3-characterizations of finite groups

3-CHARACTERIZATIONS OF FINITE GROUPS N. D. Podufalov

UDC 519.44

We shall consider only finite groups.

By a 3-characterization of agroup, we shall

mean the determination of the global properties of the group itself, starting with the structure of its 3-1ocal subgroups, or of other totally defined subgroups, whose orders are divisible by 3.

An important step in the development of the methods of 3-characterizations

was the description of the simple groups in which the centralizers of elements of order 3 have odd order [I, 2].

By the same token, the long-standing problem was solved of describing

C@~-groups, and groups containing a strongly isolatedsubgroup whose order is divisible by 3.

To begin with, these studies were made within the limits of an odd characterization of

simple groups [3], without a definite program. Recently sufficient information has accumulated on the 3-structure of simple groups, and (to first approximation) a program has been planned for 3-characterizations, analogous to the program for even characterizations.

In order to outline the basic details of the

picture formed by 3-characterlzations of groups, we shall need the following: Definition i.

Let

is called ~-fused if

X be a group, P

a prime number, and

PG3~p(Op;pC~)~ The

group

~x(P)~Op~p(~).

Now let ~ be the group we are studying, and let 5 ~ ( ~ )

(it is known [15]

wise the only insoluble compositional factors of ~ are the groups

Sz~)

that other-

). The methods of

studying ~ depend basically on which of the following two situations we are in: A) ~ has an element of order 3 whose centralizer is not 3-fused; B) all the centralizers of elements of order 3 in 6 are 3-fused. Therefore, we naturally have two different directions of study, corresponding to these situations. We consider the first case. Let ~ be an element of order 3 in 6 , and let C6(Q) portant results are obtained here when

C6(Q)has

not be 3-fused.

small 2-rank.

The most im-

The class of simple groups

with such a centralizer of an element of order 3 is interesting because it contains many sporadic groups. 3-subgroup PGA(%,~)

It was proved in [4] that if the simple group ~ has an elementary Abelian

P of order 9, and for any nonunit element Q 6 P we have or P Z ~ , g X

# is odd), then

CG(o)/~ PSL(~),

G is isomorphic, to one of the groups

PSO(4S),

PSL(3,~), M~, M~, Translated from Algebra i Logika, Vol. 18, No. 4, pp. 442-462, July-August, 1979. Original article submitted November i, 1978.

0002-5232/79/1804-0271507.50

© 1980 Plenum Publishing Corporation

271

Durakov [5] proved that if

C(o>m×PSL:z,#),~>5, and

the centralizers of elements

of order 3 not conjugate with ~ are 3-groups, then the simple group

or

G is isomorphic to

PSOcs, g>.

An interesting characterization of the group

~

is obtained in [6].

The program for studying situation A) was stated in general by Collins in [7]. Bearing in mind the results of Aschbacher and others for groups of component type, Collins imposed the completely natural additional restraint; the 2-1ocal subgroups of

~ are 2-fused.

On

the basis of this assumption, Collins proposed to obtain results analogous to Aschbacher's in groups of component type (see A. I. Kostrikin's critical review, RZhMat., 8A220, 1978). Information is also given there on the structure of centralizers of elements of order 3 in a series of known simple groups, together with some analogs of Aschbacher's theorems. The direction of study corresponding to situation B) has been developed rather more slowly (in the same way as even characterizations in the case of 2-fused 2-local subgroups). The methods of these studies basically depend on the fact whether some 2-1ocal subgroup has noncyclic 3-subgroups, or not. Without separating the cases A) and B),

Gorenstein and Lyons [8] outlined a program

for describing groups whose 2-1ocal subgroups have cyclic Sylow 3-subgroups, and made the first steps towards its realization.

We

studied the group

6 for the following hypotheses:

i.

The Sylow 3-subgroups of the 2-local subgroups of

G are cyclic.

2.

The 2-1ocal subgroups are 2-fused, without kernels and SCN/z)#~(i.e.

G has char-

acteristic type 2). 3.

The insoluble compositional factors of any subgroup o f G are known groups (from a

given list of groups). Of course the restrictions placed on the group G in hypotheses 2 and 3 are completely natural. The case when some 2-1ocal subgroup of G contains a noncyclic 3-group (for situation B))

has been considered in many articles.

group G

satisfies the above hypothesis 2 and ~&$~i~(~)

equal to 2, or ~ groups

For example, Mason [9] showed that if the simple

lies in some 2-1ocal subgroup.

P~p(q,5~) and P~p(6,Z)were

, then either the rank of R

is

Interesting characterizations of the

obtained in [i0, ii].

Up till now there is no precise program for studying situation B), but we may state a series of problems which in turn need to be solved.

These problems concern the description

of simple groups with a strongly 3-1mbedded subgroup. Definition 2.

pe~(H),

but

The proper subgroup

p~(HOH9

H of the group ~ is called strongly

for any element

Z6G

p-imbedded if

\H

Questions of the existence of strongly p-imbedded subgroups were discussed in [12], where it was shown that in case B) such subgroups occur quite frequently.

272

Excluding the description of groups with a strongly isolated subgroup whose order is divisible by 3, until now few results have been obtained on the relations of groups with a strongly 3-1mbedded subgroup.

One of the most interesting is Aschbacher's result [30].

He

proved that there does not exist a simple group with a strongly 3-imbedded 2-1ocal subgroup of 3-rank

~4, and the groups of characteristic type 2 whose compositional factors are

proper subgroups, and groups of known type. Since each simple group with cyclic Sylow 3-subgroups has a strongly 3-imbedded subgroup, it is natural to study the case of lower 3-rank separately. Problem i.

Describe the simple groups of 3-rank ~ 3

We state the following:

containing a strongly 3-imbedded

subgroup. Clearly, the infinite series of known groups with such a subgroup of 3-rank ~

Lz(3n)~U~(5~) o

exhausted by the following:

are

and groups of Rie type.

It is natural to try first to obtain a characterization of the given known groups. Here we shall prove the following results. THEOREM i. If

36~(~(Q))

Let

G contain the strongly 3-1mbedded subgroup

H, where Q6S~LL(H) .

. then one of the following statements holds:

(I) the Sylow 3-subgroup of (II) the factor-group

G is cyclic and

6 is a 3'-closed group;

G/0~,(G) contains a normal subgroup ~

where ~ is either a group of Rie type, or

of index coprime to 6,

~m~z(5~÷~)j ~

(hi) 6/0~,(61 ~Aa~(L,(89. We note that in the last case,

G/O,,[~)=zGz(3)

is the least Rie group over the field

of three elements. COROLLARY i.

If G is a simple group with a strongly 3-imbedded subgroup

H , then

~(Z(H)I COROLLARY 2.

If G

is a simple group with strongly 3-imbedded subgroup

order of the Sylow 2-subgroup of type, or is isomorphic to COROLLARY 3.

L,(q~

a) The group

H,

and the

~ is not greater than 2, then 6 is either a group of Rie . ~(z~), or

~ ( Z ~) .

~z(3"), a > I , is determined by its strongly 3-imbedded sub-

group, to within isomorphism; b) a group containing a strongly 3-imbedded subgroup isomorphic to a strongly 3-imbedded subgroup of some Rie type group

is itself a Rie type group (of the same order).

It is interesting to obtain the same characterization for the groups We shall use standard definitions and notation. Proposition i. a subgroup of

Let H be strongly

p-imbedded in

~5 (3~).

We first state the following. ~ , let ~ be a homomorphismand

~ . then:

273

a) if

Hq~G~and

p~(H~), then

H F is s t r o n g l y

and

then D is strongly

b) if D = H O ~ & M Proof. H ~.

p6~'(~),

GV

p-imbedded in

p-imbedded in

.

Let N,be the kernel of ~ and let ~ be an arbitrary nonunit

Let P

be the Sylow

~-imbedded in

satisfied.

p-subgroup of

p-subgroup of the complete inverse image of P belonging to ~ .

By Frattini's agrument, the inverse image of ~G~(~) coincides with strongly

~

6, then

~G(~)~

, i.e.,

#'~G(P)

NG~(~)~{~

Since ~

is

Thus, statement a) is

Statement b) is trivial.

Proof of Theorem i. example to the Theorem.

Suppose that Theorem 1 is false, and let ~ be the minimal counterClearly, it follows from the condition that

3~[C(~}[ such

that

..IlICH (Q)I. LEMMA i. a')

The group

~ satisfies oneof

the following conditions:

~ is a simple group;

b') ~ = 6 o A < a > , where G o is a simple group, and ~l is an element of order 3 in

C.(Q). Proof.

Since ~ is a minimal counterexample to the Theorem, then using Proposition I,

we easily obtain 0,,(G)--<'> no~mal subgroup of

Suppose that 6 is not a simple group.

6 • Clearly, ~h6oa

element of order 3 in CH(~).

If

Go is a simple group and Cd%>~,>

be a m l n i = l Let a be a n

06~o , £hen the conditions of Theorem 1 are satisfied for

Go

and its strongly 3-1mbedded subgroup

Go

is of Rie type or ~o s

morphisms of

Let %

Lz#~"),

~N6o.

till, and

Since 6 is minimal, we see that either

6~Au~(6o). Using

the properties of the auto-

60 and the existence of a strongly 3-imbedded subgroup, it is easily shown

that 91 ~: 6°I.

ZIIG:~oI

Suppose that

Au~(~ o) (I~'~oIJ6)=f ,

. Since

~ is minimal, it follows that

I~:~ol= ~ . Using the

structure of

again, we obtain a contradiction with the condition

Thus

i.e.,

of 6 • Therefore, assume that case

~

~#~o.

~o~Lz(8)

and

LEMMA 2. Q ~ Proof. to

~o

If

l.z(3z~'')

• Since

Since C(Q)

~

. then Lemma 1 is proved.

~ is minimal, we obtain , then

6=60<0>

~>{.

Go ~

Therefore, we may

~u~(Lz~g)) •

. This contradiction proves the Lamina.

Thus

~ or 6o

is isomorphic

~ must be a simple group, i.e. isomorphic to Lz(3 ~

We may assume that

t~

It is easily

~/%* f .

)

contrary

. and Lemma 2 is proved.

Q e5~/IL(6) , C.,Q(L)~,~'yE~.('6~L)).Clearly, l(Q)e3~'z(6(@)) [6 ~

such that ~({)c_H; then

contains somenonunit 3-elements of ff,then~(Q)m~(Q)'(~ fIN(Q)).

Lemma is proved.

Since in this

Here ~o is the group from the preceding Lemma.

Let there exist an involution

shown that ~ N N(Q)

274

~ = ~o

<4> .

verified that in this case

Proof.

If

6~u~(~o)

for some

LEMMA 3.

~ satisfies statement (II) of Theorem I, contrary to the choice

~=<4). then from the basic theorem in [i], either

to the choice of

311C,(~)~

contains some Sylow 2-subgroup of

~(Q)

Thus Q e S ~

I t iS now easily zC~), and the

LEMMA 4. Proof.

Let ~ be an involution in ~

Suppose that the Lemma is false.

the Lemma, and chosen so that the number the choice of

~ . C~O

~zC~,C~I~)))

IC~O!z is maximal.

Since the Sylow 3-subgroup of

~,(~¢o)'C(E)4C(g)

~Z

HOC~o

contradicting

By the minimality of G and

is noncyclic,

C~¢i)~ ~,[HOC(L)) and

Using the structure of

easily shown that

Let

Let L be an involution in Q

satisfies condition (II) or (III) of Theorem i.

loss of generality, we may assume that

3-rank

and let C(L) have 3-rank ~Z; then Cci)--cH .

511du{~)l, it is ~(i)~H

, C(~) has

and both these properties hold for the centralizer of any subgroup of

[e~z~(L)),T~

~

Clearly, the factor-group

[/5

Without

Clearly,

~(O and the fact that Thus

~e

T~%,(~(LgG~.

ToC-~

I~o;%,(~(O) ~{5)I is not divisible by 2 or 3.

Set

To.

is an elementary Abelian group

of order 4 or 8. Suppose that of index 2.

~z<~)

Clearly,

z~Z(~)nqOCT). Since ~ Thus

~

and

and let ~

~(T)~To

If

In this case,

he the subgroup of ~ containing

~(T)#<1>

~(~)~(~)

and T

then there exists an involution Hence ~(Z) has 3-rank >%, and

l~)lz~I~/>ITI=IC~oI,

~(T) =<~)

T as a subgroup

C(E) ~

. we obtain a contradiction with the choice of

is an elementary Abelian 2-group.

By an analogous approach we show that ~ O T~= <~> for

x ~ % \T

i.e. that ILl ~ E

Let ~[t) satisfy condition (III) of Theorem i, i.e. C~)/~,(~(~)) ~ ~ut(~z(~)) Proposition i of [13], we have ~g

, then

ture of F

ITol >~.

Since

~ is an elementary Abelian 2-group of order

0~,(~(~)) has a normal 2-complement.

~(~)

By

Thus,

C~T) =

it is easily seen that the factor-group

isomorphic to the Frobenius group of order 21.

Using the struc-

T ~ OCt(T))

contains a subgroup

~=~(T)/f(T)

Let ~ be an element of order 3 in ~ .

Without loss of generality, we may assume that under the natural action on T, the centralizer of the element ,_2 in [ is equal to Clearly any element in x<~>

fixed.

order of

~z)

have

~, which centralizes or inverts ~ , acting in a natural way, leaves

Using the choice of L, and the fact that is an odd number.

, we easily see that the ~ (for otherwise, we would

C(g)qH) , it is easily verified that the Sylow 3-subgroup of N

of ~(~) ~ ~(T)

in ~

normal kernel

~

has a normal 3-complement

for any

~ .

Clearly, the image

of the Frobenius group F , where the centralizer of ~

in T

is equal to

Fa in A4 is not greater

Hence we see that ~ does not contain any sections isomorphic to the group $ ~ ) ~

and thus M

is a soluble group.

Arguments analogous to those for ~ z )

show that

element /~acts regularly on some Sylow 2-subgroup of ~ Using the above properties of the subgroup in M

is cyclic, i.e. bear-

is a group of order 7, and we may assume that it coincides with the

Since 7- contains no more than 63 involutions, the index of

than 63.

g

i~>%

Since T does not belong to

ing in mind the preceding arguments, N

To •

L x <j>, where j is some involution in T,

N~(g)

has odd order.

, then I ~

Since the

is some power of 4.

M to analyze the possibilities for the index of

we see that the only possible case is

l~:~aI=l~.

In this case,

I ~ = ~ ' ~ ~.

More-

275

over, it is easily established that the Sylow 7-subgroup normal in

~

, and therefore is normal in N .

K of M is elementary Abelian and

A4~NII(C-6L(z,~) ,

Thus

which is im-

possible.

C~i)/O~,(6~i))cannot

We have shown that the factor-group Thus,

C(i) satisfies condition (If) of Theorem 1.

C(i)/O~,(C~i))

be isomorphic to Ao~(/.~(g)~ •

We first suppose that the factor-group

contains a normal subgroup isomorphic to a group of Rie type.

Since in a

group of Rie type, the centralizer of any involution is isomorphic to the direct product of the cyclic group of order 2 and the group / z(5~), for some odd involution j e t

the subgroup

~(J) has 3-rank >2.

G >f, then for an arbitrary

Since in a group of Rie type a strongly

3-imbedded subgroup coincides with the normalizer of some Sylow 3-subgroup and is therefore maximal, then C ( j ) ~ ~ for arbitrary involutions for jeZ(Tf)/~T

.

Theorem in [13], we obtain

Set

~o=C~i)

structure of

I~I=~.

As it was remarked earlier,

C(i) , we may choose

~(T)c-~ ~

IQt=Irl .

~

Let the element

Since

To~-~o . Using the

be such that Q % T .

~D(T)NI(T)NTo[~Q

proved that ~ =~o,

of ~(~) . Moreover, 3 divides the order of

~OC~E)

and

is a proper subgroup of C(~) , then it is strongly imbedded in

C(~)O~.

Therefore,

IT:QR[~
tion isomorphic to ~#.

Since

~ of To , the group

be a nonunit subgroup of To

~To

is minimal, the subgroup

Let ~ 6 ~ 2 ( ~ ( ~ ) ~ .

As before,

~(~) normalizing

, ~(~)

has 3-rank ~l .

N(~)/O~,(~/(~))~(~ ~)

~,, centralizes ~, .

find a 3-subgroup of ~(~)

By Theorem

i.e., 3

We have

is an

~(8) is an ~9 free

such that

~(~) contains a sec-

It is easily verified that

.satisfying the condition of Theorem i.

~(~) satisfies statements (II) or (III) of this Theorem.

Sylow 2-subgroup is Abelian), and ~[~)

tion shows that ~(~)

E6~o=-Q.

C(~)=-~ , which is impossible.

is a strongly 3-imbedded subgroup of N(~) 6

Moreover,

i.e., Toc ~ c T .

Suppose not, and let ~

~(8)OH

C(E)=-H ~

Hence ~ = H ~ and

we now show that for any nonunit subgroup group.

~

~ 7 - ( T ) O T o , then the structure of

However, it is obvious from the structure of C(~) and from the fact that

divides the order of

is a {%~]~-group.

Since

~(~)/0~,(N(~)) is an

~-free

has a section isomorphic to ~

which normalizes but does not centralize ~-free

Thus, any 3-subgroup group (its

, it is easy to

~, . This contradic-

group.

~ of [15], ~ has a section isomorphic to ~

• Let ~ be a 2-subgroup of

of maximal order such that ~(~) contains a section isomorphic to ~

276

C~) is equal

IT:Qol=4. Suppose that Q ~ o .

On the other hand, ~(T)=-~o . Thus the subgroup

that such a situation is impossible.

of

Using the

This contradicts the hy-

0 o 6 S y ~ ( C , ti)) and

T D- qo and

m such that or

C(E) is analogous to that Q~-H~NC(~). If ~f~C(~-)

~(g)/0j,(~i)~

T~(6).

is nonunit and contains some involution

6(E).

in particular,

Thus, the order of the Sylow 2-subgroup of

Therefore,

In this case, either

Clearly,

~ .

/~z(3m~*f~, tl>/4. and index coprime to 6.

Applying [14], we see that this case is also impossible.

pothesis that T ~ L ( ~ ) .

Since

and elements

This contradicts the choice of ~ . Thus the factor-group

has a normal subgroup isomorphic to

to 16.

jET

•

By the choice

of

, we have

Since

~=0z(N(~)), and the factor-group

~ ) = - T o and

unit subgroups of

~D(~)~ N ( ~

To,

Q~o

the subgroup

• Therefore

~ ( ~ = < I > . i.e.,

ITo:Q~=I,

C(~)~H,

and thus

Rc-Q .

~(~)/03,(C([))either contains a type, or is isomorphic to ~U~(~z(t)). In the Therefore, N(~)N ~ is a strongly 3-imbedded 1. Hence N(~) satisfies condition (II) of

N(~)~H

Now, arguing in the same way as when we were studying the structure of N(~) ,

we obtain a contradiction with the fact that there exists a section in ~q

ToC(To)/O(C(~))=-~t(i~(fOxTo

also "C(~J =-H

, which is impossible.

.

T=ToC R)

Hence

Therefore

C(~)~H

~(g)

a

{~,3,f-group and

, it is easily seen

Hence, if C ( ~ ) ~ H

, and thus

see that N(~) satisfies condition (If) or (III) of Theorem i.

N(~)/C(R) is

~(~) isomorphic to

C([)/O~,(C(£))~t(/z(~J)

. Therefore,

Using the triviality of the Schur multiplier of the group that

Since R~To.

and the factor-group

subgroup satisfying the condition of Theorem Theorem i.

~ is an elementary Abelian 2-group.

In this case we may assume that

normal subgroup isomorphic to a group of Rie first case

has no sections isomorphic to Sq .

. then by the above-proved property of normalizers of non-

Suppose that W(k) has 3-rank >~2. then

N(~)/~

~(~

We again

Since the factor-group

~[~)/R contains no sections isomorphic to

easily obtain a contradiction with the fact that

, then

~q . then we

~(~) contains a section isomorphic to ~# .

We have proved that ~(R) has 3-rank equal to i. Suppose that 3 divides the order of the facts that

~

To and

~(R)

We may assume that ~c_Q.

C(~) has 3-rank i, we easily obtain the following isomorphism:

~c~)/O~,~C(~})=~t~L~cg)). Moreover, the minlmality of G , statement e istence of a section in

N(R)~ H .

we see that

since

(I) of Theorem 1 holds for

isomorphic to

N(~) has 3-rank i, then by

N(R~ .

Therefore,

This contradicts the

C(R)

is a 3-group.

We now suppose that an element of order 3 acts nonregularly in R . contains an involution j such

31JC,~j)~.

assume that

Since

that ~)(j) Q~Lz(~) .

a strongly 3-imbedded subgroup of is minimal and 0~)

then by Lemma

C(j) , satisfying the condition of Theorem i.

Thus the elements of order 3 acts regularly in ~ . Suppose that the group

l~l'~l-z .

Since

N(R)

is 3-soluble.

[16, 17], we easily obtain

~

Since all the 2-1ocal subgroups of the factor-group

~x~/

and

C~')/1.'H is Since ~G

N[~) . we obtain a conSince ~ , is maximal,

Let ~I£S~z[N(~J)

~6

In thls

I~>4

. Hence

~-J(~)

and

The simple groups with such Sylow 2-subgroups were described in [18] and [19].

We obtain a contradiction with the choice of

~(~(~))

We may

~T(R)-~, then using the description of groups with dihedral and

polyhedral Sylow 2-subgroups

R,~$F~(O).

R

is not ~-free, we see that C(J) satisfies condition (II) or (III) of

tradiction.

case,

'

3, C~)~.H . Clearly,

Once again, as when we were studying the structure of

.

In this case,

contains a section isomorphic to 4

Theorem I.

~z(C(~))

As before, using

~(~)

. where

Thus the group N(R) is not 3-soluble.

/~(R)/~ are

has only one insoluble compositional factor.

~-O(N(~

3-soluble, then 03, ( N ~ ) ) By the above,

Using Lemma 2.1 of [20], and the fact that

N(~)

ments of order 6, we show in the same way as in [20] that the factor-group

~(~)-

has no ele-

~($)/~ (N(~))

277

Z,2(2r"). Applying

contains a normal subgroup of odd index, isomorphic to

Theorem 2 of [21],

we obtain a contradiction with the choice of Lemma 4 is proved. LEMMA 5. Proof. 3-rank

~

contains an involution ~ such that

Suppose not.

C(i)~ H

.

By the preceding Lemma, the centralizer of any involution has

~{ . i.e. either is a 3-group, or has a normal 3-complement (in the second case,

statement (I) of Theorem 1 holds for Cl~) ).

Theorem C of [15] shows that the insoluble

compositional factors of the centralizers of involutions in ~

are exhausted by the groups

SZ(9)

The fundamental result in [22] guarantees that all the 2-1ocal subgroups of ~

2-fused.

Since in G we have

~GNj(2)'/=~ (see

tains normal subgroups of odd order [24].

are

[23]), then no 2-1ocal subgroup of ~ con-

We again use the fact that ~

is not an S~-free

group. Let ~ Clearly,

be a maximal 2-subgroup whose normalizer contains a section isomorphic to ~

~=0~

(N[R)), and

we may assume that Rc~.

the factor-group

N[R)/Ris

~-free.

If ~{R)

Hence ~(~) contains an element of order 3.

fact t h a t A/(~) is 2-fused and

0{~[~))- < 4 >

has 3-rank ~ 2 ,

This contradicts the

Thus, the 3-rank of ~(~) is equal to i.

Using the structure of the centralizers of involutions, we see easily that no elements of order 6. ~,E$~2(N(~)) • closed in

~.

We first suppose that

Clearly,

and hence

J~:~J-2. ~I£$~z(~)

N(R)

Since J~J>@

is a 3-soluble group.

~(R) contains Let

, by Lemma 8.1 of [3], ~ is weakly

. By the eemma in [i], any 2-1ocal subgroup of ~

is

either soluble, or has an insoluble compositional factor whose order is divisible by 3. The basic theorem of [25] allows us to state that ~ has an insoluble 2-1ocal subgroup. Let D be a maximal 2-subgroup such that A/(P) is an insoluble group. that the 3-rank of N ~ ) order 6.

is equal to i, and that

We show again

~(P) does not contain any elements of

As at the end of the proof of Lemma 4, we easily obtain a contradiction, using

Theorem 2 of [21]. Therefore

2J{R) is not a 3-soluble group.

We now study ~/(R) using the

methods in the proof of the preceding Lemma, and obtain a contradiction with Theorem 2 of

[21]. Lemma 5 is proved. LEMMA 6. ~ 6 ~ ' z (~)' Proof.

Lemma 6 follows immediately from Lemmas 3 and 5.

LEMMA 7. Proof. ~(i)~ ~

For any involution Z6 ~, we have ~(Z)'=~C(~))I#/o~(i~) Let ~ be an involution in ~

•

If ~(~)c_~. then the Lemma is true.

By Lemma 4, ~(Z) has 3-rank equal to i.

Set

~6'~21~i~)O~).

of Theorem i, ~(~)nH

contains an element I of order 3 such that

CHnC(~>(~)=~HnC~/~ ) .

Since

~,(~)=~H~C~)(~)'

~(~) has a normal 3-complement.

278

then

~(~) ~_~

Now let

By the condition Therefore,

~C~ (~)=~C(~) (~)" Hence we see that

Thus the insoluble compositional factors of the group

C(~)

are exhausted by the groups

~)

Since the components of the 2-fibers of

C(~ are

balanced groups (for the definitions of balance, 2-fibers and components, see [24, 26]), and there exists an element of order 3 in

C(~ . centralizing some Sylow 2-subgroup of C(~

then applying Lemma 2.3 of [24], we see easily that

,

0(C(~)) contains an element of order 3.

Applying Frattini's argument, we complete the proof of the Lemma. LEMMA 8. Proof. of

N(~.

Let ~ be a nonunit subgroup of ~ . If N (R) has 3-rank We suppose not, i.e. ~ I R ) ~ H

.

Then

satisfying the condition of Theorem i.

satisfies condition (II) or (III) of Theorem i. group. 4,

Let the involution

6(L)~H

N(R) ~ H

~eZ(R).

~ 2 . then N ( R ) ~ H .

NH (R) is a strongly 3-imbedded subgroup Since ~

Therefore

is minimal, the subgroup ~(R)/O3,

I~(R~(R) is

a

It is easily seen that [(Z) has 3-rank ~Z.

Using the structure of the group

~(R) {2,3~r-

By Lemma

~(R) and the latter inclusion, we obtain

This contradicts our assumption.

The Lemma is proved. In Lemma 5 we showed that ~

contains involutions whose centralizers lie in H

Since

such involutions will play a substantial role in our further arguments, we introduce the following. Definition 3.

We call the involution i6~

exists an element ~ e ~

such that

~(L)~ H

an involution of the first type, if there

• All the remaining involutions are called in-

volutions of the second type. Clearly, the set of all involutions of the first type is normal in LEMMA 9. Proof. hand,

Let Z be an involution of the first type in H . Since ~ is of the first type,

311~H(g)I

.

Then

~(g)~H ~ for some element

Thus 311HNHzl, and hence

6 •

~(L~H ~ ~

. ~ On the other

H=H

The Lemma is proved. LEMMA i0. Proof.

Any four-subgroup of ~ contains an involution of the second type.

Suppose not, i.e.,

tions of the first type. whose centralizer in ~

~ contains a four-subgroup V which contains only involu-

We show that there exists an involution of the second type in contains a four-subgroup analogous to V .

has a normal elementary Abelian subgroup ~ of order 8. of the first type.

Let all the involutions in E be

Since there is no strongly imbedded subgroup in ~ , ~

volution j of the second type.

Clearly,

let there exist an involution L in ~ order not less than 4. require, or in

By Corollary A of [23],

Hence, either

~E(V)~cE(L)is

Suppose now that

contains an in-

~E(J) contains the same four-subgroup as

of the second type.

~E(V) contains

If V n ~ @ < 4 > .

then CE/V)

~.

Now

has

an involution of the second type, which

the required four-subgroup.

V O ~ =<~>.

If ~E(V)

is a noncycllc group, then analogously to the

above, we find the required involution of the second type.

279

CE(V) =

Let group

<~>xV

be a group of order 2.

are of the first type.

Let the involution

the involutions in the group (<~>xV)c~> <~> x <~>

We may assume that all the involutions in the

are also of the first type.

~eNE(xV)\(<~>~V) .

If all

are of the first type, then all the involutions in Then L is our required involution of the second type.

If, however, there is an involution of the second type in the group (<£~>xV)<~> , then clearly it is our required involution.

Therefore, without loss of generality, we may assume

that Q contains an involution j of the second type, such that any involution

if6 V we have

C(~)-cH

V~(j)

•

By Lemma 9, for

Now, using Lemma 7, it is easily shown that C ~ ) ~ H .

This contradicts the fact that the involution j is of the second type. Lemma i0 is proved. Denote by T the subgroup of Q generated by all the involutions of the first type lying in

@.

Clearly,

LEMMA ii.

T is weakly closed in Q with respect to

The group

6

•

T contains the subgroup To with the following properties:

IT:ToI=

i)

2) all the involutions in 3) To

To

are of the second type,

is weakly closed in ~ with respect to

Proof.

By Glauberman's

Z~-theorem,

~ . and in particular,

~4

N(Q).

Q contains two involutions of the first type.

Consider the group generated by these involutions; using the preceding Lemma, it is easily shown that

Q contains involutions of the second type, whose centralizers in ~ contain not

less than two involutions of the first type. we choose an involution

Among such involutions of the second type,

L for which the number I~(i)lz is maximal.

Without loss of generality, we may assume that Qo 6 ~ z ( C ( f ) ) • L,~H

Let < 4 ~ = ~

~/L,,/~

is a minimal

~+'/~/i

are

~/~c..ck/~/

mal natural number less than n such that

easily seen that

~+f/~

~+~/~

C(L)/~/~ = ~ + ~ / ~ K / ~

,

pL"

Let

such that

K be the maxi-

and =(C(~)OH)W~ ~C(g). ~/~ and

(~(~) f]H)/WK. The isomorphism

~/K+4 .

~o ~- ~ o ~ / ~

~ as a group of operators on the factor-group

f] (C(;)O~)/~/=<~

, i.e. , they invert

(C(L)fIN)~/~+I=C(£)

C(L)

Since ~/ is a soluble group,

Using the choice of the subgroups

in a natural way to consider Since

By Lemma 7 and the choice of

Abelian pL-groups, for some prime numbers

Such a K exists, by Lemma 7.

~/=0(~(i))~ ~o=C~(L).

be a series of normal subgroups of

normal subgroup in C(L)/~,/~, L= t,..., tt-t.

elementary

Set

the involutions of the first type in

it is allows us

~/~+~/~

~o act regularly on

each element of this factor-group.

Let ~ be the group generated by all the involutions of the first type lying in ~o By the choice of L, we have

I~I ~

tions

~/~/~

of the first type on

tions of the first type.

•

Let

Ao=C~(~+~/~)

, we show easily that

Let

D=Ao/~

.

. Using the action of involuI~:Aol =% and ~ o

Suppose that A has another subgroup

involutions of the first type.

280

.

has no involu-

A~ of index 2, containing no

• Since the factor-group

A/D

is elementary

Abelian of order 4, and ~o and ~I contain no involutions of the first type, it is easily seen that all ~he involutions of the first type lie in the same conjugacy class of the factorization of ~ by D , i.e., do not generate Thus Ao

A • This contradicts the choice of

A.

is the unique subgroup of index in A containingno involutions of the first type.

Hence ~o is weakly closed in A with respect to

G .

AoN[Qo)

Clearly,

If 0 # ~ o

•

then ,Z(~ o) N Ao contains an i n v o l u t i o n of the second type, whose c e n t r a l i z e r in Q s t r i c t l y contains

~0"

Thus Q=Qo.T=A

This contradicts the choice of L .

and

To= o is

the re-

quired subgroup of T . Lemma Ii is proved. LEMMA 12.

The normalizer of any nonunit subgroup of

~o is not contained in H, and

in particu!~r has 3-rank equal to i. Proof.

It is clear from the proof of the preceding Lemma that To centralizes some

subgroup (of odd order) not lying in not contained in LEMMA 13.

H Let

~

. Thus, the normalizer of any subgroup of T O is

• Applying Lemma 8, we complete the proof. R be a nonunit subgroup of Q

such that

N(R) contains a section iso-

morphic to $4 ~ . Then ~ contains no more than one involution of the first type. Proof.

Since

the 3-rank of

N(~) contains a section isomorphic to S~, it is easily verified that

N(~) is greater than i.

of the first type, and let type lying in

RI be the subgroup generated by all the involutions of the first

~ • Clearly, ~ I ~4 and ~ f q T

R o is weakly closed in i.e., N(~Q)

Suppose that ~ contains at least two involutions

. set ~o ~ I N ~.

It is easily shown that

~i with respect to G (see the proof of Lemma

has 3-rank greater than i.

ii). Hence ~o~N(~),

This contradicts the preceding Lemma.

The proof of Lemma 13 is complete. Let ~ be a subgroup of maximal order satisfying the condition of Lemma 13. LFJ~MA 14. Proof.

The subgroup

By Lemma 13,

~

contains precisely one involution of the first type.

~ contains no more than one involution of the first type.

pose that this Lemma is false, i.e., there is no involution of the first type in R . the maximality of R , we have

N(R)

First suppose that

~

eS/z(Oz, z(N(R))), C(R)~Oz;z(N(R))

is a 3-soluble group.

Let

and

SupBy

~eS~/~,CN(R))).

PeSE~,(O3;~(N(R)))and 5eJ~J,~,(NMcR~(Pg.

Clearly, RP$ contains some Sylow 2-subgroup of ~(~) . Since ,~(R) is 3-fused, we have ~(P) = ~ n 5 . By Theorem 3.13 of [27], P has a characteristic subgroup 9 of nilpotency class ~ 2

and exponent 3, such that

Our problem c o n s i s t s

act regularly on

}5/RNS~= i

5/~N~

o f showing t h a t

acts exactly on n

IS/~n51 =e

~ , then by Bernside's well-known

• Therefore, we may assume that ~

irregularly on ~ .

•

If all

theorem, ~

• the nonunit elements in D

is a cyclic group, and hence

contains an element a~of order 3, which acts

Since all the involutions in ~ are of the second type, we easily see

that ~ acts identically on ~ . Since the centralizers of involutions in ~ have 3-rank l, all the elements in ~ \ < x >

act regularly on ~ .

Hence the factor-group

~/<~>

acts on

281

"0/<~> is

like a group of regular automorphisms, and once more a group of exponent 3, then

~DI<~>I = 3 and ~

, we easily obtain the equation

Sylow 2-subgroup of the factor-group

N(R)/R

.

15/R~S~ =I.

~S

acts regularly on R .

~56~z(N(R)) ,

Since

Recalling that

S/RN5

acts

has order 2.

fles the conditions of Theorem 8.1 of [3], and of index 2 in

D is

Thus we have shown that the

Moreover, in any of the above cases ~ has a subgroup acts exactly, and the element ~

Since

is elementary Abelian of order 9.

It is easily seen that S/R O 5 acts identically on exactly on ~

a cyclic group.

<~> of order 3, on which S/R~$

Clearly, the group

IR~>@.

Thus

then also

R<~>5

satis-

R is a weakly closed subgroup

~S ~ $~@z(~)

Thus, any Sylow

2-group of ~ contains a subgroup of index 2, which has no involutions of the first type. By Thompson's lemma,

6 has a subgroup of index 2, which is impossible.

a 3-soluble group.

We note that the factor-group

N =N(~)/R is ~-free,

has no 2-1ocal subgroups containing sections isomorphic to ~ Suppose that N has a 2-1ocal subgroup of [15], ~

contaims subgroups

is isomorphic to ~ z ( ~ ) , teristic subgroup of

~

~o and ~I •

Let

Thus As

,

~ is maximal,

•

~ which is not 3-soluble. such that ~ o 4 ~

~(~) is not

By Corollary 7.3

and the factor-group ~/~o

U by a Sylow 3-subgroup of

~i

, and ~ a charac-

~ , whose existence is guaranteed by Theorem 3.13 of [27].

By the

same approach as in the beginning of this Lemma, we show easily that ~ has order equal to either 3 or 9, and

~(~(M)/~CM~

structure of the factor-group 3-soluble. Set

In particular,

F=S(N(~))

groups of N

~/6o • Therefore, all the 2-local subgroups of N

by

~.

Let

f~Ob,(N(R)) ,

then using the 3-solubility of 2-1ocal sub-

~ of [15], we show that

simple group at the end of Theorem ~(~)

are

~eb~(S(~(R))). since

and Theorem

~f=,3J~ we now easily obtain a contradiction with the

Xe~,(F)

greater than 1 (Be=as i0 and 4).

~(~)/FG~u~(i)

,

where

~ is the

~ of [15]. Denote the complete inverse image of ~ in If ~

has 3-rank ~l.

then the 2-rank of ~(~) is not

In this case, the factor-group

a compositional factor isomorphic to ~

•

NN(R)(X)/~N(R)(X)contains

Applying now Theorem 3.13 of [27] and arguments

analogous to the above, we again arrive at a contradiction (with the fact that ~ bedded in the group of automorphisms of

X ), and thus

X is a cyclic group.

the methods of proof of the Theorem in [20, pp. 82-83] allow us to show that

K/Oz~2(~(R))~ [/Oz~Z(~)~L. If the index of a 2-subgroup of N(R) , strictly containing

~ ~

is im-

Lemma 2.1 and ~ e ~ z ( l n)

in N(~) is an even number, it is easy to find , with normalizer containing a section iso-

morphic to $4 " Therefore the index of ~ in ~(~)

is odd.

We now apply Theorem 2 in [21]

and obtain a contradiction with the hypothesis that the Lemma was false.

Thus Lemma 14 is

proved. LEMMA 15.

Let L be the involution of the first type in

anteed by Lemma 14. there.

282

Then i belongs to Z(Q)

~ . whose existence is guar-

and is the unique involution of the first type

Proof.

Let

R,eS~z(N(R))

R

Since

is maximal, we have

Z(~,) ~-~, and since ~ is weakly closed in K , then is weakly closed in

7(R,)

Let

~z~$~2(N(R,)).

ieZ(R,) Clearly,

C@(R)-c~

.

Therefore

By the preceding Lemma,

Z(~ I) c_ ~ and

Again, i is weakly closed in Z(R z)

Analogously, we show that if

then [ is weakly closed in I(RK+ ,) •

Therefore

~6 Z(~z) .

~K~,e S~z(N(~K))

L is a central involution.

Thus

,

7(Q)

contains involutions of the first type, and they lie in Z(~,) , i.e., coincide with

L .

The uniqueness of ~ follows from the preceding Lemma. Lemma 15 is proved. LEMMA 16. Proof. ii

Z(Q)

The group ~ is not of symplectic type.

Let the involution L be as in the preceding Lemma.

contains an involution of the second type,

is not a cyclic groupo

Since

~ e 7(Q) and by Lemma

Z(Q) is not a cyclic group. Hence, Z(R)

Thus ~ has at least one noncyclic Abelian characteristic subgroup.

The lamina is proved. LEMMA 17.

Let E

be an elementary Abelian 2-group of order not less than 8 in

taining no involutions of the first type. Proof.

Set

Then

VJE=

•

Since the only insoluble compositional factors ~E(9)

result in [28], W E

N(WE) = NEE).

is a group of odd order.

has 3-rank I, and thus so does

con-

~(£) has 3-rank equal to i.

of the centralizers of involutions in E are the groups

N(~/E)

Q

Clearly,

. then by the fundamental since

~E~H,

then

N(E).

The Lemma is proved. LEMMA 18. Proof.

The group G

Let A be a characteristic elementary Abelian 2-subgroup of ~ , and

Such a subgroup of ~ exists Since the 3-rank of P

does not exist.

of order 9.

by Lamina 16.

Set ~=

N(~) is greater than i, then

Since ~

N(R)

.

contains only one involution of the first type,

Eo contains no involutions of the first type, and admits

Hence

lEol=~

I~-,(ZI(~))=E

•

i~l,(Z(~)) ~

.

and since

N(R)/6N(R)(F)contains a

N(R)/6N,R~(E)~ A3 We have obtained IR,'RI=z .

Suppose that R

N(E)/C(E) is

By the choice of

.

we may assume that

contains an involution j

.

imbedded in

of odd order.

Since N~)_D N(£o) and

tion, as in the proof of Lamina 17.

N(Eo) has Thus

3-rank

_D_,(R)=E.

iz(~)

R,RE~,(&R(~). Let R,~Q

Then X=<Eo,j

WX=~Eo

then

, and is at the same

•

> is an elementary

Abelian 2-group of order 8, containing no involutions of the first type. from the proof of the preceding Lemma, we have

.

By Lemma 17, IEol
section isomorphic to %

time ~ - f r e e , we have RI&S~,(N(R)).

P •

~=~o

Z(~) is a noncyclic group,

Using the fact that the factor-group

and the factor-group

E4 N(~},

has an elementary Abellan 3-subgroup

where

• Clearly

By our choice,

IAI m~.

Using the notation

. i.e., once again

~Eo

is a group

>-%, we easily obtain a contradic-

Suppose now that

~,6~z(6)

. Since

283

N(E)/C(E)

the factor-group of ~,~eS~z(6(E)). other hand,

Hence we easily see that


RIeS~z(~(E))

.

Thus

NO(E)=~ , .

On the

~ 1 1 [ l ( Q ) ) , and I~,(Z(Q)) is a noncyclic group containing

Let the element

2 in

contains a section isomorphic to ~3 , then by the maximality

~Q(~,)\~I

• Set ~ = ~

.

Clearly, ~

is a subgroup of index

Hence it is easily seen that Eon°P( i.e., ~O(~) is either cyclic, or a group of quaternions. Therefore ~ = E o x K , li [D)l=4 and LEfI,(D).

Thus

where

~ contains the unique involution

Z(~)

.

since

and ~@(~)~ ~

Eo

& .

Set £o = xKj>,

Em~ E. then jZ~ ~ and ~I=~A<jz>.

)=

where ~ is an involution in

It follows from the fact that j&l(~)

, that I~:CRVx)~ = %. Considering the natural action of ~(~)/~(E)

-- the

dihedral group of order 6 -- on the subgroup ~ , it is easily shown that j z does not centralize E

Thus

CEoV z) = • It is now clear that K may be chosen so that

~6~V

Thus

CRV ~) = K x
a)

E o is the mutual commutator subgroup of

and some subgroup of N(~) of order 3, which inverts a 2-element not lying in ~ . fore, without loss of generality, we may assume that _¢11(~,) = ~ V z >

= (EoA<jx>)x<~>.

Since

contains precisely two four-subgroups

~o ~ ~I"

~o and E I

Thus,

and

Hence ~ 0 Hence

~

=

L~f

Simple calculations show that

~f(~1)

has precisely two ele-

~Ix • Clearly, these subgroups are

conjugate in Q by the element ~ . Hence we see that

~f does not contain an elementary

of order 8 which is normal in Q.

elementary Abelian normal subgroup ~

of order 8 [23].

On the other hand, Q

<~>x . Thus R is a subgroup of index 2 in ~ ~

Since i~,(~)=~

and

£~Z(~,)

has an

We may assume that A?ZI(ZC@)) ~ ,

Therefore,

. then R I has 2-rank equal to 3.

[29], we obtain the final contradiction with our choice of

~

.

i.e., ~R~N~(~)=~ I.

. and we obtain a contradiction with the preceding arguments.

~,&5~z(~).

There-

~oA<jaD is the dihedral group of order 8, then it

mentary Abelian subgroups of order 8; ~ox

Abelian 2-subgroup of Q

E

Applying

.

The Lemma is proved, and this completes the proof of Theorem i. Proof of the Corollaries Corollary 1 follows easily from the fact that it holds for groups from the end of Theorem i.

To prove Corollary 2, it is sufficient to note that if the group G

the conditions of the

Corollary, then either G

satisfies

satisfies the conditions of Theorem i, or

contains no elements of order 6 (in the latter case, we apply the results of [i]). Now let the group G contain a strongly 3-imbedded subgroup subgroup of Lz(5 ~)

for some ~ >I, or of a group of Rie type.

CG(P)~P

of some Sylow 3-subgroup

P of G . Since

It is easily shown that

~ contains a simple normal subgroup

since either ~ contains no elements of order 6 or

and

Clearly,

= ~ ~2(5~).

284

H

is the normalizer

P is noncyclic, then 03,(6) = <~>. L such that

C (L) = ~I> .

~ satisfies the condition of Theorem i,

then ~ is either isomorphic to Lz(~) , or is a Rie type group. Abelian 3-group, then P is strongly isolated.

H isomorphic to such a

If P

is an elementary

It is easily seen that in this case,

If P is not an elementary Abelian 3-group, i.e.,

~ is isomorphic to a

strongly 3-imbedded subgroup of some Rie type group, then L is a Rie type group. easily seen that

P ~ L and thus L = G .

Thus,

Now it is

G is a Rie type group (of the required or-

der) and Corollary 3 is proved. LITERATURE CITED i.

N. D. Podufalov, "Finite simple groups with no elements of order 6," Algebra Logika, 16, No. 2, 200-203 (1977).

2.

L. R. Fletcher, B. Stellmacher, and W. B. Stewart, "Endliche gruppen, die kein element der ordnung 6 enthalten," Q. J. Math. Oxford, 28, No. 2, 143-154 (1977).

3.

G. Higman, "Odd characterization of finite simple groups," Lect. Notes, Univ. Michigan (1968).

4.

M. E. O'Nan, "Some characterizations by centralizers of elements of order 3," J. Algebra, 48, No. i, 113-141 (1977).

5.

B. K. Durakov, "Simple groups with fixed centralizers of elements of order 3," Sixth All-Union Symp. on Group Theory, Lecture Notes, Kiev, No. 22 (1978).

6.

R. M. Stafford, "A characterization of Janko's new simple group ~ ," Notices Am. Math. Sot., 25, No. 4, A-423 (1978).

7.

M. J. Collins, "3-structure in finite simple groups," Proc. Conf. on finite groups, W. R. Scott and G. Fletcher (eds.), New York (1976), pp. 47-61.

8.

D. Gorenstein and R. Lyons, "Finite groups of 2-1ocal 3-rank at most i," Proc. Conf. on finite groups, W. R. Scott and G. Fletcher (eds.), New York Press (1976), pp. 25-35.

9.

G. Mason, "Finite simple groups of characteristic 2,3-type," Proc. Conf. finite groups, W. R. Scott and G. Fletcher (eds.), Academic Press (1976), pp. 37-45.

i0.

A. R. Prince, "A characterization of the simple groups PSp(~.3) and PSp (6,2)," J. Algebra, 45, No. 2, 306-320 (1977).

ii.

J. L. Hayden, "A characterization of PSp (~,3~)

by the centralizer of an element of

order three," Proc. Conf. of finite groups, W. R. Scott and G. Fletcher (eds.), Academic Press (1976), pp. 85-102. 12.

N. D. Podufalov, "The existence of strongly p-imbedded subgroups of finite groups," Algebra Logika, 15, No. i, 71-88 (1976).

13.

V. D. Mazurov, "Centralizers of involutions in simple groups," Mat. Sb., 93, No. 4, 529-539 (1974).

14.

V. V. Kabanov and A. I. Starostin, "Finite groups in which the Sylow 2-subgroup of the centralizer of some involution has order 16," Mat. Zametki, 18, No. 6, 869-876 (1975).

15.

G. Glauberman, "Factorizations in local subgroups of finite groups," Conference Board of the Mathematical Sciences, Regional Conference Series in Mathematics, 33, Providence, Rhode Island (1978).

16.

D. Gorenstein and J. H. Walter, "The characterization of finite groups with dihedral Sylow 2-subgroups.

17.

I, II, III," J. Algebra, 2, 85-151, 218-270, 354-393 (1965).

J. L. Alperin, R. Brauer, and D. Gorenstein, "Finite groups with quasidihedral and wreathed Sylow 2-subgroups," Trans. Am. Math. Soc., 151, 1-261 (1970).

285

18.

D. M. Goldschmidt, "2-fusion in finite simple groups," Ann. Math., 99, No. i, 70-117

(1974). 19.

P. Chabot, "Groups whose Sylow 2-groups have c y c l i c commutator g r o u p s , " I , J. Algebra, 19, No. l , 21-30 (1971), I I , 21, No. 2, 312-320 (1972), I I I ,

20.

29, No. 3, 455-458 (1974).

N. D. Podufalov, "Finite simple groups without elements of order 6 and i0," Algebra Logika, 14, No. i, 79-85 (1975).

21.

S. A. Syskin, "The centralizers of 2-subgroups of finite groups," Algebra Logika, 17, No. 3, 316-354 (1978).

STRUCTURES OF THE DEGREES OF UNSOLVABILITY OF INDEX SETS V. L. Selivanov

UDC 517.11:518.5

In this paper the construction of a partially ordered set (~m;&m) of m-degrees of the index sets is studied.

Not very much is known about the structure of this object.

2.6 of [i] shows the richness of this structure.

Theorem

Interesting results on index sets of

finite families are obtained in [2, 3], but these relate only to a small part of the set ~m •

The question of the possibility of representation of natural operations on ( ~ m ~ & m )

is studied rather little. respect to the operation

Actually, the only thing known is the closure of ( ~ ) -

with

induced by the operation of addition of sets, and in [i] is

shown the closure of ~ m with respect to the operation of pfn-cylindrification. We study the question of the possibility of representation of operations on the partially ordered set (~)mj~), operations are known:

of ~7-degrees of unsolvability.

~m, operating

Several of such

the binary operation U, an exact upper bound; a unary operation in-

duced by the operation of addition; the binary operation x induced by the operation of Cartesian product of sets; the unary operations

pmand

fn~

induced by the operations of

fn-cylindrification and ?~-jump are introduced in [4]. The main result is the definition of an analog of an operation

b) if the element CeJf~ for which

~t--<mC

c) if the element for which

of exact upper bound in ( ~ m ~ J .

is such that

~tnC

for any

It is proved that for any

K411, then one can find an

;

Ce Jfnis such that C ~ m ~

for any

~45,

then one can find a K4fl

d4m~.

The properties a) and b) generalize the property of an exact upper bound, and c) shows a certain discreteness is ( J m; ~ m ~

From a), b), and c) it follows that any two incompar-

able elements of (Jm; ~ m ) do not have an exact upper bound. Translated from Algebra i Logika, Vol. 18, No. 4, pp. 463-480, July-August, 1979. Original article submitted October ii, 1978.

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© 1980 Plenum Publishing Corporation