Groups with the Same Set of Orders of Maximal Abelian Subgroups

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1 Filomat 8:9 (01), DOI 10.98/FIL109871A Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: Groups with the Same Set of Orders of Maximal Abelian Subgroups Neda Ahanjideh a, Ali Iranmanesh b a Department of Pure Mathematics, Faculty of Mathematical Sciences, Shahrekord University, P.O.Box 115, Shahrekord, Iran b Department of Pure Mathematics, Faculty of Mathematical Sciences, Tarbiat Modares University, P.O. Box , Tehran, Iran Abstract. Let n > 3 be an even number. In this paper, we show how the orders of maximal abelian subgroups of the finite group G can influence on the structure of G. More precisely, we show that if for a finite group G, M(G) = M(B n (q)), then G B n (q). Note that M(G) is the set of orders of maximal abelian subgroups of G. Let Γ(G) denote the non-commuting graph of G. As a consequence of our result, we show that if G is a finite group with Γ(G) Γ(B n (q)), then G B n (q). 1. Introduction For an integer z > 1, we denote by π(z) the set of all prime divisors of z. If G is a finite group, then π( G ) is denoted by π(g). The prime graph (or Gruenberg-Kegel graph) GK(G) of a group G is the graph with vertex set π(g) where two distinct primes p and q are joined by an edge (we write (p, q) GK(G)) if G contains an element of order pq. Let s(g) be the number of connected components of GK(G). A list of all finite simple groups with disconnected prime graph has been obtained in [11] and [0]. A finite group G is said to be characterizable by the set of orders of its maximal abelian subgroups, if G is uniquely determined by the orders of its maximal abelian subgroups. More precisely, a finite group G is called characterizable by the set of orders of its maximal abelian subgroups, if each finite group H with M(G) = M(H) is necessarily isomorphic to G. Recall that a simple group S is a K 3 -group if π(s) = 3. It is known that if G is any K 3 -group, the alternating group A n (where n and n are primes or n 10), PSL ( n ), Sz( m+1 ), B n (q), where n = m and any sporadic simple group, then G is characterizable by the set of orders of its maximal abelian subgroups (see [1, 6, 19]). Let M(G) = { N : N is a maximal abelian sub roup o f G}. In this paper, we have proved that: Theorem 1. Let n > 3 be an even natural number and let q be a prime power. If G is a finite group with M(G) = M(B n (q)), then G B n (q). For every n and q, the simple groups B n (q) and C n (q) have the same order. These groups are isomorphic if n = or q is even. Also, s(b n (q)) 1 if and only if n = n or n is prime and q {, 3}.. Preliminaries In this paper, fix: the subset of vertices of a graph is called an independent set if its vertices are pairwise non-adjacent. For a finite group G, we write ρ(g) (ρ(r, G)) for some independent set in GK(G) (containing 010 Mathematics Subject Classification. Primary 0D06; Secondary 0D0 Keywords. Maximal abelian subgroup, simple group of Lie type, prime graph, automorphism groups, maximal independent set. Received: 08 August 013; Revised: 1 November 013; Accepted: 08 December 013 Communicated by Francesco Belardo Research supported partially by Shahrekord University addresses: ahanjideh.neda@sci.sku.ac.ir (Neda Ahanjideh), iranmanesh@modares.ac.ir (Ali Iranmanesh)

2 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), a prime r) with a maximal number of vertices, which is named the maximal independent set and put t(g) = ρ(g) (and t(r, G) = ρ(r, G) ). Throughout this paper, by [x] we denote the integral part of x and by gcd(m, n) we denote the greatest common divisor of m and n. If m is a natural number and r is prime, the r-part of m is denoted by m r, i.e., m r = r t if r t m. The notation for groups of Lie type is according to [10]. Given α S k, we define α d to be the dk dk permutation matrix that permutes blocks of dimension d. For example, 0 I 0 (1,, 3) = 0 0 I. I 0 0 We refer to the block matrices that arise from permutations in this way as standard permutation matrices. Given X = x 1,..., x m GL d (q) and W = α 1,..., α l S k, we define the wreath product X W to be the subgroup of GL dk (q) generated the matrices Diag(x 1, I d,..., I d ),..., Diag(x m, I d,..., I d ), Diag (I d, x 1, I d,..., I d ),..., Diag(I d,..., x m ), α 1d,..., α ld (see [1]). In the whole paper, we assume that n is an even natural number such that n, q is a prime power (q = p k ) and p is prime. By GF(q), we denote the finite field with q-elements. For a finite group G, we denote the maximum element of M(G) by a(g) (see [18]) and we denote the maximum of the orders of abelian subgroups of s-sylow subgroup of G by a s (G), where s π(g). All further unexplained notations are standard and can be found in [5] and [9]. Lemma.1. [18, Table ] If S is a finite simple group of Lie type in characteristic p such that S A 1 (p α ) (where p = ), A (p α ) (where gcd(p α 1, 3) = 1), A (p α ) (where gcd(p α + 1, 3) = 1), A 3 () and F (q), then a(s) = a p (S). In particular, - if n and q is odd, then a(b n (q)) = q n(n 1)/+1 ; - a(c n (q)) = q n(n+1)/, except for C (); - for n 5, a( D n (q)) = q (n 1)(n )/+ ; - a(a n (q)) = q [(n+1) /], except for A 1 (q), where q is even and A (q), where (3, q 1) = 1. Lemma.. [3] Let G be a finite group and N G. If r G/N, r N (r is prime and r p), and if in addition p e N and p t C N (R), where R Syl r (G), then r p e t 1. Corollary.3. Let G be a finite group and N G. If r G/N, r N (r is prime and r p), and if in addition p e N and (p, r) GK(G), then r p e 1. Proof. Straightforward. Lemma.. [1, Corollary 11] Let H be a finite group such that, s π(h). If (, s) GK(H), then s-sylow subgroup of H is abelian. Lemma.5. [6] Let G and H be two finite groups such that M(G) = M(H). Then G and H have the same prime graph. Lemma.6. [1] Let n = n. If G is a finite group such that M(G) = M(B n (q)), then G B n (q). For an integer n, by ν(n), η(n) and η (n), we denote the following functions: n i f n 0 (mod ); n n if n is odd; ν(n) = i f n (mod );, η(n) = n n i f n 1 (mod ). otherwise., (1) { n if n is odd; η (n) =. n otherwise.

3 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), Lemma.7. [15, Theorem 1] Let G be a finite group with t(g) 3 and t(, G). Then the following hold: 1. There exists a finite non-abelian simple group S such that S Ḡ = G/K Aut(S) for the maximal normal solvable subgroup K of G.. For every independent subset ρ of π(g) with ρ 3 at most one prime in ρ divides the product K. Ḡ/S. In particular, t(s) t(g) One of the following holds: (a) every prime r π(g) nonadjacent to in GK(G) does not divide the product K. Ḡ/S ; in particular, t(, S) t(, G); (b) there exists a prime r π(k) nonadjacent to in GK(G); in which case t(g) = 3, t(, G) =, and S A 7 or A 1 (q) for some odd q. Lemma.8. [16, Proposition 1.1] Let G = A n be an alternating group of degree n. 1. Let r, s π(g) be odd primes. Then r and s are nonadjacent if and only if r + s > n;. let r π(g) be an odd prime. Then and r are nonadjacent if and only if r + > n. If a is a natural number, r is an odd prime and gcd(r, a) = 1, then by exp r (a) we denote the smallest natural number m such that a m 1 (mod r). Obviously by Fermat s little theorem it follows that exp r (a) (r 1). Also, if a n 1 (mod r), then exp r (a) n. If a is odd, we put exp (a) = 1 if a 1 (mod ), and exp (a) = otherwise. Lemma.9. [8, Corollary to Zsigmondy s theorem] Let a be a natural number greater than 1. For every natural number m there exists a prime r with exp r (a) = m, unless a = and m = 1, a = 3 and m = 1, and a = and m = 6. The prime r with exp r (q) = m is called a primitive prime divisor of q m 1. It is obvious that q m 1 can have more than one primitive prime divisor. We denote by r m (q) some primitive prime divisor of q m 1. If there is no ambiguous, we write r m instead of r m (q). Also, let Z m (q) denote the set of primitive prime divisors of q m 1. One can easily check the following corollary: Corollary.10. Let a, b and c be natural numbers and let s be a prime. (i) If exp s (p) = ab, then exp s (q a ) = b; (ii) if c a and gcd(c, b) = 1, then Z b (p a/c ) Z b (p a ); (iii) if a, then Z b (p a/ ) Z b (p a ). Lemma.11. [16, Propositions.1 and.] and [17, Propositions.,.5 and.7(5)] Let G = B n (q) or C n (q). Let r and s be odd primes and r, s π(g) \ {p}. Put k = exp r (q) and l = exp s (q). If 1 η(k) η(l), then r and s are nonadjacent if and only if η(k) + η(l) > n and l k is not an odd natural number. Lemma.1. [16, Proposition 3.1] Let G = B n (q) or C n (q), and let r π(g) and r p. Then r and p are nonadjacent if and only if η(exp r (q)) > n 1. Lemma.13. [16, Proposition.3] Let G = B n (q) or G = C n (q). Let r be an odd prime divisor of G, r p, and k = exp r (q). Then r and are nonadjacent if and only if η(k) = n and one of the following holds: 1. n is odd and k = (3 exp (q))n;. n is even and k = n. Lemma.1. [1, Corollary 3.3, Corollary 3.6 and the proof of Lemma 3.7] Let n be an even number and α M(B n (q)). (i) If π(α) Z n (q), then α = q n + 1 gcd(, q 1) ;

4 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), q n (ii) If π(α) Z (n 1) (q), then gcd(, q 1) α and α q(q 1)(q n 1 + 1). gcd(, q 1) q n 1 1 (iii) If π(α) Z n 1 (q), then gcd(, q 1) α and α q(q 1)(q n 1 1). gcd(, q 1) Lemma.15. Let r π(g) {p} and R Syl r (SO n+1 (q)). (i) [1, Lemma 3.18] If n = t, p and r =, then a(r) = ( q ± 1 ) t ; (ii) [, Corollary before Theorem ] let p, r = and n = r r t with r 1 <... < r t. If q r i 1 δ i (mod ), for all i {1,..., t} and R i Syl (GO ε i (q)), where ε r i i = +, if δ i = +1 and ε i =, if δ i = 1, then R R 1... R t ; ] [ n (iii) [1] let r, exp r (q) = m and n 0 = η (m) if ( 1) m 1 = 1 and ε = +, otherwise and for all i {,..., u}, S i Syl r (S r i), then. If n 0 = a 0 + a 1 r a u r u, R 1 Syl r (GO ε (q)), where ε =, η (m) R ((R 1 S )... (R 1 S ))... ((R 1 S u )... (R 1 S u )). } {{ }} {{ } a 1 times a u times The following lemma is a known fact and for an example one can extract it from [10]. Lemma.16. For the natural number m, (i) if m is odd, r Z m (q) and R Syl r (GO + m (q)), then R is abelian and R = qm 1 r ; (ii) if r Z m (q) and R Syl r (GO m (q)), then R is abelian and R = qm + 1 r. Lemma.17. Let G be a finite group such that M(G) = M(B n (q)) and let the functions η and η be defined as in (1). If r π(g) {p}, then: (i) if r =, then a r (G) ( q 1 ) n ; [ ] n (ii) if r, exp r (q) = m and n 0 = η, then a r (G) ( q η(m) + ( 1) m r ) n 0. (m) Proof. Let R Syl r (B n (q)) and R Syl r (SO n+1 (q)). Since M(G) = M(B n (q)), we conclude that a r (G) = a r (B n (q)) = a(r) which divides a(r ). (i) If r =, then we may assume that n = r r t such that r 1 <... < r t. For i {1,..., t}, put ε i = +, if δ i = +1 and put ε i =, if δ i = 1, where q r i 1 δ i (mod ). Also, let R i Syl (GO ε i (q)). Then by Lemma r i.15(ii), R R 1... R t an hence, a(r ) = a(r 1 )...a(r t ). Now Lemma.15(i) completes the proof of (i). (ii) If r, then we can assume that R Syl r (SO n+1 (q)) and n 0 = a 0 + a 1 r a u r u. Thus by Lemma.15(iii), we have R ((R 1 S )... (R 1 S ))... ((R 1 S u )... (R 1 S u )), } {{ }} {{ } a 1 times a u times where for i {,..., u}, S i Syl r (S r i) and R 1 Syl r (GO ε η (m) (q)), such that ε =, if ( 1)m 1 = 1 and ε = +, otherwise. Thus a(r) = (a(r 1 S )...a(r 1 S ))...(a(r 1 S u )...a(r 1 S u )). } {{ }} {{ } a 1 times Now we can see that for all i {1,..., u}, a(r 1 S i ) = (a(r 1 )) ri. a(r 1 ) = R 1 = q η(m) + ( 1) m r. This completes the proof of (ii). a u times But by Lemma.16, R 1 is abelian and Lemma.18. Let N be a normal subgroup of the finite group G and r, t π(n). If r ρ(t, N), r ρ(t, G) and ρ(t, G) π(g/n) =, then r π(g/n). Proof. The proof is straightforward.

5 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), Main Results We are going to prove the main theorem in the following: 3.1. On the Maximal Abelian Subgroups of the almost Simple Groups Containing B n (q) Let q and S = C n (q). We denote by φ the field automorphism of S with (a ij ) (a p ) as its map. ij Applying [10], if n, then S( φ ) = Aut(S). Let q and S = B n (q). We denote by δ the diagonal automorphism of S which is conjugate to the diagonal matrix diag(λ/λ, (λ/λ ) 1, I), where λ and λ are a square element and a non-square element of GF(q), respectively and by φ the field automorphism of S with (a ij ) (a p ) as its map. Applying [10], if ij n > is even, then S( φ δ ) = Aut(S). Note that if p =, then C n (q) B n (q). Lemma 3.1. Let G = S.T, where T Out(S) and r, r 1 π(g) such that exp r (q) = n and exp r1 (q) = (n 1). (i) If G contains a field automorphism ψ of order t, then C S (ψ) B n (q 1/t ); (ii) if q and G contains an automorphism δψ, where ψ is a field automorphism of G of order, then there is an element α M(C S (δψ)) such that (p k(n 1)/ + 1) α and if q 9, then α (p k/ 1)(p k(n 1)/ + 1) and otherwise, α 16(p k(n 1)/ + 1). Proof. (i) is a known fact (for details see [1, Proof of Lemma 3.1]) and (ii) goes back to Lemma 3.17 in [1]. Lemma 3.. Let S G Aut(S). If M is a maximal abelain subgroup of G, then [M : M S] gcd(, q 1)k. Proof. Since MS G Aut(S) and Z(S) = 1, we deduce that M M S G S Out(S). But as mentioned above, Out(S) = gcd(, q 1)k, so lemma follows. Theorem 3.3. Let n > 3 be an even number. If G is a finite group such that M(G) = M(S) and S G Aut(S), then G = S. Proof. We are going to break the proof into cases: Case 1. If G contains a field automorphism, then without loss of generality, we can assume that ψ G such that ψ is a field automorphism of the prime order t, where t k. Thus Lemma 3.1(i) implies that C S (ψ) B n (q 1/t ). Thus by Lemma.1(i,ii), C S (ψ) contains a maximal abelain subgroup M 0 of order β { (q n/t } + 1) such that β gcd(, q 1), l(q(n 1)/t + 1), where l q 1/t (q /t 1). Since M 0 is an abelain subgroup of gcd(, q 1) C S (ψ), we deduce that G contains a maximal abelian subgroup M such that M 0 ψ M, so M C G (ψ) and M 0 M S C G (ψ) S = C S (ψ), which implies that M S = M 0. Thus Lemma 3. implies that [M : M 0 ] divides gcd(, q 1)k and hence, there exits α M(G) such that β α and α β gcd(, q 1)k. We continue the proof in the following subcases: Subcase 1. If t is odd and t n, then Corollary.10(ii) forces Z n (q 1/t ) Z n (q). Let β = (qn/t + 1) gcd(, q 1). Then (q n + 1) π(α) Z n (q) and hence, by Lemma.1(i), α =. It follows that gcd(, q 1) (q n + 1) gcd(, q 1) β gcd(, q 1)k = (qn/t + 1)k. () Since n 6 is even and k 1, Lemma.9 shows that there exists s Z nk (p). Thus s (q n/t + 1) and so by (), s k. On the other hand, Fermat s little theorem shows that nk s 1, which is a contradiction.

6 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), Subcase. If t is odd and t n, then t n 1 and hence, Corollary.10(ii) forces Z (n 1) (q 1/t ) Z (n 1) (q). Let β = l(q(n 1)/t + 1) gcd(, q 1), where l q1/t (q /t 1). Then π(α) Z (n 1) (q) and hence, by Lemma.1(ii), (q n 1 + 1) α. It follows that gcd(, q 1) (q n 1 + 1) gcd(, q 1) q1/t (q /t 1)(q (n 1)/t + 1)k. (3) Since n 6 and k 1, Lemma.9 shows that there exists s Z (n 1)k (p). Thus s q 1/t (q /t 1)(q (n 1)/t + 1) and so by (3), s k. On the other hand, Fermat s little theorem shows that (n 1)k s 1, which is a contradiction. Subcase 3. If t is even, then Corollary.10(iii) forces Z (n 1) (q 1/t ) Z n 1 (q). Let β = l(q(n 1)/ + 1) gcd(, q 1), where l q 1/ (q 1). Then π(α) Z n 1 (q) and hence, by Lemma.1(iii), (q n 1 1) α. It follows that gcd(, q 1) (q n 1 1) gcd(, q 1) q1/ (q 1)(q (n 1)/ + 1)k. () Since n 6 and k are even, Lemma.9 shows that there exists s Z (n 1)k/ (p). Thus s q 1/ (q 1)(q (n 1)/ + 1) and so by (), s k. On the other hand, Fermat s little theorem shows that (n 1)k/ s 1, which is a contradiction. Case. If q and δ j ψ G, where ψ is a field automorphism of order, then k and by Lemma 3.1(ii), C Bn (q)(δ j ψ) contains a maximal abelain subgroup M 0 of order β such that (q (n 1)/ + 1) β and β l(q (n 1)/ + 1), where if q 9, then l = (p k/ 1) and otherwise, l = 16. Now, the same reasoning as in Subcase 3 in Case 1 leads us to get a contradiction. Case 3. If q and δ j G, then by Cases 1 and, G does not contain any field automorphism and δ j ψ G and hence, G = SO n+1 (q). It follows that (q n + 1) M(G), which is a contradiction. These contradictions show that G = B n (q). 3.. Proof of the Main Theorem Theorem 3.. If G is a finite group such that M(G) = M(B n (q)), then G B n (q). Proof. If n = n, then Lemma.6 completes the proof. Thus we may assume that n n. This allows us to assume that n 6. Since M(G) = M(B n (q)), we have GK(G) = GK(B n (q)), considering Lemma.5. [ ] 3n + 5 Therefore, π(g) = π(b n (q)), t(g) = t(b n (q)) = 5 and ρ(, G) = ρ(, B n (q)) = {, r n (q)}, using [16, Tables, 6 and 8]. It follows by Lemma.7(1) that there is a finite non-abelian simple group S such that S Ḡ = G/K Aut(G) for the maximal solvable subgroup K of G such that t(s) t(g) 1. We continue the proof in the following steps: Step I) K = 1. This implies that S G Aut(S). Proof. Put ρ = {r nk (p), r (n 1)k (p), r (n 1)k (p)}. By Corollary.10(i), exp rmk (p)(q) = m and by [16, Table 8], if [ ] n + 1 (n, q) = (6, ), then ρ(g) = ρ(b n (q)) = {7, 11, 13, 17, 31} and otherwise, ρ(g) = ρ(b n (q)) = {r i (q) : [ ] n i n} {r i (q) : < i n, i 1 (mod )}. These imply that ρ ρ(g) and hence, ρ is independent. Thus by Lemma.7(), there is a prime z {r (n 1)k (p), r (n 1)k (p)} π(s) such that z π(k). Also, r nk (p) ρ(, S) and hence Lemma.7(3) forces r nk (p) π(s) and r nk (p) π(k). Let R Syl rnk (p)(s) and R 1 Syl z (S). We have that R and R 1 act coprimely on K. We claim that K p = 1. If not, then we deduce that K has an R-invariant p-sylow subgroup P 1 and an R 1 -invariant p-sylow subgroup P. Thus Z(P 1 )R and Z(P )R 1 are subgroups of G. Since exp rnk (p) (q) = n, we have (p, r nk(p)) GK(B n (q)) = GK(G), by Lemma.1. It follows

7 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), that by Corollary.3, r nk (p) p t 1, where Z(P 1 ) = Z(P ) = p t and hence, nk t. If C Z(P )(R 1 ) = p e, then there is α M(G) = M(B n (q)) such that zp e α. It follows by Lemma.1(ii,iii) that p e p k. Also, exp z (p) {(n 1)k, (n 1)k} and by Lemma., z p t e 1 and hence, (n 1)k t e. Since nk t, we conclude that there is a natural number a such that t = nka. Therefore, (n 1)k nka e = (n 1)ka+ka e, so (n 1)k ka e. Since e k, we have that ka e and hence, (n 1)k ka e ka. It follows that (n 1) a, so t n(n 1)k. But Z(P ) is an abelian subgroup of G and hence, Z(P ) = p t a(g) = a(b n (q)), which is a contradiction, because if p is even, then a(b n (q)) = q n(n+1) and otherwise, a(b n (q)) = q n(n 1) +1, by Lemma.1. Now, we show that K = 1. If this is not the case, then there is a prime s π(k). Since a(b n (q)) M(G) is a power of p, we may assume that there is an abelian p-subgroup P of G such that P = a(g). Also, K p = 1, so P acts coprimely on K and hence, we can see that K has a P-invariant s-sylow subgroup S 0. So Z(S 0 )P is a subgroup of G. We may assume that Z(S 0 ) is a s-elementary abelian subgroup of G and Z(S 0 ) = s α. But P is abelian and P = a(g). This implies that C PZ(S0 )(Z(S 0 )) is abelian and hence, C PZ(S0 )(Z(S 0 )) = s α p β < P. Also, N PZ(S0 )(Z(S 0 )) C PZ(S0 )(Z(S 0 )) Aut(Z(S 0)) = GL α (s). Thus GL α (s) has an abelian subgroup of order P /p β. On the other hand, similar to the proof of Lemma.17 we can see that a p (GL α (s)) < s α. Therefore, P /p β < s α, which is a contradiction. It follows that K = 1. Thus by Lemma.7(1), S G Aut(S). Step II) G S p < q n n. Proof. Let p G S. Since t(s) t(g) 1, [16, Tables 3,8], S A 7, A 1 (q). Also, since (, r n (q)) GK(G), Lemma.7(3)(a) forces r n (q) G S and so, Syl r n (q)(g) = Syl rn (q)(s). Let R Syl rn (q)(g). It follows by Frattini s argument that G S p N G (R). Thus there is a p-subgroup Q of G such that QR is a subgroup of G and G S p Q. Since (p, r n (q)) GK(B n (q)) = GK(G), the action of Q on R is Frobenius. Therefore, G S p R 1. Also, (, r n (q)) GK(G) and hence by Corollary., R is abelian. Thus R = a(r 1 ), where R 1 Syl rn (q)(b n (q)). But since n = m n, B n (q) rn (q) = q n + 1. So, G q m S + 1 p < q n n. rn (q) Step III) S is not isomorphic to a sporadic simple group. Proof. If S is isomorphic to a sporadic simple group, then since Z(S) = 1, we have by Step I, G S Out(S). But Out(S), using [10, page 171, Table 5.1.c]. Therefore, ρ(g) = ρ(s), by Lemma.18. So, t(s) = t(g). [ ] [ ] 3n + 5 3n + 5 Also, t(s) 11 and t(g) = by [16, Tables and 8]. Therefore, since 11 if and only if n 13, we conclude that n {6, 10, 1}. Thus, we have the following cases: [ ] 3 a) If n = 6, then t(s) = t(g) = = 5. It follows that S {Fi 3, Fi, F 3} (up to isomorphism), considering [16, Table ]. On the other hand, exp rnk (p) (q) = n and hence by Lemma.13, r nk(p) ρ(, G). Thus since by Lemma.7(3)(a), ρ(, G) ρ(, S), we conclude that r nk (p) ρ(, S) and hence, Fermat s little theorem implies that there is an element z ρ(, S) such that 1k = nk z 1, which is a contradiction, considering the elements of ρ(, Fi 3 ), ρ(, Fi ) and ρ(, F 3) (see [16, Table ]). [ ] 35 b) If n = 10, then t(s) = t(g) = = 8. It follows that S F, considering [16, Table ]. Similar to the previous argument, we can assume that there is an element z ρ(, S) such that 0k = nk z 1, which is impossible, considering the elements of ρ(, F ). [ ] 1 c) If n = 1, then t(s) = t(g) = = 10, which is impossible, because there does not exist any sporadic simple group S with t(s) = 10.

8 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), Step IV) S is not isomorphic to the alternating group A x of degree x. Proof. If S is isomorphic to A x, then similar to the previous argument, we can see that r n (q) ρ(, S). Since n 6, we have t(g) 5 and hence, t(s). Therefore, [16, Table 3] implies that x 7 and hence, G Aut(A x ) = S x. But by Lemma.8(), ρ(, S) = {s π(a x ) : x 3 s x} {}, so x 3 r n (q) x. It q n + 1 follows that r n (q) M(G), r n (q) M(G) or 3r n (q) M(G). Thus by Lemma.1(i), gcd(, q 1) = dr n(q), where d {1,, 3}. This forces q n + 1 {1,, 3}, which is impossible. Step V) S is isomorphic to the simple group of Lie type in characteristic p. Proof. Using Steps (II,III,IV) and the classification theorem of finite simple groups, we conclude that S is a simple group of Lie type in characteristic p. If p p, then since by Lemma.1, a(g) = a(b n (q)) is a power of p, we can see that G a(g) a p (S) S. (5) p On the other hand, a p (S) a(s). We continue the proof in the following cases: Case 1. Let S F ( n+1 ), for n 1. Since t(s) t(g) 1, we may assume that S is not isomorphism to A 1 (p e ), A (p e ) with (p e 1, 3) = 1, A 3 () and A (p e ) with (p e + 1, 3) = 1, using [16, Table 8]. Therefore, Lemma.1 implies that a(s) = a p (S). Also, since p p, we obtain by Lemma.17 that a p (B n (q)) < q n. But a p (S) a p (G), so a(g) a p (G) G/S p < ( ) q n q n n = q 3n n n(n 1), using (5,6,7) and Step II. It follows that by Lemma.1, either + 1 k < 3nk n k ( ) n(n + 1) or k < 3nk n k. This forces n < 6, which is a contradiction. Case. Let S F ( m+1 ), for n 1. Fix q = m+1. Then [16, Table 5] implies that ρ(, S) = {, s 1, s, s 3 }, for some s 1 π((q 3 + 1)/(q + 1)) and s, s 3 π((q 6 + 1)/(q + 1)). Without loss of generality, we can assume that s 1 Z 3(m+1) () and s, s 3 Z 6(m+1) (). Thus Fermat s little theorem shows that m + 1 s i 1, for i {1,, 3}. It is known that Out(S) Z m+1, so G/S Out(S) Z m+1 shows that π(g/s). Thus by Lemma.7(3), ρ(, G) π(g/s) =. On the other hand, by [16, Tabeles,6], t(, G) =, so Lemma.18 forces to exist 1 j 3 such that s j π(g/s) π(out(s)) = π(m+1). This implies that s j m+1, contradicting (8). Step VI) S B n (q) or C n (q). Proof. By Step V, S is a simple group of Lie type in characteristic p. Since r nk (p) ρ(, G), Lemma.7(3)(a) forces r nk (p) π(s). Now, we consider all simple groups of Lie type in characteristic p one by one: a) Let S B m (p e ) or S C m (p e ). Then max{exp s (p) : s π(g) {p}} = nk and π(s) π(g). Thus by (9), max{exp s (p) : s π(s) {p}} = nk. On the other hand, B m (p e ) = C m (p e ) = p me (p e 1)...(p me 1) and hence, max{exp s (p) : s π(s) {p}} = me. It follows that nk = me. If r (n 1)k (p) π(s), then r (n 1)k (p) π( G S ). But Z(S) = 1 and G Aut(S). So, G S Out(S). Since Out(S) e (see [10, Propositions.. and.6.3]), we have r (n 1)k (p) e. Also, nk = me and hence, r (n 1)k (p) nk. But Fermat s little theorem implies that (n 1)k r (n 1)k (p) 1, which is a contradiction. Otherwise, r (n 1)k (p) π(s). Thus (n 1)k = max{exp s (p) : s π(s) (Z nk (p) {p})} = (m 1)e. It follows that e = k and m = n. Therefore (6) (7) (8) (9)

9 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), S B n (q) or S C n (q). b) Let S D m (p e ). Applying the same argument as that of in Step VI(a) shows that nk = me and since by [10, Proposition.8.], Out(S) 3 e, we get that r (n 1)k (p) π(s). Thus (n 1)k = (m 1)e. It follows that e = k and m = n. It is evident that a p (G) a p (S) G S p and by Lemma.1, q n(n 1) +1 a p (G) and a p (S) = q (n 1)(n ) +. n(n 1) (n 1)(n ) Therefore, (II) implies that + 1 < + n, which is impossible. [ ] 3m + 1 c) Let S D m (p e ). Since t(s) = and t(s) t(g) 1, by [16, Table 8] and Lemma.7(), respectively, we have m 5 and hence, max{exp s (p) : s π(s) {p}} = (m 1)e and max{exp s (p) : s π(s) (Z (m 1)e (p) {p})} = (m )e. On the other hand, [10, Proposition.7.3] implies that Out(S) 8e and hence, arguing as in Step VI(a) shows that nk = (m 1)e and (n 1)k = (m )e. Therefore, m 1 = n and e = k. But r m (q) π(s) and hence r n+1 (q) π(g) = π(b n (q)). Since B n (q) = q n (q 1)...(q n 1)/(, q 1), there exists a natural number f such that 1 f n and r n+1 (q) q f 1. It follows that n + 1 f. Moreover, n is even and hence, n + 1 f, which is a contradiction. [ ] m + 1 d) Let S A m 1 (p e ). Using Lemma.7() and [16, Table 8], t(s) = t(g) 1. Thus m 7 and hence, max{exp s (p) : s π(s) {p}} = me. Also, max{exp s (p) : s π(s) (Z me (p) {p})} = (m 1)e. Thus arguing as in Step VI(a) shows that me = nk and (m 1)e = (n 1)k. [ Therefore, ] m = n and e = k. But a p (S) a p (G) q n(n+1) (n + 1) and a p (S) = p [(m+1)e/] (k) n(n + 1)k, by Lemma.1. Therefore,, which is impossible. e) Let S A m 1 (p e ). We denote max{exp s (p) : s π(s) {p}} by α. Arguing as in Step VI(a) shows that nk = α = { me m 1 (mod ); (m 1)e otherwise. and if m 1 (mod ), then max{exp s (p) : s π(s) (Z me (p) {p})} = (m )e and otherwise, max{exp s (p) : s π(s) (Z (m 1)e (p) {p})} = (m 3)e. On the other hand, max{exp s (p) : s π(s) (Z nk (p) {p})} = (n 1)k. Therefore, we can see that if m 1 (mod ), then m = n and e = k, and hence m is even, which is a contradiction. Also, if m 0 (mod ), then we can assume that m 1 = n and e = k, and hence m is odd, which is a contradiction. f) If p =, e = f + 1 and S F (p e ), then similar to the previous argument and by the order of F (p e ) we can see that 1e = nk and 6e = (n 1)k. It follows that k = 3e and n =, which is a contradiction. g) If S E 7 (p e ), then Out(S) = (, q 1)e, considering [10, page 170, Table 5.1.B] and hence similar to the previous argument, we can see that r (n 1)k (p) π(s). Also, by the order of E 7 (p e ) we can see that max{exp s (p) : s π(s) {p}} = 18e and max{exp s (p) : s π(s) (Z 18e (p) {p})} = 1e. Again, similar to the previous argument we can conclude that 18e = nk and 1e = (n 1)k. It follows that n = 9, which is impossible. h) If S E 8 (p e ), then Out(S) = e, considering [10, page 170, Table 5.1.B]. Similar to the previous argument, we may assume that 30e = nk and (n 1)k = e. It follows that n = 5, which is a contradiction. i) If S {F (p e ), E 6 (p e ), E 6 (p e ), B ( e+1 ), G (3 e+1 )}, then the same argument as that of in the previous case shows that {exp s (p) : s π(g) {p}} = nk and {exp s (p) : s π(g) (Z nk (p) {p})} = (n 1)k. Therefore, we can see that n <, which is a contradiction. j) If S {G (p e ), 3 D (p e ), F () }, then [16, Table 9] implies that t(s) 3, which is a contradiction. Therefore, we conclude that S B n (q) or S C n (q). Step VII) S B n (q). Proof. By Step VI, it is enough to show that if p, then S C n (q). If not, then a(s) = a(c n (q)) a(g) = a(b n (q)), because M(G) = M(B n (q)). Therefore, by Lemma.1, q n(n+1) q n(n 1) +1, which is impossible and hence S C n (q). If p =, then C n (q) B n (q). It follows that S B n (q). Step VIII) G = S B n (q).

10 N. Ahanjideh, A. Iranmanesh / Filomat 8:9 (01), Proof. Using Theorem 3.3 and the previous steps, we conclude that G = S B n (q), as claimed. AAM s Conjecture. Given an arbitrary non-abelian group H, associate a graph Γ(H) to H which is called the non-commuting graph of H. The vertex set V(Γ(H)) is H Z(H) and the edge set E(Γ(H)) consists of (x, y), where x and y are distinct non-central elements of H such that xy yx. AAM s conjecture implies that if S is a non-abelian finite simple group and H is a group such that Γ(H) Γ(S), then H S. Lemma 3.5. If S is a finite simple group and H is a finite group such that Γ(S) Γ(H), then 1. [7] Z(H) = 1;. [, Theorem.5] M(H) = M(S). In [13], authors prove that AAM s conjecture holds for finite simple groups. As a consequence of the main theorem, we prove the following corollary. It is worth mentioning that our proof is different from [13]. Corollary 3.6. Let n > 3 be an even natural number and let q be a prime power. If G is a finite group such that Γ(G) Γ(B n (q)), then G B n (q). Proof. By Lemma 3.5, Γ(G) Γ(B n (q)) gives that M(G) = M(B n (q)). Therefore, Theorem 3. completes the proof. Acknowledgment. The first author would like to thank Shahrekord University for financial support. Partial support by the Center of Excellence of Algebraic Hyperstructures and its Applications of Tarbiat Modares University (CEAHA) is gratefully acknowledged by the second author (AI). References [1] N. Ahanjideh, A. Iranmanesh, A characterization of B n (q) by the set of orders of maximal abelian subgroups, Int. J. Algebra and Computation 19 () (009) [] N. Ahanjideh, A. Iranmanesh, On the relation between the non-commuting graph and the prime graph, Int. J. Group Theory 1 (1) (01) 5 8. [3] J. Bi, A characterization of L n (q) by the normalizers orders of their Sylow subgroups, Acta Math. Sinica (New Ser.) 11 (3) (1995) [] R. Carter, P. Fong, The Sylow -subgroup of the classical groups, J. Algebra 1 (196) [5] R.W. Carter, Finite groups of Lie type: conjugacy classes and complex characters, Wiley, New York, [6] G.Y. Chen, A characterization of alternating groups by the set of orders of their maximal abelian subgroups, Siberian Math. J. 7 (3) (006) [7] M.R. Darafsheh, Groups with the same non-commuting graph, Discrete applied math. 157 (009) [8] W. Feit, On large Zsigmondy primes, Proc. Amer. Math. Soc. 10 (1) (1988) [9] B. Huppert, Endliche Gruppen, Springer-Verlag, [10] P. Kleidman, M. Liebeck, The subgroup structure of finite classical groups, Cambridge University Press, [11] A.S. Kondrat ev, Prime graph components of finite simple groups, Math. USSRSb. 67 (1) (1990) [1] G. Malle, A. Moreto, G. Navarro, Element orders and Sylow structure of finite groups, Mathematische Zeitschrift 5 (1) (006) [13] R.M. Solomon, A.J. Woldar, Simple groups are characterized by their non-commuting graphs, J. Group Theory 16 (013) [1] M. Stather, Constructive Sylow theorems for the classical groups, J. Algebra 316 (007) [15] A.V. Vasil ev, I.B. Gorshkov, On recognition of finite simple groups with connected prime graph, Siberian Math. J. 50 () (009) [16] A.V. Vasil ev, E.P. Vdovin, An adjacency criterion for the prime graph of a finite simple group, Algebra and Logic (6) (005) [17] A.V. Vasil ev, E.P. Vdovin, Cocliques of maximal size in the prime graph of a finite simple group, Algebra and Logic 50() (011) [18] E.P. Vdovin, Maximal orders of abelian subgroups in finite Chevalley groups, Matematicheskie zametki 69 () (001) [19] L. Wang, A new characterization of simple K 3 groups, Northeast. Math. J. 17 () (001) [0] J.S. Willams, Prime graph components of finite groups, J. Algebra 69 () (1981) [1] A.J. Weir, Sylow p-subgroups of the classical groups over finite fields with characteristic prime to p, Proc. Amer. Math. Soc. 6 (1955)