation into PSL2(12). The corresponding factor group G* of G will then have a non-abelian free subgroup and hence so will G. In case (d) we have G2 =< a~, aa; 4 = a~~ = R~(a~, a3) = 1 >. Suppose G2 admits a suitable representation. Since m = 4 ¢ 2 this representation can be extended to an essential representation p of G such that p(G) is non-elementary. Therefore p(G) and hence G will contain a nonabelian free subgroup. Suppose then that Ge has an essential representation p~ with respective images A~, A3 such that tr([A~, A3]) = 2. Then as before we may choose p2 so that< A~, A3 > is infinite and metabelian with

9.3.2

LINEARITY FOR GENERALIZEDTETRAHEDRON GROUPS

281

tr(A3) ~ -1-2 if e3 -- 0. In G introduce the relation R~ -- 1 to obtain a factor group G*. Then p2 {considering G~. with respect to G*} may be extended to an essential representation p* : G* -+ PSL2 (C) such that p* (G*) is non-elementary. Therefore G* and hence G contains a non-abelian free subgroup. This completes case (i) when k = 1. Nowsuppose that k _> Choose an essential representation pl : GI -* PSL~(C) with respective images A~, A2 and assume that < A~, A2 > is elementary otherwise G would automatically have a non-abelian free subgroup from Lemma9.3.2 If < Aa, A2 > = As, the alternating group, then we may change the generators of Ga to get an image group of G1 in PSL2 (C) which is infinite. Hence we may assume as before that < A~, A2 > is infinite and metabelian and tr([A1, A2]) = 2. Construct A~ to get an essential representation of If either tr([A1, A3]) ¢ 2 or tr([A~, A3]) ~ 2 then < A1, A~, A3 > is nonelementary and G then contains a non-abelian free subgroup. Therefore we may assume that tr([A~, A3]) --- 2 and tr([A2, A3]) = 2. Using Lemma9.3.3 and the results above for k = 1 we can reduce to the case where m = 2. The condition tr([A~, A~]) --- 2 gives very strong restrictions on the exponents a~, fl,-..,~,fk occurring in R~(a~,a2). There would be analagous restrictions for the exponents occurring in R~(al, a3) and R3(a2, a3) since tr([Aa, A3]) = tr([A2, A~]) = 2. Despite these restrictions and the fact there are manyreductions from Lemma9.3.3 and the results for k = 1 the proof for this case still splits into several subcases to consider. Recall that we have the conditions 2 _< el ~ e2 and (1/O) + (l/e2) + (l/m) typical subcase is then e~ = e2 = 6, e3 = 2, m= 2, p = q = 3 and k

k

k

E o~, =-- O(mod6), Eft’ ~ O(raod6),2 ---- - 0(ro od6), i-~ l

i= l

R2(al, a3) al a3 and R3(a2, a3 ) = a2a3. Inthi s sub case we int roduce the relation a2a = 1 to obtain a factor group G* which has an image group in PSL2(C) which is non-elementary. Hence G* has a free subgroup of rank 2 and therefore G does also. The other subcases follow in a similar but sometimes lengthier manner. The details of the arguments and statements which are involved but not covered by the above example follow in the samer manner as in Chapter 7. Finally we consider case (ii) where e~ = If e2 = 0 then G~ has a suitable representation for any m> 2 so assume then that e2 > 2. Consider first e2 > 3. If m > 3 then from our previous results G contains a non-abelian free subgroup. An analagous statement holds if p > 3. Therefore if e2 > 3 we may reduce to the case where m = p = 2. By the same arguments we can reduce to e3 > 2 and we consider first e3 > 3. From Theorem9.3.2 if (l/e2) ÷ (1/e3) + (1/2) + (1/2) + (l/q) < 2 SQ-universal and must contain a non-abelian free subgroup. Therefore we

282

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

may assume (l/e2) + (l/e3) + (1/2) + (1/2) + (l/q) _> 2. Then e2 = e3 = q = 3 because e2, e3 >_ 3 and q ~ 2. Again from the techniques used in Chapter 7 and using the arguments in a symmetric fashion we see that G has a free subgroup of rank 2 via a suitable representation into PSL2(C). Nowlet el = 0, e2 _> 3, m= p --- 2 and e3 -- 2. If G1has a suitable representation into PSL2 (C) then G has a non-abelian free subgroup as before so assume that G1 has no such suitable representation. Using the analysis as in Chapter 7 we get that e2 = 2, k~ _> 4 and R~(a~,a2) = ak22T~(a1,a2) with T~(a~, a2) in the commutator subgroup of the free product on a~ and an. Nowwe argue in a symmetric fashion. If R2(al, a3) is conjugate in the free product on a~,a3 to some a~a3,n ~ 0 ,then we may extend each essential representation of G~ into PSL:(C) with tr(p~(al)) and infinite, non-abelian image to a suitable representation of G. If G2 has a suitable representation then G also has a free subgroup of rank 2. Hence we are left with the situation R2(al,a3) = a3T2(a~,a3) with T2(al,a3) in the commutator subgroup of the free product on al and a3. If (1/ee) ÷ (1/2) ÷ (1/2) ÷ (1/2) ÷ (l/q) < 2 then G is SQ-universal and must contain a non-abelian free subgroup. Therefore we may assume (l/e2) + (1/2) ÷ (1/2) ÷ (1/2) + (l/q) _> 2. Altogether then we al, a2, an; a~~ -~ a~3-- (ak2~Tl(al, a2))2 ---- (a3T2(al, a~))2 --~ R~(a~,an) -- 1 with q -- 4, e2 -- 4, k2 -- 2 or q -- 3, e2 -- 6, k2 : 3 or q -- 3, e2 -- 4, k2 ~- 2. If q = 4, e2 ~- 4, k2 -- 2 we introduce the relation a3 : 1 to obtain the factor group G* ~-< al, a2; a~ --- (a~T~(a~, a2)) 2 -- 1 > which has a free subgroup of rank 2. If q --- 3, e2 = 6, k2 -- 3 introduce the relations aa -- 1 and a23 -- 1 to obtain the factor group G* =< a~,a2; aa~ = (T~(a~,au)) 2 ~- 1 > which hasa free subgroup of rank 2. Nowlet q-- 3, e2 = 4, k~ = 2. We have tr([A~, A3]) ?~ 2 for any essential representation of G3 into PSL~(C). Hence if we choose an essential representation of G1 with tr(p~(a~)) ~ ±2 and infinite non-abelian image this may be extended to a suitable representation of G. Nowlet e~ ---- 2. By symmetrical arguments to the above we can reduce finally to the case where e2 ----- e3 --~ m ---- p = 2. Choose an essential representation of G~ with respective images A~,A~ and tr(A1) ~ ±2 as before. If tr([A~, A2]) -- 2 then we may assume as before that < A~, A2 is infinite metabelian. Since q ~ 2 the representation of G~ extends to G and G must contain a non-abelian free subgroup as before. If tr([A~, A3]) ¢ the only possibility where < A~, A: > does not contain a non-abelian free subgroup is if < A1, A~ > -- Z2 * Z2 the infinite dihedral group. In this case G has the factor group G*=
q~ :1>

9.3.2

LINEARITY FOR GENERALIZEDTETRAHEDRON GROUPS

283

with 3 _< q _< qlWorking with G*, G* will contain a non-abelian free subgroup as above or G* will have the factor group G** G**=< a,, a2, a3; a~ = a] = (ala~.) 2 = (a,a3) 2 = (a2aa)q’ = 1 >. Now G** does contain a free subgroup of rank two. Starting with < AI,A: >= Z: * Z~ and tr(A~) ¢ +2 the infinite dihedral group we may construct Aa so that < A1, A2, Aa > is non-elementary. This can be done since 3 _< q _< ql and Z2 * Z2 has no element of finite order _> 3. Therefore G** and hence G contains a non-abelian free subgroup completing the proof of lemma9.3.3. In Theorem 9.3.4(2) we assumed that at least one of p, q is not equal to 2. This assmnption allowed us to extend essential representations Pi : G~ -~ PSLe(C) with pi(Gi) non-abelian. If p = q = 2 we also have some results. Using the same type of arguments as in Theorem9.3.4 we have the following. THEOREM 9.3.5. Let G be a generaBzed tetrahedron group wi~h p = q = 2. Suppose that either of the following two conditions hoM (1) el = and (e:, m) 7~ (2, 2) (2) 2 _< el _< e2 and (1/el) -t- (l/e2) -t- (l/m) Then G has a free subgroup of rank two with the possible exceptions ea = 2 and(el, e2, m)= (3, 8, 2), (3, 10, 2), (4, 5, 2), (4, 6, 2), (4, 8, 2) or PROOF.The proof is done in the same manner as the proofs of the lemmas for Theorem 9.3.4. We give a sketch and show how the possible exceptions can arise. If there is an essential representation p~ : G1 --~ PSL2(C) with tr ([pl (a~), p2 (a~)]) ¢ 2 then we mayalways find an extension to an essential representation of G. Hence if ea ¢ 2 and p = q = 2 we argue essentially as before, using the argument also in a symmetric fashion, to get a free subgroup of rank two, possibly passing to a factor group of G if necessary. Especially if et ¢ e2 we obtain that G has a free subgroup of rank 2. If e~ = e2 then as before we may reduce the problem to the group 2 G=
2 =(a2aa) 2 =(ala3)

- 1>

with e~ = e2 _> 5 and ea ¢ 2. But nowsince e~ = e2 _> 5, G~ has a suitable representation into PSL2 (C) and so G has a free subgroup of rank 2. Now suppose that e~ = p = q = 2. Suppose also that 2 <_ et <_ c2 and (1/e~) + (1/e~) + (l/m) < 1. If m _> 3 everything goes through analagously as in the proof of Theorem9.3.4 by considering all the possible cases and finding free subgroups in each one. This is purely computational

284

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

but lengthy. If m = 2 we can reduce to the case where all possible representations of G1 axe metabelian. Then as already mentioned in the proof of Theorem9.3.4, the restriction on the exponents occurring in Rl(al, allow us to consider factor groups by introducing a relation a{1 if or a~2 if e2 = 2f2. This can be done since ea = p = q = m = 2. Then we get a free subgroup of ra~k 2 as long as (1/fi) + (1/e2) + 1/2 < 1 or spectively (1/fu) + (1/el) + 1/2 < 1. These restrictions,together with considerations as in section 7.3,give the exceptions listed in the statement of the theorem. If on the other hand el = 0 and (e~, m) = (2, 2) then can be i~ffinite solvable. 9.3.3

A Freiheitssatz

and the Virtual

Torsion-Free

Property

A generalized tetrahedron group G is a triangular product of the generalized triangle groups G~, G2, Ga . In general it is not clear when the vertex groups in a triangulax product inject into the product however below we give conditions for this to occur in a generalized tetrahedron group. From this we can determine conditions for such a group to be virtually torsion-free. For a triangular product of groups, Gersten and Stallings [G-S] have defined the concept of an angle. Using this concept they proved that if the resulting triangle is non-positively curved then the vertex groups inject. Using this we obtain the following. THEOREM9.3.6. Let G be a generalized tetrahedron group and its generalized triangle group factors. Suppose that (l/m) (l/p) + (l/q) _< 1. Then: (1) The natural maps from Gi to G for i = 1, 2, 3 are injections (2) If H is a finite subgroup of G then H is conjugate in G to a finite subgroup of some Gi , i = 1, 2,3 G1, G2, G3

PROOF.Using the theory of essential representations of Gi, i = 1, 2, 3 into PSL~(C) we can control the behavior of the non-trivial words of short length in the free product of the generators with respect to the canonical epimorphisms onto the Gi. To see this consider G~. Let S(al,a2) = ~*1 ~"~(~2-’~’-~" < 79 < ""~’1 ~’2 with 1 <-- r and 7~ ¢ O,~ij # 0 and 1 -el if el _> 2,1 _< tj < euife2 _> 2 in the free product on al,a2. Let r <_ m- 1 if 2 _< m _< 5 and r = 5 ifm >_ 6. By direct matrix calculations we can find an essential representation p~ : Gt --~ PSL2(C) with plS(a~, a.)) ¢ 1. Therefore we can measure the Gersten-Stallings angle see [Sta]}. Together with (l/m) + (l/p) + (i/q) _< 1 we can directly the results of Gersten and Stallings on non-positively cttrved triangles of groups to obtain the theorem. {See also the work Pride [P 2] on groups

9.3.3

FREIHEITSSATZANDVIRTUALTORSION-FREEPROPERTY 285

in which every defining relator involves only two types of generators The application of Gersten-Stallings results to the generalized tetrahedron groups was suggested by M. Lustig. Wenote that there exist generalized tetrahedron groups where the factors do not inject and we present an example. Let G=< al, a2, a3; a~ = (ala2a I a2 ) 2 --1 --1 m m _> 2 and let G~ =< al,a2; (alaual a 2 ) -- 1 >. Assu~ne that G1 injects into G. Then we have a3(a~a~a-~la~l)a~ 1 -- a-~2a~la~a2. Hence (a~2a~lala2) "~ = 1 in G1.. From Theorem 4.11 in [M-K-S] we must have 1 is conjugate in the free group on a~, a2 to k(a-~2a~lala~) that a~a~a~la~ with k -- ±1. But from theorem 1.3 also in [M-K-S] this does not hold so G~ does not inject into G. Anyfinitely generated linear group is Virtually torsion-free - that is contains a normal torsion-free subgroup of finite index. The following corollary gives sufficient conditions for this to be true in the case of a generalized tetrahedron group.

COROLLARY 9.3.3. Let G be a generalized te~rahedron group and G~, G2, G3 its generalized triangle group factors. Suppose that (l/m) (l/p) + (l/q) _< Suppose &rt her ~hat s < e~ if e~ >_2, l <_ t < e2 ire2 >_ 2 or that m _> 4 and R1 is not conjuga$e in the Tree product on al, a~ to a product X1Y1 for some X1 and of orders xl >_ 2, yl >_ 2 respec~i~ly where (l/m) ÷ (l/x1) ÷ (l/y1) R2 = a~a~,with w ~ O,u ¢ O,l _< W < el if e~ >_2, l <_ u < e3 if ea >_2 or that p >_ 4 and R~ is not conjugate in the Tree product on al, a3 to a product X2Y2 for some X2 and Y~ of orders x~ >_ 2, y2 >_ 2 respectively where (l/p) + (1/xu) + (l/y2) > 1; R3 = a~a~,with v ?~ O, z ¢ O, 1 <_ v < e2 if e2 ~_ 2, 1 <_ z < e3 if e3 >_ 2 or that q >_ 4 and R3 is not conjugate in the Tree product on a2, a3 to a product X3Y3 for some X3 and Y3 of orders x3 _> 2, y3 _> 2 respectively where (l/q) + (l/x3) + (1/y3) Then: order in G is conjugate to a power of (1) Any element of finite al,a2,a3,R1,R2 or R3. (2) G is virtually torsion-free. ( G contains a torsion-Tree normal subgroup of finite index.}

PROOF.From Theorem9.3.6 since (l/m) + (l/p) ~- (l/q) _< I each triangle group factors G~ inject into G and any finite subgroup is conjugate to a finite subgroup in one of the factors. The conditions on the relators guarantee that any element of finite order in a factor is conjugate to a power of either R~ or a powerof one of the two generators it involves. This handles

286

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

part (i). Nowfrom theorem 9.3.1 G admits an essential representation into PSL2 (C). However form the preceding statement this implies that an essential representation is essentially faithful. Therefore the remainder of the result follows from Theorem6.2.1. Wecan also extend Corollary 9.3.3 to the case where 3 _< m, p, q and no letter in Ri, i = 1, 2, 3, has order two, especially if all ei are odd. THEOREM 9.3.7. Let G be a generalized tetrahedron group and G1,G2,G3 its generalized triangle group factors. If Gi , i -- 1 or 2 or 3, is non-metabelian and admits a faithful representation into PSL2(C) then Gi injects into G. PROOF.Given a faithful representation ~ of G1 into PSL~(C) there exists an essential representation p of G which extends 7r since G1 is assumed non-metabelian. Since p maps G into PSL2(C) and p is faithful on G1, G1 must inject into G. Wenote that Theorem 9.3.7 also holds if Gi is metabelian and admits a faithful representation into PSL2 (C). There are some difficulties with the case whenG1 is the (3, 3, 3) triangle group (~1 ----( a~, a~; a3~ 3 --~ (ala2) a ---- 1 > . In this case the representation techniques reduces the problem of extending the faithful representation to the case where each Gi is a (3, 3, 3) triangle group. The calcuations are quite lengthy and we omit them. Wenext give a result as to when a generalized tetrahedron group actually admits a splitting as a non-trivial free product with amalgamation. THEOREM 9.3.8. Let G be a generalized tetrahedron group and Gi its first generalized triangle group factor. Suppose e~ -~ 0 and at least one of p, q is not equal to 2. Suppose further that t~ ~ a~a~T(a~, a2) with s ~ O, 0 ~_ ~ ~ e2 ire2 ~_ 2 and T(ai, a2) in the commutator subgroup in the free produc~ on ai, a2. Then G admits a splitting as a non-trivial free product with amalgamation. PROOF.Passing to a suitable factor group if necessary we may assume that e2 _~ 2 and e3 >_ 2. Given an essential representation pl : G1 --~ PSL2 (C) with tr(p~ (al)) ~ ±2 and pl (G~) non-abelian, since at least of p, q is not equal to 2 this can be extended to an essential representation p : G --~ PSL2 (C). NowH. Bass in [Ba 1] proved that if K is a finitely generated subgroup of GL2(C) then one of the following cases must occur. (1) There is an epimorphism f : K -~ Z such that f(U) -- fo r al l unipotent elements U in K (2) K is an amalgamated free product K = Ko *~ Ki with Ko ~ H ~ K1 and such that every finitely generated unipotent subgroup of K is contained in a conjugate of Ko or K1

9.3.3

FREIHEITSSATZANDVIRTUALTORSION-FREEPROPERTY 287

and d roots of unity (4) K is conjugate to a subgroup of GLe(A) where A is a ring of algebraic integers. Under the conditions on el and R1 the essential representation pl of G1 can be constructed so that the image < A1, Au > does not satisfy conditions (1),(3) or (4) of Bass’ theorem. Nowp: G-~ PSL~(C) be th e e ssential representation of G extending pl. The image p(G) cannot satisfy conditions (1), (3) or (4) of Bass’ theorem since then pl(G1) would also. It that p(G) must satisfy condition (2) and therefore p(G) is a non-trivial free product with amalgamation. However from a result of Zieschang [Z] groups which have non-trivial amalgamated free products as epimorphic images axe themselves non-trivial amalgamated free products and so G is a non-trivial free product with amalgamation. Finally we close by giving an extension of a representation results and applications to an extended class of three generator groups which includes the generalized tetrahedron groups. THEOREM 9.3.9. Let G =< a,b,c;a r = b R~(a,b) = Sm(a,c) = Tn(b,c) Oorr >_ 2, s = Oors >_ 2, t = Oort _> 2 and RI,...,Rk, S, T axe nontrivial cyclically reduced words in the free products on the generators they involve. Assume further that Ri(a,b), i -- 1,... ,k involves both a and b, S(a, c) involves both a and c and T(b, c) involves both b and c. G1 --< a,b;a r = bs = Rl(a,b) .... = Ra(a,b) = (1) If there exists a representation Pl : G1 --~ PSL2(C),a -~ A,b --~ with < A, B > non-metabelian then G ~ {1} and in fact G is non-metabelian. If there exists a representation Pl : G1 -~ PSL2(C), a -~ A, b --~ (2) with < A, B > non-elementary then G has a free subgroup of rank 2 (3) If GI is non-metabelian and if there is a faithful representation pi : G1 -~ PSL2(C) then GI embeds into PROOF.(1) Let Pl : G~ -~ PSL2(C),a --~ A,b --~ with < A, B > nonmetabelian. First suppose that G has the factor group G*--< a, b, c; a~" -b~ ~- c 2 = R~(a,b) ..... Rk(a,b) = ~ = (b e) ~ = 1 >. < A,B > embeds into a factor group G** of G* since there is a D in PSL2(C) with DAD-~ -~ A-~,DBD-1 = B-1. Therefore G ~ (1) because A, B > is non-~netabelian. Nowsuppose that, G* is not a factor group of G. Then we may extend

288

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

Pl to a representation p : G -~ PSL2(C) as before because < A, B > is non-metabelian. Therefore in this case also G ?~ {1}. (2)Let Pl : G1 -~ PSL2(C),a -~ A,b --~ with < A, B > nonelementary . Then tr([A, B]) ¢ 2 and we may extend pl to a representation p : G -~ PSL2 (C). Therefore G has a non-abelian free subgroup since < A, B > has one. The proof of (3) is entirely analagous to that for generalized tetrahedron groups (see Theorem9.3.7). 9.3.4

Euler Characteristic

for Generalized

Tetrahedron Groups

In section 7.6 the construction of a rational Euler characteristic for the generalized triangle groups was presented. A similar construction of Euler characteristic was given in section 8.3 for groups of F-type. M.Stille [St] has constructed a similar rational Euler characteristic for the generalized tetrahedron groups. In this final section we briefly review his results. If G =< a, b; ap = bq = Rm(a, b) = 1 > is a generalized triangle group, where as usual, R is a cyclically reduced word in the free product on a, b involving both a and b and m _> 2, then we say that the relator R has property E if either m >_ 4 and R is conjugate within the free product on a, b to a product of the form XY for some elements X, Y of respective orders x >_ 2, y >_ 2 where (l/m) + (l/x) + (l/y) > 1 or m -- 3 and there ~re letters of order involved in R. Wethen obtain. THEOREM 9.3.10. Let G be a generalized sentation of ~he form (9.3.2) so tha$

tetrahedron

group with pre-

1G ~-( al,a2,a3;a with the standard conditions on exponents and relators and let G~, G2, G3 be its generalized triangle group factors. Suppose that (l/m) + (l/p) (l/q) <_ and none of Ri, R2, Ra has property E. T henG hasa rati onal Euler characteristic X satisTying: 3

(1) x(G) =-2+ ~ + ~ +1 ~ I+ ~=i a, wi~h a, = O if (2) ~rther

ei = O or ai

~1if ei _>2 ~ x(G) =I+~=~(X()-x(A~)) G where A~ =< a~ > ~or i =1,2,3 are reg~ded as sub~oups of G. vcd(G) ~

PROOF.Under the given conditio~ it follows ~om Theorem 9.3.6 and Corollary 9.3.3 that G is virtually torsion-~. As in the construction of the Euler ch~acteristic for generalized triangle groups (s~ the proof of Theorem 7.6.1) construct the Cayley complex for the given presentation.

9.3.5

FINITE GENERALIZEDTETRAHEDRON GROUPS

289

From results of Pride [Pr 2,3] the relation module splits as a direct sum of cyclic submodules. Therefore the complex X is a two-dimensional contractible G-complexsuch that every isotropy group is trivial or finite cyclic and X has only finitely many cells mod G. This implies then that X has an equivariant Euler characteristic xa(X). Because X is contractible and G is virtually torsion-free we have that x(G) = xa(X). Because of the decomposition of the relation modulethe cellular chain complex of X is a free resolution of Z over ZH for every torsion-free subgroup H of G. Hence G is of homological type WFL(see [Brn]) and vcd(G) _< 2. The computations follow as in the proof of Theorem7.6.1. COROLLARY 9.3.4. Let G be a generalized tetrahedron group as in Theorem 9.3.10. If G has a faithful representation p : G -~ PSL2(C) such that p(G) is a Kleinian group of finite covolume then: 3 I+I 1 with ~ = O if e~ = O or a~ = 1~ if e~ _> 2. PROOF.In these cases G has a rational Euler characteristic as in Theorem 9.3.10. If p(G) has finite eovolume then from work of Ratcliffe [Ra] X(G)_> 9.3.5

Finite

Generalized

Tetrahedron Groups

Recall that a complete classification of the finite generalized triangle groups has been given (see section 7.3.4). Analogous work has been started on the classification of the finite generalized tetrahedron groups. Recall that Coxeter gave a criterion for an ordinary tetrahedron group to be finite while this was extended by Tsaranov to generalized tetrahedron groups where each relator is of the form Ri(u, v) = uavb. Nowlet G be a general generalized tetrahedron group with a presentation of the form G ----< 1al,a2,a3, . a~1 = a;~ = a~ = RF(~I, ~) = R~(~, a~) = R~(a~, a~) >. FromTheorem9.3.3 it follows directly that G is infinite if 1 1 1 1 1 1 --+--+--+--+-+-<2. e~ e2 e3 m p q Notice how this generalizes the non-Euclidean infinite case for generalized triangle groups. Purther conditions on the infiniteness of G follow from Theorems 9.3.4 and 9.3.5. From Theorem 9.3.6 if ~ + ~ + ~ < 1 then each triangle group factor will inject into G. HenceG will be infinite if any of the factors are infinite. Edjvet,Rosenberger, Stille and Thomas [E-R-S-T] have proved that this condition suffices for G to be infinite.

290

ALGEBRAICGENERALIZATIONSOF DISCRETE GROUPS

THEOREM 9.3.11. Let G be a generalized sentation of the form

tetrahedron

G--
group with pre-

-- R~(a2, a3)=1>

if-~ + ~ + 1<1. ~_

Using similar techniques they further prove THEOREM 9.3.12. Let G be a generalized sentation of the form G=
tetrahedron

group with pre-

=aae3 =RT(a~,as) =R~(a~,aa) =R~(a2, a3)=1>

Then:

Of course by sym~netry in part (a) of the theorem, G wifl be infinite either 1 1 1 e2 e3 p or 1 1 1 --+--+-<1. e3 e~ q There are similar symmetric statements for part (b). Further work on the finite generalized tetrahedron groups has been done by Stille [Sti 1] and Runge[Ru].

.

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SUBJECT

A-thin triangles 27 adjunction problem 112-113 amalgamated free product 11 arboreal group theory 31-34 automatic groups 34-38 automatic structure 36,37 asynchronous automatic structure 37 base of an HNNgroup 13 Bass-Serre theory 16-18 Baumslag-Shalen decomposition 58 Baumslag-Solitar group 64 biautomatic group 37 character variety 212 cocompact Fuchsian group 88 Cohen-Lyndon theorem 120 commutative transitive 46 confluence 36 conjecture F 128 conjugacy problem 26 conj ugacy pinched one relator groups 59-64 conjugacy separable 56 context-free group 36 context-free language 35 cyclically pinched one relator groups 53-59 cyclically reduced in a free group 6 in a free product 12 decidable theory 65 deficiency 50 Dehn algorithm 27,28 delta-thin triangles 28

INDEX

Dirichlet region 88 discontinuous group 80 discrete groups 177-110 elementarily equivalent 66 elementarily free group 66 elementary Nielsen transformation 19 elementary subgroup 81 ele~nentary theory 65 elementaxy embedding- 65 elliptic transformatiori 78 equations over groups 112-113 essential representation 139 essentially faithful representation 139 Euler characteristic for Fuchsian groups 2,90 for one-relator products 216-220 for groups of F-type 237 for generalized tetrahedron groups 288 extended deficiency 139 extension of centralizers 68 F-group 2,91,222 factor in a free product 9 in a one-relator product 1 Fenchel-]~bx theorem 96 FHS(see Freiheitssatz) finite generalized tetrahedron groups 289-290 finite generalized triangle groups 201-208 finite state automaton 35 309

ALGEBRAIC GENERALIZATIONS OF DISCRETEGROUPS fixed point 78 Ford domain 83 free actions 17 free exponential group 68 free group 6 free group rank 6 tree length 6 free product 9 free product with amalgamation11 freely reduced word 6 Freiheitssatz (FHS) general 40 for generalized tetrahedron groups 284 for groups of F-type 226 for groups of SN-type 255 for one-relator amalgamated products 149-152 for one-relator groups for one-relator products locally indicable factors 113-117 high-powered relators 121-130 of cyclic groups 144-148 for small cancellation products 125 Freiheitssatz (FHS)factor Fuchsian groups 81-103 algebraic analysis 89-103 Fuchsian representation 169 fully residually free 66 fundamental domain groups acting on trees 17 Fuchsian groups 82 generalized tetrahedron groups 266-290 generalized triangle groups 166-220 generalized triangle group factor 268 generalized word problem 248 genus 90 genus problems 118

geometric group theory 25-38 geometric rank 106 Gersten conjecture 31 Gersten-Stallings angle 284 group amalgam 11 groups of F-type 221-252 groups of SN-type 254-266 groups of special NEC-type254-266 HNNextension 13 HNNgroup 13 Howie’s tower proof 48 Hurwitz groups 171 hyperbolic groups 25-31 hyperbolic transformation 78 isomorphism problem 26 Kervaire conjecture 113,130-136 Klyachko’s theorem 131 Kurosh theorem 10 Lambda-tree 33 Lambda-free group 34 Left 56 limit points 80,237 linear isoperimetric inequality 28 linear properties groups of F-type 231-248 groups of SN-type 258-266 tetrahedron groups 274-289 linear fractional transformation 77 locally indicable 53,1 t3-117 loop 18 loxodromic transformation 78 LRS monster 202 Lyndonlength function 33 Lyndon Van-Kampen diagram 72 Lyndon’s Identity theorem 119 Magnus method 42 Magnusrelators 250 Magnus subgroups 76

SUBJECTINDEX maximumtnodulus principle 73 Moebiustransformation 77 multipler 78 n-free 54 NECgroups 103-110 Nielsen transformation 18-25 Nielsen equivalence 19 Nielsen realization problem 76 Nielsen reduction 20-23 Nielsen-Schreier theorem 7 non-Euclidean crystallographic group 103-110 normal form in a free group 6 in a free product 10 in an amalgamatedfree product 11 in an HNNgroup 13 one-relator amalgamatedproducts 149 one-relator groups 1,39-76 one-relator products 1,111-136 one-relator products of cyclics 2,137-220 linear properties 159-220 Tits alternative 160-166 ordinary tetrahedron groups 267 ordinary triangle groups 91,167-171 Papakyriapopolous towers 46 parabolic transformation 78 parafree group 62 pictures over groups 126 Poincare polygon 87 Poincare presentation 2,89 polygonal product 268 power conjugacy problem 248 quasiconvex 29 R-diagram 73 R-tree 32

31~

Ree-Mendelsohnpairs 167,194 Ree-Mendelsohn theorem 50 regular language 35 R-free group 32 reduced sequence free product 10 free product with amalgamation 11 residually finite 56 residually free 66 restricted Gromov(RG) group rewriting rule 25 Riemann-Hurwitz formula Fuchsian groups 97 one-relator products 217 groups of F-type 237 Rips complex 31 Rips theorem 32 RM-pairs 167 Schreier system 7 Schreier transversal 7 Selberg’s lemma95 separated HNNextension 69 signature Fuchsian group 89 NECgroup 105 sixth group 71 small cancellation products 125 small cancellation theory 70-74 Spelling theorem 237 for one-relator groups 45 for one-relator products 118 spherical pictures 126 SQ-universal 15 stable letters 13 staggered presentation 41-42 subgroup separable 53,56,113-117

312

ALGEBRAIC GENERALIZATIONS

surface group 53 syllable length I0 Tarski problem 64-70 Tits alternative 49 for one-relator groups 49 for one-relator products 160-220 trace polynomial 176 tree product 15 tree-free group 34 treed HNNgroup triangle group 41,167-171 triangular product 268 type FL 216 type WFL216 universal sentence 64 universal theory 65 universally free group 66 vertex groups 16 virtual property 75 virtually torsion-free 51 word hyperbolic 28 word problem 26

OF I)ISCRETE

GROUPS

INDEX

OF

NAMES

Aab, M. 56,64,20 Allenby, R. 56,98,239,240,241 Alperin, R. 32,34,60,66,96 Anshel, I. 49 Bagherzadeh,G.H. 76,120 Bass,H. 17,32,34,52,60,66,67,98,214,216,237,286,287 Baumslag, B. 50,66,67,114,140,173,260,276 Baumslag, G. 4,37,39,46,49,54,55,56,58,62,75,138,140,145,147,171,203,228,249 Beardon,A. 79,80 Bencsath, K. 75 Bestvina, M. 30,58,251 Bieri,R. 58 Bogley, W. 47,49 Boyer, S. 147 Britton, W. 71 Brodskii,S.D. 53,11~4,117,121 Brown, K. 97,216,217,237,238 Brunner,A. 56,64,239 Burns, R. 56,64,239 Cannon,J. 34,35 Chandler,B. 62 Chiswell,I. 31,32,33,34,67,98,154,216,237 Cohen, D.E. 120 Cohen, M. 30 Coldeway, H. 104 Collins,D. 19,21,124,126,210,211 Conder,M. 171,203,221 Corson,J. 126 Coxeter,H.S.M. 253,267,289 Culler,M. 4,51,160,212 Dehn, M. 26,27,46,70 Dicks, W. 6 Duncan, A. 112,118,128,131,211,217 313

314

ALGEBRIACGENERALIZATIONSOF DISCRETE GROUPS

Dunwoody,213 Dyer,J. 56,98,241 Eckmarm,B. 76 Edjvet,M. 117,120,164,260,289 Epstein,D.B.A. 34 Feign,M. 30,58,251 Fenchel,W. 51 Fine,B. 19,25,30,55,56,58,60,61,62,66,68,75,98,106,203,238,249 Fischer,J. 51,140,208 Fox,R. 51,96 Gaglione, A. 25,55,56,66,67,68 Gersten,S. 31,37,132,269,284 Gerstenhaber,M. 132 Gilman,J. 107 Gitik, R. 64 Gonzalez-Acuna,F. 112,126 Greendlinger,M. 71,72 Gromov,M.25,27,28,29,45,58 Gurevich,G.A. 118 Hagelberg,M. 153 Harrison, N. 34 Helling, H. 4,51,153 Hempel, A. 229 Higman, G. 14,96,113,140,171 Hoare,A.H. 76,97,98,104 Holt, D. 34 Howie,J. 3,46,47,48,53,112,114,117,118,120,121,128,131,132,160,175,189, 193,202,203,208,210,211,217 Johnson,D. 7 Juhasz,A. 30,45,57,58,124,248,151 Kalia,R.N. 19,21 Karrass,A. 6,15,16,49,51,76,97,98,104,140,208,251,268 Katok,S. 79 Kaufmann, 106,253,264,266 Kerckhoff S.P., 76 Kervairs,M. 113.131 Kharlamapovich, O. 30,58,66,67,69,251 Kim,A. 153 Kim, G. 56,63,64 Klyachko,A. 113,121,131 Klymenko,E. 109 Lev~i,L. 160,202,203

INDEX OF NAMES Levin,F. 67,96,113,132,163,192,203 Lipschutz, S. 57,248 Liriano, S. 62 Litz, S. 152 Long, D. 213 Lossov,K.I. 191 Lustig, M. 30,285 Lyndon, R. 6,39,40,49,71,72,73,97,114,119,131,222 Lysenok,I.G. 27,28 Macbeath, M. 104,171 Macalachlan, C. 152,213 Magnus, W. 6,39,40,41,42,43,49,62,70,74,114,118,119,155,168,248,251 Mahmud, R.M.S. 126 Majeed, A. 109,162,198 Malcev, A.I. 232 Martinez, E. 104 Maskit, B. 87,107 McCool, J. 41,114 Mendelsohn, N.S. 4,50,144 Mennicke, J. 153 Metafsis, V. 160,202,203,208 Moldavanskii,D.I. 41,114 Morgan, J. 4,31,32,138,140,145,147,203 Muller,D.E. 36,76,171 Murasugi,K. 46,75 Myasnikov, A. 30,58,66,67,68,69,251 Neumann, B.H. 14,131,221 Neumann,H. 14,140 Neumann, P.M. 96 Newman, B.B. 45,51,117,140,248 Niblo,G. 56,63,64,239,240 Nielsen,J. 96 Olshanskii,Y. 124 Patterson,M.S. 34 Peczynski,N. 19,23,25,60,106 Peluso,A. 58 Perraud,F. 124,126,210,211 Pietrowski,A. 268 Pride, S. 45,47,49,50,114,117,118,140,173,249,260,276,284 Purzitsky, N. 106 Ratcliffe,J. 154 Ree,R. 50,144

315

316

ALGEBRIACGENERALIZATIONSOF DISCRETE GROUPS

Reid,A. 213 Reiwer,W. 19,23,25,60 Remeslennikov,V. 66,67,69 Rips,E. 31,32,124 Rohl,F. 25,60,61 Rosenberger,G. 4,19,21,23,25,55,56,60,61,62,63,66,67,68,98,106,153,160 192,202,203,228,230,238,249,251,269,289 Rothaus,O. 132 Rourke,C.P. 126 Runge,C. 290 Saeerdote,G.S. 53,166 Sageev,Y. 213 Sakuma,M. 109 Sasse,S.L. 63 Scheik,H. 71 Schreier,O. 41 Schupp,P.E. 6,36,39,40,41,49,53,71,97,103,114,118,166,222 Scott,G.P. 6,56,238,239 Seal,Z. 29,67 Selberg,A. 51,95,208 Serre,J.P. 6,15,17,129 Shalen,P. 4,28,31,32,37,53,55,58,138,140,145,147,152,153,160,171,203,212 Shapiro,M. 28 Short,H. 37,49,112,114,126,132 Solitar,D. 6,15,49,51,56,64,76,97,104,140,208,239,251 Souvignier,B. 160,202,203 Spellman,D. 25,55,56,66,67,68 Stallings,J. 269,284 Stebe,P. 56,98 Stille,M. 19,62,249,289 Tang, C.Y. 56,98,239,240,249,250 Tarksi,A. 65 Tartakovskii,V.A. 71 Taylor,T. 46,75 Thomas,R. 140,143,160,202,203,208,289 Thurston,w. 34 Tits,J. 31,160 Tretkoff, M. 6,64 Tsaranov,S.V. 268,289 Urbanski,M. 34 Van Kampen,E.R. 73 Vinberg,E.B. 14,149,153,253,267,273,276

INDEX OF NAMES Vogt H. 104 Wall,C.T.C. 6,98,216,237 Wehrfritz,B.A.F. 57,95,96 Weidmann,R. 106 Weinbaum,C.M. 73 Whitehead, J.H.C. 75 Wilkie,H.C. 104 Wise,D. 63 Zamboni,L. 34 Zieschang,H. 19,21,23,60,76,104,106,253,264,266

317