A Characterization of PSU3(q) for q>5

Southeast Asian Bulletin of Mathematics (2002) 26: 33–44

Southeast Asian Bulletin of Mathematics : Springer-Verlag 2002

A Characterization of PSU3 (q) for q I 5 A. Iranmanesh, B. Khosravi and S.H. Alavi Department of Mathematics, Tarbiat Modarres University, P.O. Box: 14115-137, Tehran, Iran and Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran E-mail: [email protected]

AMS Subject Classification: 20D05, 20D60 Abstract. Based on the prime graph of a finite simple group, its order is the product of its order components (see [4]). We prove that the simple groups PSU3 ðqÞ are uniquely determined by their order components. Our result immediately implies that the Thompson’s conjecture and the Wujie Shi’s conjecture [16] are valid for these groups. Keywords: Prime graph, order component, finite group, simple group.

1. Introduction Let G be a finite group. The prime graph of G is constructed as follows: the set of vertices is pðGÞ consisting of prime numbers dividing the order of the group, two vertices p and q are joined by an edge if and only if G contains an element of order pq. Denote the connected components of the graph by p1 ; p2 ; . . . ; pt . Then jGj can be expressed as a product of m1 ; m2 ; . . . ; mt , where mi are positive integers with pðmi Þ ¼ pi . These mi ’s are called the order components of G (see [6], [2]). We use OCðGÞ to denote the set of order components of G. If G is of even order, then we always assume that 2 is a member of p1 and we call m2 ; . . . ; mt
the odd  order components of G. Let GðGÞ denote the prime graph of G and let t GðGÞ denote the number of order components of G. The order components of non-abelian simple groups having at least three prime graph components are obtained by G.Y. Chen [2, Tables 1, 2, 3], and we present the order components of non-abelian simple groups with two order components in Table 1 by [12, 18]. The following groups are uniquely determined by their order components: Suzuki-Ree groups [7], Sporadic simple groups [4], PSL 2 ðqÞ [2], E8 ðqÞ [8], G 2 ðqÞ where q 1 0 ðmod 3Þ [3], PSLð3; qÞ where q is an odd prime power [11], Ap where p and p  2 are primes [9] and F4 ðqÞ where q is even [11]. In this paper, we prove that PSU3 ðqÞ where q > 5 are also uniquely determined by their order components, that is we have: The Main Theorem. Let G be a finite group and M ¼ PSU3 ðqÞ where q > 5 and OCðGÞ ¼ OCðMÞ. Then G G M.

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A. Iranmanesh, B. Khosravi, and S.H. Alavi

2. Preliminary Results In order to prove the main theorem, first we bring some lemmas. Lemma 2.1. ([18, Theorem A]) If G is a finite group with its prime graph having more than one component, then G is one of the following groups: (a) (b) (c) (d) (e)

a Frobenius or 2-Frobenius group; a simple group; an extension of a p1 -group by a simple group; an extension of a simple group by a p1 -solvable group; an extension of a p1 -group by a simple group by a p1 -group.

Lemma 2.2. ([18, Corollary]) If G is solvable with at least two prime graph components, then G is either a Frobenius or 2-Frobenius group and G has exactly two prime graph components, one of which consists of the prime dividing the lower Frobenius complement. The next lemma follows from Theorem 1 in [1]: Lemma 2.3. ([1]) Let G be a Frobenius group of even order
andH, K be Frobenius complement and Frobenius kernel of G respectively. Then t GðGÞ ¼ 2, and the prime graph components of G are pðHÞ, pðKÞ and G has one of the following structures: (a) 2 A pðKÞ and all Sylow subgroups of H are cyclic. (b) 2 A pðHÞ, K is an abelian group, H is a solvable group, the Sylow subgroups of odd order of H are cyclic groups and the 2-Sylow subgroups of H are cyclic or generalized quaternion groups. (c) 2 A pðHÞ, K is an abelian group and there exists H0 a H such that jH : H0 j a 2, H0 ¼ Z  SLð2; 5Þ, ðjZj; 2:3:5Þ ¼ 1 and the Sylow subgroups of Z are cyclic. The next lemma follows from Theorem 2 in [1] and Lemma 3 in [18]:
 Lemma 2.4. ([1]) Let G be a 2-Frobenius group of even order. Then t GðGÞ b 2 and G has a normal series 1 t H t K t G such that (a) p1 ¼ pðG=KÞ U pðHÞ and pðK=HÞ ¼ p2 ; (b) G=K and K=H are cyclic, jG=Kj divides jAutðK=HÞj, ðjG=Kj; jK=HjÞ ¼ 1 and jG=Kj < jK=Hj; (c) H is nilpotent and G is a solvable group.
 Lemma 2.5. [6, Lemma 8] Let G be a finite group with t GðGÞ b 2 and N is a normal subgroup of G. If N is a pi -group for some prime graph component of G and m1 ; m2 ; . . . ; mr are some of order components of G but not a pi -number, then m1 m2 . . . mr is a divisor of jNj  1. Lemma 2.6. Let M ¼ PSU3 ðqÞ where q > 5. Suppose DðqÞ ¼ ð3; q þ 1Þ. Then:

q 2 qþ1 , k

where k ¼

A Characterization of PSU 3 ðqÞ for q > 5

35

(a) If p A pðMÞ, then jSp j a q 3 where Sp A Sylp ðMÞ;
 (b) If p A p1 ðMÞ and p a j jMj, then p a  1 2 0
mod DðqÞ . (c) If p A p1 ðMÞ and p a j jMj, then p a þ 1 1 0 mod DðqÞ if and only if p a ¼ q 3 or q ¼ 7 and
p a ¼ 2 7 .  (d) If k ¼ 3 and DðqÞ  1 j jMj, then q ¼ 8. ðq 2 qþ1Þ

Proof. (a) Observe that jMj ¼ q 3 ðq þ 1Þ 2 ðq  1Þ k and ðq  1; q þ 1Þ ¼ 1 or 2. Thus if q is even, the factors are coprime and if q is odd and p a j jGj, then p a jq 3 or p a j 2ðq þ 1Þ 2 or p a j 4ðq  1Þ. Therefore, (a) follows. a a
For proof  (b), we assume that there exists p A p1 ðMÞ, p j jMj and p  1 1 0 mod DðqÞ . Then we obtain a contradiction. We consider the following two cases: Case 1. q is even:
 q 2 qþ1 (1.1) If p a j ðq  1Þ then p a a q  1 but p a  1 1 0 mod DðqÞ , hence k < q 2 qþ1 p a and so k < q  1, which is a contradiction. ðqþ1Þ 2 (1.2) If p a j ðq þ 1Þ 2 then for k ¼ 1 since 2 < q 2  q þ 1, p a must be equal to ðq þ 1Þ 2 . Thus p a  1 ¼ q 2  q þ 1 þ 3q  1, hence q 2  q þ 1 ¼ 3q  1 which has ðqþ1Þ 2 q 2 qþ1 no solution. If k ¼ 3, since q b 8 and 5 < 3 , p a must be equal to ðq þ 1Þ 2 or 2 2 ðqþ1Þ q qþ1 and similarly 3 divides 3q  1 or q  1 respectively, which is a contradic3 tion since q > 5. q2 (1.3) If p a jq 3 then for k ¼ 1, since 2 < q 2  q þ 1, q 2 jp a and hence p a ¼ q 2 2 s ð0 a sÞ. But p a  1 ¼ ðq 2  q þ 1Þt, thus q j t þ 1 and p a  1 a q 3  1 < q 3 þ 1. Therefore q  1 a t < q þ 1, hence t ¼ q  1 or q. An easy calculation shows q2 q 2 qþ1 q2 that it is a contradiction. Similarly for k ¼ 3, since 4 < 3 we have 2 j p a and ðq 2 qþ1Þ t where q  3 a t < 3ðq þ 1Þ. An easy investigation shows that pa  1 ¼ 3 this is a contradiction. Case 2. q is odd: q 2 qþ1 (2.1) If p a j 4ðq  1Þ, then p a a 4ðq  1Þ but if q 0 11, then k > 4ðq  1Þ
 and hence p a  1
2 0 modDðqÞ , and if q ¼ 11 then DðqÞ ¼ 37 and 4ðq  1Þ ¼ 40 i.e., p a  1 2 0 mod DðqÞ which is a contradiction. (2.2) If p a j 2ðq þ 1Þ 2 , then for k ¼ 1, p a must be equal to ðq þ 1Þ 2 or 2ðq þ 1Þ 2 2ðqþ1Þ 2 and for k ¼ 3, p a must be equal to s
where s ¼ 1; 2; 4; 6; 8 or 9. Easy calculations show that in any case p a  1 2 0 mod DðqÞ and it is a contradiction.
 (2.3) If p a jq 3 then similar to (1.3) we conclude that p a  1 2 0 mod DðqÞ and it is a contradiction. a a 3 a 7 (c) Let p A p1 ðMÞ and
p j jMj.If p ¼ q or q ¼ 7 and p
¼ 2 then it is obvious that p a þ 1 1 0 mod DðqÞ . Conversely if p a þ 1 1 0 mod DðqÞ , since p A p1 ðMÞ and p a j jMj, then there are two cases: Case 1. q is even: q 2 qþ1 (1.1) If p a j ðq  1Þ then p a aq  1 < k a p a þ 1 i.e., p a ¼ q  1 and p a þ 1 ¼ 2 2 q qþ1 q qþ1 , hence q ¼ k which is impossible. k ðqþ1Þ 2

(1.2) If p a j ðq þ 1Þ 2 then for k ¼ 1, since 3 þ 1 < q 2  q þ 1, p a must be equal to ðq þ 1Þ 2 . Hence p a þ 1 ¼ q 2 þ 2q þ 2 ¼ tðq 2  q þ 1Þ, hence q ¼ 4 and p a ¼ 5 2

36

A. Iranmanesh, B. Khosravi, and S.H. Alavi ðqþ1Þ 2

ðq 2 qþ1Þ

while by assumption q > 5. If k ¼ 3 then since 5 þ 1 < , p a must be equal 3 ðqþ1Þ 2 2 to ðq þ 1Þ or 3 . An easy calculation shows that such equalities are impossible. (1.3) If p a jq 3 then similar to (1.3) in part (b), p a must be equal to q 3 . Case 2. q is odd: q 2 qþ1 (2.1) If p a j 4ðq  1Þ and q 0 11 then p a a 4ðq  1Þ < k a p a þ 1 and hence 4q  3 ¼ DðqÞ which is impossible. ðqþ1Þ 2 (2.2) If p a j 2ðq þ 1Þ 2 then for k ¼ 1, p a must be equal to 2 or ðq þ 1Þ 2 or 2 2ðqþ1Þ 2ðq þ 1Þ 2 and for k ¼ 3, p a must be equal to where s ¼ 1; 2; 4; 6; 8 or 9. s a 7 Easy calculations show that p ¼ 2 and q ¼ 7 and in the other cases p a þ 1 2 0
 mod DðqÞ . (2.3) If p a jq 3 then similar to (1.3), p a must be equal to q 3 .
 q2 (d) If DðqÞ  1 j jMj then 3 j q 3 ðq þ 1Þðq  1Þ. But ðq  2; q  1Þ ¼ 1, q2 q2 3 ð 3 ; q þ 1Þ ¼ 1, thus 3 jq and for q ¼ p n , p n  2 ¼ 3p s where 0 a s a 3n, n1 therefore p ¼ 2. Since q > 5 and 2ð2 3 1Þ j2 3n , it is immediate that q ¼ 8. 9 Lemma 2.7. Let G be a finite group, and M ¼ PSU3 ðqÞ where q > 5 and OCðGÞ ¼ OCðMÞ. Then G is neither a Frobenius group nor a 2-Frobenius group. Proof. G is not a Frobenious group otherwise by lemma 2.3 OCðGÞ ¼ fjHj; jKjg where H and K are Frobenious kernel and Frobenious complement q 2 qþ1 of G respectively. If 2 j jHj then jKj ¼ k , jHj ¼ q 3 ðq þ 1Þ 2 ðq  1Þ. If P is a p-

q 2 qþ1 j jPj  1 k q 2 qþ1 and jKj ¼ k

sylow subgroup of H then since H is nilpotent, P x G and hence

which is a contradiction by Lemma 2.6(b). If 2 j jKj then jHj ¼ q 3 ðq þ 1Þ 2 ðq  1Þ. Now if P is a p-sylow subgroup of H, then jPj < jKj, but jKj j ðjPj  1Þ, which is a contradiction. Therefore G is not a Frobenious group. Let G be a 2-Frobenius group and q is even. By Lemma 2.4 there is a normal q 2 qþ1 series 1 t H t K t G such that jK=Hj ¼ k < ðq þ 1Þ 2 and jG=Kj < jK=Hj. Thus there exists a prime p such that p j ðq þ 1Þ 2 and p j jHj. If P is a p-sylow subgroup of H, since H is nilpotent, P must be a normal subgroup of K with P J H and q 2 qþ1 q 2 qþ1 jHj. Therefore, j ðjPj  1Þ by Lemma 2.5 and hence p a  1 1 0 jKj ¼ k k
 mod DðqÞ , which is a contradiction by Lemma 2.6. If q is odd then we consider 2ðq þ 1Þ 2 instead of ðq þ 1Þ 2 and proceed similarly. 9 Lemma 2.8. Let G be a finite group. If the order components of G are the same as those of M ¼ PSU3 ðqÞ, where q > 5, then G has a normal series 1 t H t K t G such that H and G/K are p1 -groups and K/H is a simple group. Moreover, the
odd order  component of M is equal to some of those of K/H, in particular, t GðK=HÞ b 2. Proof. The first part of the lemma follows from the above lemmas since the prime graph of M has two prime graph components. For primes p and q, if K=H has an element of order pq, then G has one. Hence, by the definition of prime graph component, the odd order component of G must be an odd order component of K=H. 9

A Characterization of PSU 3 ðqÞ for q > 5

37

3. Proof of the Main Theorem By Lemma 2.8, G has a normal series 1 t H t
K t G such  that H and G=K are p1 -groups, K=H is a non-abelain simple group, t GðK=HÞ b 2 and the odd order component of M is an odd order component of K=H. We summarize the relevant information in Tables 1–3 below: We now proceed the proof in the following steps: Step 1. If K=H G An where n ¼ p, p þ 1, p þ 2 and p b 5 is a prime number. Then if k ¼ 1 with q 2  q þ 1 ¼ p or if q 2  q þ 1 ¼ p  2 with p > 5, then q 2  q  1 ¼ p  2 or q 2  q  1 ¼ p  4, respectively, but ðq 2  q  1; p1 Þ ¼ 1, thus a contradiction. If q ¼ 5 then q 2  q þ 1 ¼ 3 which has no solution in Z. If k ¼ 3 then DðqÞ ¼ p or p  2 and hence DðqÞ  1 j jGj. By lemma 2.6(d), q ¼ 8 and hence DðqÞ ¼ 19. Now if DðqÞ ¼ p then 17 j jA19 j but 17 a 8 3 ð8 þ 1Þ 2 ð8  1Þ, and, if DðqÞ ¼ p  2 then p ¼ 21 which is not a prime number. Step 2. If K=H G Ar ðq 0 Þ then we distinguish the following 5 cases: 2.1. K=H G Ap 0 1 ðq 0 Þ where ð p 0 ;
q 0 Þ 0 ð3; 2Þ; ð3; 4Þ or K=H G Ap 0 ðq 0 Þ where 0 0 ðq  1Þ j ðp 0 þ 1Þ. Then q 0p  1 1 0 mod DðqÞ which is not possible by lemma 2.6(b). q 0 1 2.2. K=H G A1 ðq 0 Þ, where 4 j ðq 0 þ 1Þ. If DðqÞ ¼ 2 , then q 0  1 1 0
 mod DðqÞ which is a contradiction by 2.6(b). If DðqÞ ¼ q 0 and k ¼ 1 then q 0 ¼ q 2  q þ 1 and hence jK=Hj ¼ jA1 ðq 0 Þj ¼ qðq 2  q þ 1Þðq 2  q þ 2Þðq  1Þ=2 2

q qþ2 a jGj. If k ¼ 3 and DðqÞ ¼ q 0 then but this is 2
 a contradiction since DðqÞ  1 j jGj and thus by Lemma 2.6(d) we must have q ¼ 8 and DðqÞ ¼ 19, hence q 0 ¼ 19 which is a contradiction since 5 j jA1 ð19Þj but 5 a jPSUð3; 8Þj. 2.3. K=H G A1 ðq 0 Þ where 4 j ðq 0  1Þ. Since0 the possibility DðqÞ
¼ q 0 is dis q þ1 cussed in case 2.2, we assume that DðqÞ ¼ 2 . Then q 0 þ 1 1 0 mod DðqÞ . Using Lemma 2.6(c), it follows that q ¼ 7 and q 0 ¼ 2 7 (which is impossible since 4 a ð2 7  1Þ) or q 0 ¼ q 3 . If q 0 ¼ q 3 then jK=Hj ¼ jA1 ðq 0 Þj ¼ q 3 ðq 3  1Þðq 3 þ 1Þ=2 and thus we must have ðq 2 þ q þ 1Þ j ðq þ 1Þ 2 , which is a contradiction. 2.4. K=H G A1 ðq 0 Þ where 4jq 0 . DðqÞ can not be equal to q 0  1, thus DðqÞ ¼ 0 q þ 1. By Lemma 2.6(c), q 0 ¼ q 3 (which is discussed in case 2.3) or q ¼ 7 and q 0 ¼ 2 7 . But 127 j jA1 ð2 7 Þj and 127 a jGj. 2.5. K=H G A2 ð2Þ or K=H G A2 ð4Þ then DðqÞ must be equal to 3, 5, 7, 9 which is impossible.

Step 3. If K=H G 2 Ar ðq 0 Þ then we consider 3 cases:
 0 3.1. K=H G 2 Ap 0 1 ðq 0 Þ. Then q 0p þ 1 1 0 mod DðqÞ . By Lemma 2.6(c) 0 0 p0 0 0 q 0p ¼ q 3 or q ¼ 7 and q 0p ¼ 2 7 . If q 0 ¼ q 3 then since q 0ð p ð p 1Þ=2Þ j jK=Hj by Lemma 2.6(a) p 0 ¼ 3 and q ¼ q 0 , hence K=H ¼ PSU3 ðqÞ. Then jGj ¼ jPSU3 ðqÞj ¼ jK=Hj ¼ jKj=jHj which implies that jHj ¼ 1 and jKj ¼ jGj ¼ jPSU3 ðqÞj. There0

fore, K ¼ PSU3 ðqÞ and hence G ¼ PSU3 ðqÞ. If q 0p ¼ 2 7 then q 0 ¼ 2 and p 0 ¼ 7

38

A. Iranmanesh, B. Khosravi, and S.H. Alavi

Table 1: The order components of simple groups1 with tðGÞ ¼ 2 Group

Orcmp1

Orcmp2

Ap , p 0 5; 6 p and p  2 not both prime Apþ1 , p 0 4; 5 p  1 and p þ 1 not both prime Apþ2 , p 0 3; 4 p and p þ 2 not both prime Ap1 ðqÞ, ð p; qÞ 0 ð3; 2Þ; ð3; 4Þ

3  4 . . . ð p  3Þð p  2Þð p  1Þ

p

3  4 . . . ð p  2Þð p  1Þð p þ 1Þ

p

Ap ðqÞ, q  1 j p  1 2

Ap1 ðqÞ

2

Ap ðqÞ, q þ 1 j p þ 1 ð p; qÞ 0 ð3; 3Þ; ð5; 2Þ 2 A3 ð2Þ Bn ðqÞ, n ¼ 2 m b 4, q odd Bp ð3Þ

3  4 . . . ð p  1Þð p þ 1Þð p þ 2Þ p1 i q pð p1Þ=2 Pi¼1 ðq  1Þ p1 i q pð pþ1Þ=2 ðq pþ1  1Þ Pi¼2 ðq  1Þ  p1
i pð p1Þ=2 q Pi¼1 q  ð1Þ i
 p1 q pð pþ1Þ=2 ðq pþ1  1Þ Pi¼2 q i  ð1Þ i

2

Cp ðqÞ, q ¼ 2; 3

p1 2i q p ðq p þ 1Þ Pi¼1 ðq  1Þ 2

p1 2i q pð p1Þ Pi¼1 ðq  1Þ

Dp ðqÞ, p b 5, q ¼ 2; 3; 5

1 q pð pþ1Þ ðq p ð2; q1Þ

Dn ð2Þ, n ¼ 2 þ 1 b 5 Dp ð3Þ, p 0 2 m þ 1, p b 5

2

Dn ð3Þ, n ¼ 2 m þ 1 0 p, m b 2 G 2 ðqÞ, q 1 e ðmod 3Þ, e ¼ G1, q>2 3 D4 ðqÞ F4 ðqÞ, q odd

2

2

nðn1Þ

n2 2i ð2 þ 1Þð2 n1  1Þ Pi¼1 ð2  1Þ p1 2i pð p1Þ 3 Pi¼1 ð3  1Þ n

n2 2i 1Þð3 n1  1Þ Pi¼1 ð3  1Þ q ðq  eÞðq 2  1Þðq þ eÞ

1 nðn1Þ n ð3 þ 23 6 3

q 12 ðq 6  1Þðq 2  1Þðq 4 þ q 2 þ 1Þ q 24 ðq 8  1Þðq 6  1Þ 2 ðq 4  1Þ

2

q n þ1 2 3 p 1 2 q n þ1 ð2; q1Þ q p 1 ð2; q1Þ q p 1 q1 q p 1 ð2; q1Þ q n þ1 ð2; qþ1Þ n1

þ1

3 p þ1 4 3 n1 þ1 2

q 2  eq þ 1 q4  q2 þ 1 q4  q2 þ 1

F4 ð2Þ 0 E6 ðqÞ

2 11  3 3  5 2 q 36 ðq 12  1Þðq 8  1Þðq 6  1Þðq 5  1Þðq 3  1Þðq 2  1Þ

E6 ðqÞ, q > 2

q 36 ðq 12  1Þðq 8  1Þðq 6  1Þðq 5 þ 1Þðq 3 þ 1Þðq 2  1Þ

q 6 þq 3 þ1 ð3; q1Þ q 6 q 3 þ1 ð3; qþ1Þ

M12 J2 Ru He Mcl Co1 Co3 Fi22 F5 ¼ HN

26  33  5 27  33  52 2 14  3 3  5 3  7  13 2 10  3 3  5 2  7 3 27  36  53  7 21 2  3 9  5 4  7 2  11  13 2 10  3 7  5 3  7  11 2 17  3 9  5 2  7  11 2 14  3 6  5 6  7  11

11 7 29 17 11 23 23 13 19

2

2

p1 2i þ 1Þðq pþ1  1Þ Pi¼1 ðq  1Þ

n1 2i q nðn1Þ Pi¼1 ðq  1Þ

Dn ðqÞ, n ¼ 2 m b 4

2

m

5

n

n1 2i q n ðq n  1Þ Pi¼1 ðq  1Þ

2

q 1 ðq1Þð p; q1Þ q p 1 q1 q p þ1 ðqþ1Þð p; qþ1Þ q p þ1 qþ1

26  34 n1 2i q ðq  1Þ Pi¼1 ðq  1Þ 2 p1 2i 3 p ð3 p þ 1Þ Pi¼1 ð3  1Þ n2

C n ðqÞ, n ¼ 2 m b 2

Dpþ1 ðqÞ, q ¼ 2; 3

p p

1 p is an odd prime number.

13

q4  q2 þ 1 pffiffiffiffiffiffiffiffi q 2 þ 2q 3 pffiffiffiffiffi þ q þ 2q þ 1 q2  q þ 1 pffiffiffiffiffi q þ 3q þ 1

73 q4 þ 1 pffiffiffiffiffiffiffiffi q 2  2q 3 pffiffiffiffiffi þ q  2q þ 1 q2 þ q þ 1 pffiffiffiffiffi q  3q þ 1

E 7 ð3Þ

757

127

2p þ 1

q 12 ðq 4  1Þðq 3 þ 1Þ  ðq 2 þ 1Þðq  1Þ q 6 ðq 2  1Þ 2 q 3 ðq 2  1Þ 23 2  3 63  5 2  7 3  11 2  13 3  19  37  41  61  73  547

2 pþ1 þ 1

ð3 p1 þ 1Þ=2

2  3 pð p1Þ ð3 p1  1Þ p2 2i  Pi¼1 ð3  1Þ pð pþ1Þ p 2 ð2  1Þ p1 2i  Pi¼1 ð2  1Þ 2 63  3 11  5 2  7 3  11  13  17  19  31  43 q 24 ðq 6  1Þ 2 ðq 4  1Þ 2

F4 ðqÞ 2jq, q > 2 2 F4 ðqÞ q ¼ 2 2nþ1 > 2 G 2 ðqÞ, 3jq 2 G 2 ðqÞ, q ¼ 3 2nþ1

ð3 p þ 1Þ=4

q q qþ1 3 5 p7ffiffiffiffiffi q  2q þ 1

qþ1 q1 q 8 26 2 15  3 6  5 q2

1093

ðq  1Þ=2 ðq þ 1Þ=2 q1 7 7 11 p ffiffiffiffiffi q þ 2q þ 1

p

p2

3  4 . . . ð p  3Þð p  1Þ

Orcmp 3

Ap , p and p  2 are primes A1 ðqÞ, 4 j q þ 1 A1 ðqÞ, 4 j q  1 A1 ðqÞ, 2jq A2 ð2Þ A2 ð4Þ 2 A5 ð2Þ 2 B2 ðqÞ q ¼ 2 2nþ1 > 2 2 Dp ð3Þ p ¼ 2 n þ 1, n b 2 2 Dpþ1 ð2Þ p ¼ 2 n  1, n b 2 E 7 ð2Þ

Orcmp 2

Orcmp 1

Group

Table 2: The order components of simple groups1 with tðGÞ b 3

q1

9

Orcmp 4

Orcmp 5

Orcmp 6

A Characterization of PSU 3 ðqÞ for q > 5 39

2

36

9

2

Orcmp 1 2

 3  5  7  11 24  32 27  32 7 2  32  5  7 2 10  3 3  5  7 23  3  5 27  35  5 2 21  3 3  5  7  11 3 29  32  53 13 2  37  52  7 29  34  5  73 8 2  3 7  5 6  7  11 2 18  3 6  5 3  7 18 2  3 13  5 2  7  11  13 2 21  3 16  5 2  7 3  11  13 2 46  3 20  5 9  7 6  11 2  13 3  17  19  23  29  31  47 2 41  3 13  5 6  7 2  11  13  17  19  23 2 15  3 10  5 3  7 2  13

1 p is an odd prime number.

F3 ¼ Th

F2 ¼ B

E6 ð2Þ M11 M22 M23 M24 J1 J3 J4 HS Sz ON Ly Co2 F23 0 F24 F1 ¼ M

2

Group

Table 2: (continued)

19

31

13 5 5 11 11 7 17 23 7 11 11 31 11 17 17 41

Orcmp 2

31

47

17 11 7 23 23 11 19 29 11 13 19 37 23 23 23 59

Orcmp 3

29 71

31 67

31

19

11

19

Orcmp 4

37

Orcmp 5

43

Orcmp 6

40 A. Iranmanesh, B. Khosravi, and S.H. Alavi

A Characterization of PSU 3 ðqÞ for q > 5

41

Table 3: The order components of E8 ðqÞ E8 ðqÞ, q 1 0; 1; 4 ðmod 5Þ

Group Orcmp 1 Orcmp 2 Orcmp 3 Orcmp 4 Orcmp 5

q

120

ðq

18

 1Þðq

14

 1Þðq 12  1Þ 2 ðq 10  1Þ 2 ðq 8  1Þ 2 ðq 4 þ q 2 þ 1Þ q8 þ q7  q5  q4  q3 þ q þ 1 q8  q7 þ q5  q4 þ q3  q þ 1 q8  q6 þ q4  q2 þ 1 q8  q4 þ 1

Group

E8 ðqÞ, q 1 2; 3 ðmod 5Þ

Orcmp 1

q 120 ðq 20  1Þðq 18  1Þðq 14  1Þðq 12  1Þðq 10  1Þðq 8  1Þðq 4 þ 1Þ  ðq 4 þ q 2 þ 1Þ q8 þ q7  q5  q4  q3 þ q þ 1 q8  q7 þ q5  q4 þ q3  q þ 1 q8  q4 þ 1

Orcmp 2 Orcmp 3 Orcmp 4

since p 0 is an odd prime. But 2 15 j j 2 A6 ð2Þj and 2 15 > 7 3 ¼ q 3 , which is a contradiction by Lemma 2.6(a). 3.2. K=H G 2 Ap 0 ðq 0 Þ where ðq 0 þ 1Þ j ð p 0 þ 1Þ and ð p 0 ; q 0 Þ 0 ð3; 3Þ; ð5; 2Þ then 0

0

0

0

0

q 0p ¼ q 3 or q ¼ 7 and q 0p ¼ 2 7 . If q 0p ¼ q 3 since p 0 is an odd prime, q 0ð p ð p þ1Þ=2Þ > 0 0 q 02p > q 3 , which is a contradiction. If q 0p ¼ 2 7 then q 0 ¼ 2 and p 0 ¼ 7, but q 0 þ 1 ¼ 0 3 a 8 ¼ p þ 1. 3.3. K=H G 2 A3 ð2Þ or K=H G 2 A5 ð2Þ. Then DðqÞ must be equal to 5, 7, 11 which is impossible. Step 4. If K=H G Br ðq 0 Þ then we consider 2 cases: 0 t 4 and q 0 is odd. Then q 0r þ 1 1 0
4.1. K=H  G Br ðq Þ where r 0r¼ 2 b 3 mod DðqÞ . By Lemma 2.6(c) q ¼ q or q ¼ 7 and q 0r ¼ 2 7 , which is a contradiction since q 0 is odd. If q 0r ¼ q 3 then
q 12 j jGj, which is again a contradiction.  4.2. K=H G Bp ð3Þ. Then 3 p  1 1 0 mod DðqÞ which is impossible by Lemma 2.6(b). Step 5. If K=H G Cr ðq 0 Þ then we consider 3 cases:
 5.1. K=H G Cr ðq 0 Þ where r ¼ 2 t b 2. Then q 0r þ 1 1 0 mod DðqÞ . By Lemma 2.6(c) q ¼ 7 and q 0r ¼ 2 7 (which is impossible since r is not a power of 2) or q 0r ¼ q 3 . Now since ðq 0r  1Þðq 0r þ 1Þ j jK=Hj we conclude that ðq 3  1Þðq 3 þ 1Þ j jGj and hence q 2 þ q þ 1 j q 3 ðq þ 1Þ which is impossible.
 0 5.2. K=H G Cp 0 ðq 0 Þ where q 0 ¼ 2; 3. Then q 0p  1 1 0 mod DðqÞ , which is a contradiction by Lemma 2.6(b). 5.3. K=H G Dr ðq 0 Þ where ðr; q 0 Þ ¼ ðp 0 ; q 0 Þ ð p 0 b 5; q 0 ¼ 2; 3; 5Þ or ðr; q 0 Þ ¼ ðp 0 þ 1; q 0 Þ ðq 0 ¼ 2; 3Þ. Here we get a contradiction similarly. Step 6. If K=H G 2B2 ðq 0 Þ where q 0 ¼ 2 2tþ1 > 2, then we consider 3 cases: 6.1. DðqÞ ¼ q 0  1. ffiffiffiffiffiffiffi get a contradiction by
Lemma 2.6(b). pWe  6.2. DðqÞ ¼ q 0  2q 0 þ 1. Then q 02 þ 1 1 0 mod DðqÞ . Therefore, q 02 ¼ 2 7

42

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(which is impossible since 2 7 is not a square) or q 02 ¼ q 3 and hence q is even. If k ¼ 1 then 2 2tþ1  2 tþ1 ¼ qðq  1Þ and since ðq; q  1Þ ¼ 1, we conclude that q ¼ 2 tþ1 and q  1 ¼ 2 t  1, which is a contradiction. If k ¼ 3 and we assume that q ¼ 2 m . Then 2ð3:2 2t  3:2 t þ 1Þ ¼ 2 m ð2 m  1Þ which implies that m ¼ 1, which is a contradiction since >ffi 5. pqffiffiffiffiffiffi 0 6.3. DðqÞ ¼ q þ 2q 0 þ 1 ¼ 2 2tþ1 þ 2 tþ1 þ 1. We proceed similarly. Step 7. If K=H G 2Dr ðq 0 Þ then we consider 6 cases: 7.1. K=H G 2Dr ðq 0 Þ where r ¼ 2 t > 2. Then q 0r ¼ 2 7 (which is impossible, since r 0 2 t ) or q 0r ¼ q 3 . Since r  1 b 3 we have q 9 j jGj which is a contradiction by Lemma 2.6(a). 7.2. K=H G 2Dr ð2Þ where r ¼ 2 t þ 1 b 5. Then 2 r1 ¼ 2 7 (which is impossible since r ¼ 8) or 2 r1 ¼ q 3 . Since r b 5 we have q 15 j jGj, which is a contradiction by Lemma 2.6(a). 7.3. K=H G 2Dp ð3Þ where 5 a p 0 2 r þ 1. Then 3 p ¼ q 3 and since p is an odd prime, p ¼ q ¼ 3, but q > 5. 7.4. K=H G 2Dr ð3Þ where r ¼ 2 t þ 1 0 p, t b 2. Then 3 r1 ¼ q 3 , hence 3 r rðr1Þ ðq Þ j3 which is a contradiction by Lemma 2.6(a). 7.5. K=H G 2Dp ð3Þ where p ¼ 2 t þ 1; t b 2. Then we proceed similarly. 7.6. K=H G 2Dpþ1 ð2Þ where p ¼ 2 r  1; r b 2. If DðqÞ ¼ 2 p þ 1 then 2 p ¼ q 3 . Since 4 a p þ 1 we have q 12 j jGj which is impossible by Lemma 2.6(a) or 2 p ¼ 2 7 , hence p ¼ q ¼ 7 and r ¼ 3, but 127 j j 2 D8 ð2Þj and 127 a jGj. Similarly if DðqÞ ¼ 2 pþ1 þ 1 then 2 pþ1 ¼ q 3 and hence q 9 j jGj which is impossible, or, 2 pþ1 ¼ 2 7 which is again a contradiction since p ¼ 6. Step 8. If K=H G G 2 ðq 0 Þ then we consider 3 cases: 8.1. K=H G G 2
ðq 0 Þ where 2 < q 0 1 1 ðmod 3Þ. Then DðqÞ ¼ q 02  q 0 þ 1 and hence q 03 þ 1 1 0 mod DðqÞ so q 03 ¼ 2 7 (which is impossible) or q 03 ¼ q 3 . Therefore, q ¼ q 0 and hence q 6 j jGj which is impossible. 8.2. K=H G G 2 ðq 0 Þ where 2 < q 0 1 1 ðmod 3Þ. Then DðqÞ ¼ q 02 þ q 0 þ 1 and
 hence q 03  1 1 0 mod DðqÞ which is impossible. 8.3. K=H G G 2 ðq 0 Þ where 3jq 0 . Then q 02 G q 0 þ 1 ¼ DðqÞ. This is similar to cases 8.1 and 8.2. Step 9. If K=H G E 7 ð2Þ or E 7 ð3Þ or 2E6 ð2Þ or 2 F4 ð2Þ 0 then DðqÞ must be equal to 13, 17, 19, 73, 127, 757, 1093 which has no solution in Z except q ¼ 8 in 2E6 ð2Þ and q ¼ 9 in E 7 ð2Þ, but 2 36 j j 2E6 ð2Þj and 127jE 7 ð2Þj which is impossible since 127 a jGj and 2 36 > q 3 . Step 10. If K=H G 3D4 ðq 0 Þ then DðqÞ ¼ q 04  q 02 þ 1, and hence q 06 þ 1 1 0
mod DðqÞ which implies that q 06 ¼ 2 7 (which is impossible) or q 02 ¼ q and q 6 j jGj, which is again impossible. Step 11. If K=H G F4 ðq 0 Þ then we consider 2 cases. 11.1. If DðqÞ ¼ q 04  q 02 þ 1 then we proceed similar to step 10. 11.2. If DðqÞ ¼ q 04 þ 1, then q 04 ¼ 2 7 (which is impossible) or q 04 ¼ q 3 and 18 q j jGj which is again impossible.

A Characterization of PSU 3 ðqÞ for q > 5

43

pffiffiffiffiffiffiffiffi 2 0 Step q 0 ¼ 2 2rþ1 > 2 then DðqÞ ¼ q 02 G 2q 03 þ q 0 G pffiffiffiffiffiffiffi12. If K=H G F4 ðq Þ where
 2q 0 þ 1, hence q 06 þ 1 1 0 mod DðqÞ then q 06 ¼ 2 7 (which is impossible) or q 06 ¼ q 3 and hence q ¼ q 02 , so q 012 ¼ q 6 j jGj which is a contradiction by Lemma 2.6(a). pffiffiffiffiffiffiffi 2 Step 13. If K=H G G 2 ðq 0 Þ where q 0 ¼ 3 2rþ1 then DðqÞ ¼ q 0 G 3q 0 þ 1. In this
is impossible) or q 03 ¼ q 3 . case, q 03 þ 1 1 0 mod DðqÞ and thus q 03 ¼ 2 7 (which pffiffiffiffiffi 0 2rþ1 Therefore, q ¼ q ¼ 3 and hence k ¼ 1. But q G 3q þ 1 ¼ q 2  q þ 1 means 3rþ1 r that 3  2  3 G 1 ¼ 0 which is impossible. 0 2 0 06 03 0 Step 14. If K=H G
E6 ðq Þ or E6 ðq Þ then DðqÞ ¼ ðq þ q þ 1Þ=ð3; q  1Þ and 09 hence q  1 1 0 mod DðqÞ which is impossible by Lemma 2.6(b).

Step 15. If K=H is a sporadic simple group then DðqÞ must be equal to 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 59, 67, 71, which has a solution when DðqÞ ¼ 19; 37; 43. But if DðqÞ ¼ 19; 37; 43, then q ¼ 8; 11; 7 respectively. If q ¼ 8 then jGj ¼ 2 9 :7:3 4 :19. By the table of sporadic simple groups 19 is an odd order component of F5 , J1 , J3 or ON. But 5 divides jF5 j, jJ1 j, jJ3 j or jONj and 5 a jGj and since q ¼ 8 then q 3 ¼ 2 9 < 2 14 and 2 14 j jF5 j which is a contradiction. Again by the table of sporadic simple group 37 is an odd order component of Ly and J4 . But 31 a j 2 A2 ð11Þj and 31 j jLyj and 31 j jJ4 j. Similarly if DðqÞ ¼ 43 then 31 j jJ4 j but 31 a j 2 A2 ð7Þj. Step 16. If K=H GE8 ðq 0 Þ then since all odd order components are less than or equal to q 09 we have q < q 09 or q 10 < q 090 which is a contradiction by Lemma 2.6(a). The proof of the main theorem is now completed. 9 Remark 3.1. It is a well known conjecture of J.G. Thompson that if G is a finite group with ZðGÞ ¼ 1 and M is a non-abelian simple group satisfying NðGÞ ¼ NðMÞ where NðGÞ ¼ fn j G has a conjugacy class of size ng, then G G M. We can give a positive answer to this conjecture by our characterization of the groups under discussion. Corollary 3.2. Let G be a finite group with ZðGÞ ¼ 1, M ¼ PSU3 ðqÞ where q > 5 and NðGÞ ¼ NðMÞ, then G G M. Proof. By [2, Lemma 1.5] if G and M are two finite groups satisfying the conditions of Corollary 3.2, then OCðGÞ ¼ OCðMÞ. So the main theorem implies this corollary. 9 Remark 3.3. Wujie Shi and Bi Jianxing in [16] put forward the following conjecture: Conjecture. Let G be a group, M a finite simple group, then G G M if and only if (i) jGj ¼ jMj, and, (ii) pe ðGÞ ¼ pe ðMÞ, where pe ðGÞ denotes the set of orders of elements in G.

44

A. Iranmanesh, B. Khosravi, and S.H. Alavi

This conjecture is valid for sporadic simple groups [13], groups of alternating type [17], and some simple groups of Lie types [14, 15, 16]. As a consequence of the main theorem, we prove the validity of this conjecture for the groups under discussion. Corollary 3.4. Let G be a finite group and M ¼ PSU3 ðqÞ where q > 5. If jGj ¼ jMj and pe ðGÞ ¼ pe ðMÞ, then G G M. Proof. By assumption we must have OCðGÞ ¼ OCðMÞ, then the corollary follows by the main theorem. 9 Acknowledgment. The first author would like to thank the Institute for Studies in Theoretical Physics and Mathematics (IPM) Tehran, Iran for the financial support (No. 81200019).

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