ON FINITE GROUPS WITH THE SAME PRIME GRAPH AS THE PROJECTIVE GENERAL LINEAR GROUP PGL(2, 81)

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1 Transactions on Algebra and its Applications 2 (2016), ISSN: ON FINITE GROUPS WITH THE SAME PRIME GRAPH AS THE PROJECTIVE GENERAL LINEAR GROUP PGL(2, 81) ALI MAHMOUDIFAR Abstract. Let G be a finite group. The prime graph of G is denoted by Γ(G). Also G is called recognizable by prime graph if for every finite group H with Γ(G) = Γ(H), we conclude that G = H. Until now, it is proved that if k is an odd number and p is an odd prime number, then PGL(2, p k ) is recognizable by prime graph. In this paper, we generalize this result and we prove that the finite group PGL(2, 81) is recognizable by prime graph. Key Words: Projective general linear group, prime graph, element order Mathematics Subject Classification: 20D05, 20D60, 20D Introduction Let N denotes the set of natural numbers. If n N, then we denote by π(n), the set of all prime divisors of n. Let G be a finite group. The set π( G ) is denoted by π(g). Also the set of element orders of G is denoted by π e (G). We denote by µ(s), the maximal numbers of π e (G) under the divisibility relation. The prime graph of G is a graph whose vertex set is π(g) and two distinct primes p and q are joined by an edge (and we write p q), whenever G contains an element of order pq. The prime graph of G is denoted by Γ(G). A finite group G is called recognizable Received: 4 June 2016, Accepted: 10 November Communicated by Ahmad Yousefian Darani; Address correspondence to A. Mahmoudifar; alimahmoudifar@gmail.com. c 2016 Transactions on Algebra and its Applications. 43

2 44 A. Mahmoudifar by prime graph if for every finite group H such that Γ(G) = Γ(H), then we have G = H. In [6], it is proved that if p is a prime number which is not a Mersenne or Fermat prime and p 11, 19 and Γ(G) = Γ(PGL(2, p)), then G has a unique nonabelian composition factor which is isomorphic to PSL(2, p) and if p = 13, then G has a unique nonabelian composition factor which is isomorphic to PSL(2, 13) or PSL(2, 27). In [3], it is proved that if q = p α, where p is a prime and α > 1, then PGL(2, q) is uniquely determined by its element orders. Also in [1], it is proved that if q = p α, where p is an odd prime and α is an odd natural number, then PGL(2, q) is uniquely determined by its prime graph (for more results, see [9, 10, 11]). In this paper as the main result we consider the recognition by prime graph of almost simple group PGL(2, 81). 2. Preparation of manuscript Lemma 2.1. ([12]) Let G be a finite group and N G such that G/N is a Frobenius group with kernel F and cyclic complement C. If ( F, N ) = 1 and F is not contained in NC G (N)/N, then p C π e (G) for some prime divisor p of N. Lemma 2.2. ([5]) Let G be a finite group and π(g) 3. If there exist prime numbers r, s, t π(g), such that {tr, ts, rs} π e (G) =, then G is non-solvable. Lemma 2.3. (see [14]) Let G be a Frobenius group with kernel F and complement C. Then every subgroup of C of order pq, with p, q (not necessarily distinct) primes, is cyclic. In particular, every Sylow subgroup of C of odd order is cyclic and a Sylow 2-subgroup of C is either cyclic or generalized quaternion group. If C is a non-solvable group, then C has a subgroup of index at most 2 isomorphic to SL(2, 5) M, where M has cyclic Sylow p-subgroups and ( M, 30) = 1. Using [8, Theorem A], we have the following result:

3 On finite groups with the same prime graph Lemma 2.4. Let G be a finite group with t(g) 2. Then one of the following holds: (a) G is a Frobenius or 2-Frobenius group; (b) there exists a nonabelian simple group S such that S G:= G/N Aut(S) for some nilpotent normal subgroup N of G. Lemma 2.5. ([15]) Let G = L ε n(q), q = p m, be a simple group which acts absolutely irreducibly on a vector space W over a field of characteristic p. Denote H = W G. If n = 2 and q is odd then 2p π e (H). 3. Main Results Lemma 3.1. Let G be a finite group and N be a nilpotent normal subgroup of G. Also let Ḡ := G/N be an almost simple group related to an abelian simple group S. Then one of the following assertions holds: (1) S C G (N)N/N and each p π(s) \ π(n) is adjacent to every q π(n), in Γ(G). (2) If G contains a Frobenius subgroup F := H C, with H as its kernel and C as its complement, then each p π(n) \ π(h) is adjacent to every q π(c), in Γ(G). Proof. Since N is a normal subgroup of G, C G (N)N/N is a normal subgroup of G/N. This shows that if C G (N)N/N is nontrivial, then S C G (N)N/N, since S G/N Aut(S), which implies (1). Therefore, we assume that C G (N)N/N is trivial. Let p π(n)\π(h). Also Let N p be the Sylow p-subgroup of N. Since N is nilpotent, N = O p (N) N p. Also since N is normal in G and (O p (N) Φ(N p )) is characteristic in N, we get that (O p (N) Φ(N p )) is normal in G. Hence we conclude that: G N = G/(O p (N) Φ(N p)) N/(O p (N) Φ(N p )) & N O p (N) Φ(N p ) = N p Φ(N p ). So without loose of generality, we may assume that N is an elementary abelian p- group and also C G (N)N/N is trivial. Now we obtain (2) by Lemma 2.1. This completes the proof.

4 46 A. Mahmoudifar Theorem 3.2. If G is a finite group such that Γ(G) = Γ(PGL(2, 81)), then G = PGL(2, 81), i.e. the almost simple group PGL(2, 81) is recognizable by prime graph. Proof. By [13, Lemma 7], it follows that: µ(pgl(2, 81)) = {3, 80, 82}. Therefore, the prime graph of PGL(2, 81) has two connected components which are {3} and π(3 8 1). Also we obtain that the subsets {2, 3} and {3, 5, 41} are two independent subsets of Γ(G). First of all, we prove that G is neither a Frobenius nor a 2-Frobenius group. Step 1. Let G be a Frobenius group with kernel K and complement C. By Lemma 2.3, we know that K is nilpotent and π(c) is a connected component of the prime graph of G. Hence we conclude that π(k) = {3} and π(c) = {2, 5, 41}, since 3 is the singleton vertex in Γ(G). If C is non-solvable, then by Lemma 2.3, C consists a subgroup isomorphic to SL(2, 5). This implies that 3 π(sl(2, 5)) π(c), which is a contradiction since π(c) = {2, 5, 41}. Therefore, C is solvable and so it contains a {5, 41}-Hall subgroup, say H. Since K is a normal subgroup of G, KH is a subgroup of G. Also we have π(kh) = {3, 5, 41}. Thus KH is a subgroup of odd order and so is a solvable subgroup of G. On the other hand, in the prime graph of G, the subset {3, 5, 41} is independent. Hence KH is a solvable subgroup of G such that its prime graph contains no edge, which contradicts to Lemma 2.2. Therefore, by the above argument we get that G is a Frobenius group. Step 2. Let G be a 2-Frobenius group with the normal series 1 H K G, where K is a Frobenius group with kernel H and G/H is a Frobenius group with kernel K/H. We know that G is a solvable group. This implies that G contains a {3, 5, 41}-Hall subgroup, say T. Now similar to the previous step, we deduce that T is solvable and Γ(T ) is an empty graph which is a contraction by Lemma 2.2.

5 On finite groups with the same prime graph Step 3. By Steps 1 and 2, the finite group G is neither Frobenius nor 2-Frobenius. So by Lemma 2.4, we conclude that there exists a nonabelian simple group S such that: S G := G/K Aut(S) in which K is the Fitting subgroup of G. Since {2, 3} is an independent subset of Γ(G), by Lemma 2.4, we conclude that 3 π(s). Also we know that π(s) π(g). Since π(g) = {2, 3, 5, 41}, so by [7, Table 8], we get that S is isomorphic to A 5, A 6, PSU 4 (2), S 4 (3 2 ) or PSL 2 (3 4 ). Now we consider each possibility for the simple group S. Step 3.1. Let S be isomorphic to the alternating group A 5 or A 6. Since π(s) π(out(s)) = {2, 3, 5}, we conclude that 41 π(k). We know that S consists a Frobenius subgroup 2 2 : 3. Hence since 41 π(k), by Lemma 3.1, we deduce that 41 3, which is a contradiction. Step 3.2. Let S be isomorphic to the simple group PSU 4 (2). By [4], the finite group S contains a Frobenius group 2 2 : 3, so similar to the above argument we get a contradiction. Step 3.3. Let S be isomorphic to the simple group S 4 (9). By [2], in the prime graph of the simple group S, we have 2 3. Hence 2 3 in Γ(G), which is a contradiction. Step 3.4. Let S be isomorphic to PSL 2 (81). Hence PSL 2 (81) Ḡ Aut(PSL 2(81)). Let π(k) contains a prime r such that r 3. Since K is nilpotent, we may assume that K is a vector space over a field with r elements (analogous to the proof of Lemma 3.1). Hence the prime graph of the semidirect product K PSL 2 (81) is a subgraph of Γ(G). Let B be a Sylow 3-subgroup of PSL 2 (81). We know that B is not cyclic. On the other hand K B is a Frobenius group such that π(k B) = {r, 3}. Hence B should be cyclic which is a contradiction. Let π(k) = {3}. In this case, by Lemma 2.5, we get that there is an edge between 2 and 3 in the prime graph of G which is a contradiction.

6 48 A. Mahmoudifar Therefore, by the above discussion, we deduce that K = 1. Also this implies that PSL 2 (81) G Aut(PSL 2 (81)). We know that Aut(PSL 2 (81)) = Z 2 Z 4. Since in the prime graph of PSL 2 (81) there is not any edge between 41 and 2, we get that G = PSL 2 (81). Also if G = PSL 2 (81) : θ, where θ is a field automorphism, then we get that 2 and 3 are adjacent in G, which is a contradiction. If G = PSL 2 (81) : γ, where γ is a diagonal-field automorphism, then we get that G does not contain any element with order 2 41 (see [3, Lemm 12]), which is contradiction, since in Γ(G), This argument shows that G = PGL 2 (81), which completes the proof. Acknowledgments The author wish to express his thanks to the referee for his(her) valuable remarks and comments that improved the presentation of the paper. References [1] Z. Akhlaghi, M. Khatami and B. Khosravi, Characterization by prime graph of PGL(2, p k ) where p and k are odd, International Journal of Algebra and Computation 20 (7), , [2] A. A. Buturlakin, Spectra of Finite Symplectic and Orthogonal Groups, Siberian Advances in Mathematics, 21 (3), , [3] G. Y. Chen, V. D. Mazurov, W. J. Shi, A. V. Vasil ev and A. Kh. Zhurtov, Recognition of the finite almost simple groups P GL 2(q) by their spectrum, J. Group Theory 10 (1), 71-85, [4] M. A. Grechkoseeva, On element orders in covers of finite simple classical groups, J. Algebra, 339, , , [5] G. Higman, Finite groups in which every element has prime power order, J. London Math. Soc., 32, , [6] B. Khosravi, n-recognition by prime graph of the simple group P SL(2, q), J. Algebra Appl., 7 (6), , [7] R. Kogani-Moghadam and A. R. Moghaddamfar, Groups with the same order and degree pattern, Sci. China Math., 55 (4), , [8] A. S. Kondrat ev, Prime graph components of finite simple groups, Math. USSR- SB., 67 (1), , 1990.

7 On finite groups with the same prime graph [9] A. Mahmoudifar, Recognition by prime graph of the almost simple group P GL(2, 25), Journal of Linear and Topological Algebra, accepted. [10] A. Mahmoudifar and B. Khosravi, On quasirecognition by prime graph of the simple groups A + n (p) and A n (p), J. Algebra Appl., 14 (1), (12 pages), (2015). [11] A. Mahmoudifar and B. Khosravi, On the characterization of alternating groups by order and prime graph, Sib. Math. J., 56 (1), , [12] V. D. Mazurov, Characterizations of finite groups by sets of their element orders, Algebra Logic, 36 (1), 23-32, [13] A. R. Moghaddamfar and W. J. Shi, The number of finite groups whose element orders is given., Beiträge Algebra Geom., 47 (2), , [14] D. S. Passman, Permutation groups, W. A. Bengamin, New York, [15] A. V. Zavarnitsine, Fixed points of large prime-order elements in the equicharacteristic action of linear and unitary groups, Sib. Electron. Math. Rep., 8, , Ali Mahmoudifar Department of Mathematics, Department of Mathematics, Tehran North Branch, Islamic Azad University Tehran, Iran alimahmoudifar@gmail.com