CHARACTERIZATION OF PROJECTIVE GENERAL LINEAR GROUPS. Communicated by Engeny Vdovin. 1. Introduction

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1 International Journal of Group Theory ISSN (print): , ISSN (on-line): Vol. 5 No. 1 (016), pp c 016 University of Isfahan CHARACTERIZATION OF PROJECTIVE GENERAL LINEAR GROUPS ALIREZA KHALILI ASBOEI Communicated by Engeny Vdovin Abstract. Let G be a finite group π e (G) be the set of element orders of G. Let k π e (G) s k be the number of elements of order k in G. Set nse(g):={s k k π e (G)}. In this paper, it is proved if G = PGL (q), where q is odd prime power nse(g) = nse(pgl (q)), then G = PGL (q). 1. Introduction If n is an integer, then we denote by π(n) the set of all prime divisors of n. If G is a finite group, then π( G ) is denoted by π(g). We denote by π e (G) the set of orders of its elements. It is clear that the set π e (G) is closed partially ordered by divisibility, hence it is uniquely determined by µ(g), the subset of its maximal elements. The prime graph Γ(G) of a group G is defined as a graph with vertex set π(g) in which two distinct primes p, q π(g) are adjacent if G contains an element of order pq. Let t(g) be the number of connected components of Γ(G) π 1,..., π t(g) be the connected components of Γ(G). If π(g), then we always suppose that π 1. Then π 1 is called the even component of Γ(G) π,..., π t(g) are called the odd components of Γ(G). Let p be a prime. A group G is called a C pp if p π(g) p is an isolated vertex of the prime graph of G. In the other words, the centralizers of its elements of order p in G are p-groups. Given a finite group G, we can express G as a product of integers m 1, m,..., m t(g), where π(m i ) = π i for each i. These numbers m i are called the order components of G. In particular, if m i is odd, then we call it an odd order component of G (see [1]). According to the classification theorem of finite simple groups [7, 17, 19], we can list the order components of finite simple groups with disconnected prime graphs as in Tables 1-3 in [8]. MSC(010): Primary: 0D0; Secondary: 0D60. Keywords: Element order, set of the numbers of elements of the same order, projective general linear group. Received: November 013, Accepted: 16 June

2 18 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei Set s i = s i (G) := {g G the order of g is i} nse(g) := {s i i π e (G)}. In fact, s i is the number of elements of order i in G nse(g) is the set of sizes of elements with the same order in G. Throughout this paper we denote by ϕ the Euler s totient function. If G is a finite group, then we denote by P q a Sylow q subgroup of G. All other notations are stard we refer to [16], for example. For the set nse(g), the most important problem is related to Thompson s problem. In 1987, J. G. Thompson posed a very interesting problem related to algebraic number fields as follows. For each finite group G each integer d 1, let G(d) = {x G x d = 1}. We say that the groups G 1 G are of the same type if G 1 (d) = G (d), for all d = 1,, 3,.... Thompson s problem. Suppose G 1 G are of the same order type. If G 1 is solvable, is G necessarily solvable? (see [0, Problem 1.37]) It is easy to see that if G H are of the same order type, then nse(g) = nse(h) G = H. W. J Shi in [1] made the above problem public in Unfortunately, no one can solve it or give a counterexample until now, it remains open. The influence of nse(g) on the structure of finite groups was studied by some authors (see [3,, 5, 9]). In [], it is proved that L (p), where p is prime, is characterizable by nse(g) its order. In [1], it is proved that PGL (p), where p > 3 is prime number, is characterizable by nse(g) p π(g). In this paper it is proved that PGL (q), where q > 3 is odd prime power, is characterizable by nse(g) its order. In fact the main theorem of our paper is as follows: Main Theorem. Let G be a group such that nse(g) = nse(pgl (q)), where q > 3 is odd prime power G = PGL (q). Then G = PGL (q). We note that there are finite groups G which are not characterizable even by nse(g) G. For example see the Remark in [9].. Preliminary Results We first quote some lemmas that are used in deducing the main theorem of this paper. Lemma.1. [1] Let G be a finite group n be a positive integer dividing G. If M n (G) = {g G g n = 1}, then n M n (G). Lemma.. [1] Let G be a group such that nse(g) = nse(pgl (p)), where p > 3 is prime divisor of G but p does not divide G. Then G = PGL (p). Lemma.3. [18, Theorem 3] Let G be a finite group. Then the number of elements whose orders are multiples of n is either zero, or a multiple of the greatest divisor of G that is prime to n. Lemma.. [13] Let G be a Frobenius group of even order with H K its Frobenius kernel Frobenius complement, respectively. Then t(g) = T (G) = {π(k), π(h)}.

3 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei 19 Lemma.5. ([10, Theorem 10:3:1], [11, Theorem 18:6], [15]) Let G be a Frobenius group with kernel F complement C. Then the following assertions are true. (a) F is nilpotent. (b) F 1 (mod C ). (c) Every subgroup of C of order p q, with p, q (not necessarily distinct) primes, is cyclic. In particular, every Sylow subgroup of C of odd order is cyclic a Sylow -subgroup of C is either cyclic or a generalized quaternion group. If C is non-solvable, then C has a subgroup of index at most isomorphic to SL (5) M, where M has cyclic Sylow p-subgroups ( M, 30) = 1 in particular, 15, 0 / π e (G). If C is solvable O(C) = 1, then either C is a -group or C has a subgroup of index at most isomorphic to SL (3). A group G is a -Frobenius group if there exists a normal series 1 H K G such that K G/H are Frobenius groups with kernels H K/H, respectively. Lemma.6. [13] Let G be a -Frobenius group of even order which has a normal series 1 H K G such that K G/H are Frobenius groups with kernels H K/H, respectively. Then the following assertions hold. (a) t(g) = T (G) = {π 1 (G) = π(h) π(g/k), π (G) = π(k/h)}. (b) G/K K/H are cyclic, G/K divides Aut(K/H), ( G/K, K/H ) = 1. (c) H is nilpotent G is solvable. Lemma.7. [17] Let G be a finite group with t(g), then one of the following assertions is true: (a) G is a Frobenius or -Frobenius group; (b) G has a normal series 1 H K G such that H G/K are π 1 -groups K/H is simple, where H is a nilpotent group G/K Aut(K/H). Moreover, any odd order component of G is also an odd order component of K/H. Lemma.8. The set nse(pgl (q)) consists of the numbers 1, q 1 q together with all of the numbers of the form ϕ(r)q(q 1)/ all of the numbers ϕ(t)q(q + 1)/, where r > is a divisor of q + 1 t > is a divisor of q 1. Proof. The group PGL (q), where q = p n, has one conjugacy class of size q 1, which is related to elements of order p. So s p (PGL (q)) = (q 1). Also, this group has two conjugacy classes of sizes q(q 1)/ q(q + 1)/, which are related to elements of order. So s (PGL (q)) = q. Suppose that < r (q + 1). By [6, p. 6] we have µ(pgl (q)) = {q 1, p, q + 1}. Then r π e (PGL (q)). To find s r (PGL (q)), let H be a cyclic subgroup of order r of PGL (q) = T. We know T : C T (H) is the size of the conjugacy class of H. The group PGL (q) has (q 1)/ conjugacy classes of order q(q 1) (q 3)/ conjugacy classes of order q(q + 1). Since r > divides q + 1, we have T : C T (H) = q(q 1). Now we will show the number of conjugacy classes of such subgroups H is ϕ(r)/. Since r > divides q +1, we have each element of order r lies in a unique, up to conjugation, subgroup R of order q +1 of PGL (q)=t. Now, N T (R) = R C, is a dihedral group of order (q +1). So all elements of order r of R C lie in a unique subgroup of order r of R. Therefore there are

4 0 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei ϕ(r) elements of order r in N T (R). Now every element in R is conjugate to its inverse, so there are ϕ(r)/ classes of elements of order r in N T (R), hence there are ϕ(r)/ classes of elements of order r in PGL (q). Therefore, s r (PGL (q)) = ϕ(r)q(q 1)/. Also if t > divides q 1, then by µ(pgl (q)), t π e (PGL (q)) s t (PGL (q)) = ϕ(t)q(q + 1)/. Let s n be the number of elements of order n. We note that s n = kϕ(n), where k is the number of cyclic subgroups of order n in G. Also we note that if n >, then ϕ(n) is even. If n G, then by Lemma.1 the above notation we have ϕ(n) s n ( ) n d n s d In the proof of the main theorem, we apply ( ) the above comments. 3. Proof of the Main Theorem Let q = p n > 3, where p is an odd prime. Let G be a group such that G = PGL (q) = q(q 1) nse(g) = nse(pgl (q)). As q is the only odd number in nse(g) = nse(pgl (q)), by recalling that s (G) is odd one deduces that s (G) = q. We now prove the main theorem of this paper. To continue we need to the following Lemmas. Lemma 3.1. s p (G) = s p (PGL (q)) = (q 1). Proof. By ( ) we know that 1 + s p (G) is divisible by p, so s p (G) 1 (mod p). By nes(g), the only number m in nse(g) that m 1 (mod p) is q 1, so we must have s p (G) = q 1. Lemma 3.. rp / π e (G) for every r π(g). Proof. First we show that if r π(g)\{p}, then s r q 1. Suppose that s r = q 1. By ( ) we know that 1 + s r (G) is divisible by r, so r q, a contradiction. Thus by nse(g) if r π(g)\{p}, then q s r. Now we show that if rp π e (G) where r π(g)\{p}, then q s rp. Suppose rp π e (G), by ( ), rp 1 + s r + s p + s rp. We know that 1 + s p is divisible by q, on the other h, q s r. Therefore q s rp. Also if p i π e (G) where i, then since ϕ(p i ) s p i, s p i q 1 by nse(g), q s p. By Lemma.3, the number of elements whose orders are multiples of p is either zero, or a multiple of the greatest divisor of G that is prime to p. By Lemma 3.1, s p (G) = q 1 that is the greatest divisor of G prime to p, so the number of elements whose orders are multiples of p is (q 1) l, where l N. If l = 1, then since s p (G) = q 1, p is the only element of G whose order is multiples of p. Thus rp / π e (G) for every r π(g). Let l 1, let there exists r π(g) such that rp π e (G). Then the number of elements whose orders are multiples of p is (q 1)k+q(q+1)t+q(q 1)s = (q 1) l, where k, s t are non-negative integers(we note that except p, h π e (G) may be such that s h = q 1,

5 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei 1 so we write (q 1) k). Thus q(q + 1)t + q(q 1)s = (q 1) (l k). So q (q 1) (l k). Then q (l k). Therefore (q 1) l is greater than order of group G, a contradiction. Lemma 3.3. r π e (G) for every prime r distinct from p. Proof. Let r / π e (G) for some prime divisor r of G, with r distinct from p. Then the group P r acts fixed point free on the set of elements of order. Therefore P r s = q, a contradiction. Lemma 3.. OC(G) = OC(PGL (q)). Proof. By Lemma , we have t(g) = T (G) = {(q 1), {p}}, which implies that OC(G) = {q 1, q} = OC(PGL (q)). Lemma 3.5. G is neither Frobenius nor -Frobenius. Proof. If q is prime, then by Lemma., G = PGL (q). Thus G is neither Frobenius nor -Frobenius. Let q = p n, where p is prime n > 1. Assume G = NH is a Frobenius group with Frobenius kernel N Frobenius complement H. By Lemma., we have T (G) = {π(n), π(h)} = {π(q 1), {p}}. Since G = q(q 1) H ( 1) by Lemma.5(b), it follows that = q 1 H = q, which is impossible because q (q ) if q. Let G be a -Frobenius group. By Lemma.6, G has a normal series 1 H K G such that K G/H are Frobenius groups with Frobenius kernels H K/H, respectively. By Lemma.6(a), π (G) = π(k/h) = {p} by Lemma.6(b), K/H is cyclic. Thus K/H has an element of order p n. By Lemma 3., we get a contradiction. Lemma 3.6. G is isomorphic to PGL (q). Proof. By Lemma 3.5.7, G has a normal series 1 N G 1 G such that N is a nilpotent π 1 -group, G/G 1 is a solvable π 1 -group, G 1 /N is a simple C pp -group. By the definition of the prime graph component, the odd order component q of G is of a certain odd component of G 1 /N since G is a simple C pp -group. In particular, t(g 1 /N). Furthermore, G 1 /N G/N Aut(G 1 /N) by Lemma.7. Now using the classification of finite simple groups the results in Tables 1 3 in [8], we consider the following steps. Step 1. We prove that G 1 /N can not be an alternating group A n. If G 1 /N = A n, then since the odd order components of A n are primes, say p or p, we conclude that q = p or q = p. In both cases, q is a prime number. By Lemma., G is isomorphic to PGL (q), a contradiction. Step. If G 1 /N = A r (q ), then we distinguish the following six cases..1. Suppose G 1 /N = A p 1 (q ), where (p, q ) (3, ), (3, ), p is an odd prime q is a prime power. Then q = q p p (p 1) 1 (q 1)(p,q q Π p 1 1) i=1 (qi 1) (q 1). Thus q = (q p 1) p [(q 1)(p,q 1)] q

6 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei q p (p 1) < q p (p 1) Π p 1 i=1 (qi 1) q 1 < q. Hence q p (p 1) < q p. Therefore p (p 1) < p, which implies that p < 5. Since p is an odd prime, p = 3. Now it follows that q = (q 3 1) [(q 1)(3,q 1)] q 3 (q 1)(q 1) q 1. Therefore q = q (q ) q (q +) = (q +q ) > (q +q +1) (q 3 1) = [(q 1)(3,q 1)] q > q 3 (q 1)(q 1). It follows that q > (q 1)(q 1) = q 3 q q + 1 > q 3 q q. Thus > q q 1, which shows that q q 5 < 0. Hence q =, a contradiction... Suppose G 1 /N = A p (q ), where (q 1) (p + 1), p is an odd prime q is a prime p power. Then q = q 1 q p (p +1) < q p q 1 (p +1) q p (p +1) (q p +1 1)Π p 1 p i= (qi 1) (q 1). Thus q = (q 1) (q p +1 1)Π p 1 i= (qi 1) q 1 < q. Hence q p (p +1) < q p (q 1) q p. Therefore p (p +1) < p, which implies that p < 3, a contradiction..3. G 1 /N = A 1 (q ), where (q + 1) q is a prime power. Then q = q or q 1. Moreover, q (q 1) q(q 1) in both cases. If q = q, then q (q 1) q(q 1), which implies that = 1 or. If = 1, then A 1 (q) G Aut(A 1 (q)). It follows that G = PGL (q), where (q + 1) q is a prime power. If =, then G/N = A 1 (q). Since G/C G (N) Aut(N) = C 1, it follows that G = C G (N). Hence N Z(G), which implies that p π e (G), a contradiction. If q = q 1, then q = q + 1. Since q (q 1) q(q 1), we have that (q+1)[(q+1) 1] q(q 1). It follows that (q+1)[(q+1) 1] q(q 1), which implies that 7q 1, a contradiction... G 1 /N = A 1 (q ), where (q 1) q is a prime power. Then q = q or q +1. Moreover, q (q 1) q(q 1) in both cases. If q = q, then q (q 1) q(q 1), which implies that = 1 or. If = 1, then A 1 (q) G/N Aut(A 1 (q)). It follows that G = PGL (q), where (q 1) q is a prime power. If =, then G/N = A 1 (q). Since G/C G (N) Aut(N) = C 1, it follows that G = C G (N). Hence N Z(G), which implies that p π e (G), a contradiction. If q = q +1, then q = q 1. Since q (q 1) q(q 1), we have that (q 1)[(q 1) 1] q(q 1). It follows that (q 1)[(q 1) 1] q(q 1), which implies that q 1, a contradiction..5. G 1 /N = A 1 (q ), where q q is a prime power. Then q = q + 1 or q 1, q (q 1) q(q 1). If q = q + 1, then q = q 1. It follows that (q 1)[(q 1) 1] q(q 1), which implies that q + 1 = (q )t, where t is a natural number. Hence (q )(t 1) = 3, which shows that q = 1 or q = 3. Because q, we get a contradiction. If q = q 1, then q = q + 1. Since q (q 1) q(q 1), we have that (q + 1)[(q + 1) 1] q(q 1). It follows that q(q + 1)(q + ) q(q 1), which implies that q + < q 1, a contradiction..6. G 1 /N = A (). Then q must be equal to 3, 5 or 7. Since q > 3 so q = 5 or 7 by Lemma., G =PGL (5) or PGL (7), a contradiction. Step 3. If G 1 /N = A r (q ), then we distinguish the following three cases.

7 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei Suppose G 1 /N = A p 1 (q ), where p is an odd prime q is a prime power. Then q q = p +1 (q +1)(p,q +1) q p (p 1) < q p (p 1) q p (p 1) Π p 1 i=1 (qi ( 1) i ) (q 1). Thus q = (q p +1) (p +1) [(q +1)(p,q +1)] q +1). Therefore p (p 1) Π p 1 i=1 (qi ( 1) i ) q 1 < q. Hence q p (p 1) < q (p < (p + 1), which implies that p < 7. Since p is an odd prime, we have that p = 3 or 5. If p = 3, then q = (q 3 +1) [(q +1)(3,q +1)] (q q + 1) < q q < q 3 (q + 1) < q 3 (q + 1)(q 1) (q 1) < q. Thus q < q < q, a contradiction. If p = 5, then q = (q 5 +1) [(q +1)(5,q +1)] (q q 3 +q q +1) < q 8 q 10 < q 10 Π i=1 (qi ( 1) i ) (q 1) < q. Thus q 10 < q < q 10, a contradiction. 3.. Suppose G 1 /N = A p (q ), where (q + 1) (p + 1) such that p is an odd prime, q is a prime p power (p, q ) (3, 3), (5, ). Then q = q +1 q +1 p Thus q = (q +1) q p (p +1) (q +1) q (p +1) q p (p +1) < q p q (p +1) p (p +1) (q p +1 1)Π p 1 i= (qi ( 1) i ) (q 1). Π p 1 i=1 (qi ( 1) i ) q 1 < q. Hence < q (p +1). Therefore p (p +1) < (p +1), which implies that p <. Since p is an odd prime, we have that p = 3. Therefore, q 6 (q 1)(q 1) (q 1) < q = (q3 +1) contradiction Suppose G 1 /N = A 3 (), A 3 (3) or A 5 (). (q +1) = (q If G 1 /N = A 3 (), then q = 5 by Lemma., G = PGL (5), a contradiction. q + 1) < q, a If G 1 /N = A 3 (3), then q = 5 or 7 by Lemma., G =PGL (5) or PGL (7), a contradiction. If G 1 /N = A 5 (), then q = 7 or 11 by Lemma., G =PGL (7) or PGL (11), a contradiction. Step. If G 1 /N = B r (q ), then we consider the following two cases..1. Suppose G 1 /N = B r (q ), r = t q is an odd prime power. Then q = qr +1 q r (q r 1)Π r 1 i=1 (qi 1) (q 1). Thus q = (qr +1) q (r+1) q r r < q (q r 1)Π r 1 i=1 (qi 1) q 1 < q. Hence q r < q (r+1). Therefore r < (r + 1), which implies that r < 3, a contradiction... Suppose G 1 /N = B p (3), where p is an odd prime. Then q = 3p 3 p (3 p +1)Π p 1 i=1 (3i 1) (q 1). Thus q = (3p 1) 3 p 3 p (3 p +1)Π p 1 i=1 (3i 1) (q 1) < q. Hence 3 p < 3 p. Therefore p < p, which implies that p <, a contradiction. 1 Step 5. If G 1 /N = C r (q ), then we consider the following two cases Suppose G 1 /N = C r (q ), r = t q is an odd prime power. Then q = qr +1 (,q 1) q r (q r 1)Π r 1 i=1 (qi 1) (q 1). Thus q = (qr +1) (,q 1) q(r+1) q r < q r (q r 1)Π r 1 i=1 (qi 1) q 1 < q. Hence q r < q (r+1). Therefore r < (r + 1), which implies that r < 3. Since r = t, we have r =. Now q = q +1 (,q q(q 1) < (q + 1). Therefore q < 3, which is a 1) contradiction since q is a prime power.

8 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei 5.. Suppose G 1 /N p = C p (q ), where p is an odd prime q = or 3. Then q = q 1 (,q 1) p q p (q p +1)Π p 1 i=1 (qi 1) (q 1). Thus q = (q 1) (,q 1) qp q p < q p (q p +1)Π p 1 i=1 (qi 1) (q 1). Hence q p < q p. Therefore p < p, which implies that p <, a contradiction. Step 6. If G 1 /N = D r (q ), then we consider the following two cases Suppose G 1 /N = D p (q ), where p 5 is an odd prime q =, 3 or 5. p Then q = q 1 q p (p 1) q 1 Π p p 1 i=1 (qi 1) (q 1). Thus q = (q 1) q p q p (p 1) (q 1) < q p (p 1) Π p 1 i=1 (qi 1) (q 1) < q. Hence q p (p 1) < q p. Therefore p (p 1) < p, which implies that p < 3, a contradiction. 6.. Suppose G 1 /N = D p +1 (q ), where p is an odd prime q = or 3. p Then q = q 1 (q p 1) (,q 1) qp (,q 1) 1 (,q 1) q p (p +1) (q p + 1)q p +1 1)Π p 1 i=1 (qi 1) (q 1). Thus q = q p (p +1) 1 (p +1) < (,q 1) qp (q p + 1)(q p +1 1)Π p 1 i=1 (qi 1) q 1 < q. Hence q p (p +1) < q p. Therefore p (p + 1) < p, which implies that p < 1, a contradiction. Step 7. If G 1 /N = B (q ), where q = t+1 >, then we distinguish the following three cases Suppose q = q 1. Then q = q + 1. Since q (q q + 1)(q + q + 1) (q 1), it follows that (q + 1) [(q + 1) + 1] q 1 < q, a contradiction. 7.. Suppose q = q q + 1. Since q (q 1)(q + q + 1) (q 1) q >, it follows that q (q q + 1)(q + q + 1) q (q 1)(q + q + 1) q 1 < q = (q q + 1). Therefore q (q + q +1) < q q +1 < q + q +1, which shows that q < 1, a contradiction Suppose q = q + q + 1. Since q (q 1)(q q + 1) (q 1), it follows that q (q q + 1) q (q 1)(q q + 1) q 1 < q = (q + q + 1). Therefore q (q q ) < q (q q + 1) < q + q + 1 < q + q, which shows that q (q q ) < q + q. Thus q (q q ) < q + < 3 q. Hence q q < 3. It follows that 7 < q < + 7, which shows that 1 < q < 7. This is a contradiction since q = t+1 8. Step 8. If G 1 /N = D r (q ), where q is a prime power, then we distinguish the following six cases Suppose that G 1 /N = D r (q ), where r = t. Then q = qr +1 (,q +1) qr(r 1) Π r 1 i=1 (qi 1) (q 1). Thus q = (qr +1) q(r+1) q r(r 1) < q r(r 1) Π r 1 (,q +1) i=1 (qi 1) q 1 < q. Hence q r(r 1) < q (r+1). Therefore r(r 1) < (r + 1), which implies that 0 < r <, a contradiction. 8.. Suppose that G 1 /N = D r (), where r = t Then q = r r(r 1) ( r + 1)( r 1 1)Π r i=1 (i 1) (q 1). Thus q = ( r 1 + 1) r r(r 1) < r(r 1) ( r + 1)( r 1 1)Π r i=1 (i 1) q 1 < q. Hence r(r 1) < r, which implies that r < 3, a contradiction.

9 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei Suppose that G 1 /N = D p (3), where 5 p t + 1, p is an odd prime. Then q = 3p +1 3 p (p 1) Π p 1 i=1 (3i 1) (q 1). Thus q = (3p +1) 16 3 (p +1) 3 p (p 1) < 3 p (p 1) Π p i=1 (3i 1) q 1 < q. Hence 3 p (p 1) < 3 (p +1), which implies that p (p 1) < (p +1). Thus 0 < p <, a contradiction. 8.. Suppose that G 1 /N = D r (3), where r = t + 1 p such that t p is an odd prime. Then q = 3r r(r 1) (3 r + 1)(3 r 1 1)Π r i=1 (3i 1) (q 1). Thus q = (3r 1 +1) 3 r 3 r(r 1) < 1 3r(r 1) (3 r + 1)(3 r 1 1)Π r i=1 (3i 1) q 1 < q. Hence 3 r(r 1) < 3 r, which implies that r(r 1) < r. Thus r < 3, a contradiction Suppose that G 1 /N = D p (3), where p = t + 1, t p is an odd prime. Then q = 3p +1 or 3p +1) If q = 3p, then 3 p (p 1) (3 p 1 1)(3 p 1 + 1)Π p i=1 (3i 1) (q 1). Thus q = (3p 16 3 (p +1) 3 p (p 1) < 3 p (p 1) (3 p 1 1)(3 p 1 + 1)Π p i=1 (3i 1) q 1 < q. Hence 3 p (p 1) < 3 (p +1), which implies that p (p 1) < (p + 1). Thus 0 < p <, a contradiction ) If q = 3p, then 1 3p (p 1) (3 p 1 1)(3 p + 1)Π p i=1 (3i 1) (q 1). Thus q = (3p 3 p 3 p (p 1) < 1 3p (p 1) (3 p 1 1)(3 p + 1)Π p i=1 (3i 1) q 1 < q. Hence 3 p (p 1) < 3 p, which implies that p (p 1) < p. Thus p < 3, a contradiction. Step 9. If G 1 /N = G (q ), where q is a prime power, then we distinguish the following three cases Suppose G 1 /N = G (q ), where < q 1 (mod 3). Then q = q q + 1 q 6 (q 3 1)(q 1)(q + 1) (q 1). Thus q = (q q + 1) q q 6 < q 6 (q 3 1)(q 1)(q + 1) q 1 < q. Hence q 6 < q, which implies that q < 1, a contradiction. 9.. Suppose G 1 /N = G (q ), where < q 1 (mod 3). Then q = q + q + 1 q 6 (q 3 + 1)(q 1)(q 1) (q 1). Thus q = (q + q + 1) (q 3 1) q 6 q 6 (q 1) < q 6 (q 3 + 1)(q 1)(q 1) q 1 < q. Hence q 6 (q 1) < q 6, which implies that q <, a contradiction Suppose G 1 /N = G (q ), where 3 q. Then q = q + q + 1 or q q + 1. If q = q + q + 1, then q 6 (q 1) (q q +1) (q 1). Thus q = (q +q +1) (q 3 1) q 6 q 6 (q 1) < q 6 (q 1) (q q + 1) q 1 < q. Hence q 6 (q 1) < q 6, which implies that q <, a contradiction. If q = q q + 1, then q 6 (q 1) (q + q + 1) (q 1). Thus q = (q q + 1) q q 6 < q 6 (q 1) (q +q +1) q 1 < q. Hence q 6 < q, which implies that q < 1, a contradiction. Step 10. If G 1 /N = E 7 (), E 7 (3), E 6 () or F (). If G 1 /N = E 7 (), then G 1 /N = E 7 () = q = 73 or 17. By Lemma., G =PGL (73) or PGL (17), a contradiction. If G 1 /N = E 7 (3), then G 1 /N = E 7 (3) = q = 757 or By Lemma., G =PGL (757) or PGL (1093), a contradiction.

10 6 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei If G 1 /N = E 6 (), then G 1 /N = E 6 () = q = 13, 17 or 19. By Lemma., G =PGL (13), PGL (17) or PGL (19), a contradiction. If G 1 /N = F (), then G 1 /N = F () = q = 13. By Lemma., G =PGL (13), a contradiction. Step 11. If G 1 /N = 3 D (q ), where q is a prime power, then q = q q + 1 q 1 (q 6 1)(q 1)(q +q +1) (q 1). Thus q = (q q +1) < q 8 q 1 < q 1 (q 6 1)(q 1)(q +q +1) q 1 < q. Hence q 1 < q 8, which implies that q < 1, a contradiction. Step 1. If G 1 /N = F (q ), where q is a prime power, then we distinguish the following two cases Suppose G 1 /N = F (q ), where q is a an odd prime power. Then q = q q + 1 q (q 8 1)(q 6 1) (q 1) (q 1). Thus q = (q q + 1) q 8 q < q (q 8 1)(q 6 1) (q 1) q 1 < q. Hence q < q 8, which implies that q < 1, a contradiction. 1.. Suppose G 1 /N = F (q ), where q q >. Then q = q + 1 or q q + 1. If q = q + 1, then q (q 6 1) (q 1) (q q + 1) (q 1). Thus q = (q + 1) < q 10 q < q (q 6 1) (q 1) (q q + 1) q 1 < q. Hence q < q 10, which implies that q < 1, a contradiction. If q = q q + 1, then q (q 6 1) (q 1) (q + 1) (q 1). Thus q = (q q + 1) < q 8 q < q (q 6 1) (q 1) (q + 1) q 1 < q. Hence q < q 8, which implies that q < 1, a contradiction. Step 13. If G 1 /N = F (q ), where q = t+1 >, then q = q ± q 3 + q ± q + 1 q 1 (q 1)(q 3 +1)(q +1)(q 1)(q ± q 3 +q ± q +1) (q 1). Thus q = (q ± q 3 +q ± q +1) q 10 q 1 < q 1 (q 1)(q 3 +1)(q +1)(q 1)(q ± q 3 +q ± q +1) q 1 < q. Hence q 1 < q 10, which implies that q < 1, a contradiction. Step 1. If G 1 /N = G (q ), where q = 3 t+1 > 3, then q = q ± 3q + 1 q 3 (q 1)(q ± 3q + 1) (q 1). Thus q = (q ± 3q + 1) [(q + 1) 3q ] = (q q + 1) < q q 3 (q 1) < q 3 (q 1)(q ± 3q + 1) q 1 < q. Hence q 3 (q 1) < q, which implies that q <, a contradiction. Step 15. If G 1 /N = E 6 (q ) (q ) or E 6 (q ) (q > ), where q is a prime power. Then q = q6 ±q 3 +1 (3,q ±1) q 36 (q 1 1)(q 8 1)(q 6 1)(q 5 ±1)(q 3 ±1)(q 1) (q 1). Thus q = (q6 ±q 3 +1) (3,q ±1) (q 9 1) q 18 q 36 < q 36 (q 1 1)(q 8 1)(q 6 1)(q 5 ± 1)(q 3 ± 1)(q 1) q 1 < q. Hence q 36 < q 18, which implies that q < 1, a contradiction. Step 16. If G 1 /N is a sporadic simple group, then q = 5, 7, 11, 13, 17, 19, 3, 9, 31, 37, 1, 3, 7, 59, 67 or 71. By Lemma., G is isomorphic to PGL (5), PGL (7), PGL (11), PGL (13),

11 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei 7 PGL (17), PGL (19), PGL (3), PGL (9), PGL (31), PGL (37), PGL (1), PGL (3), PGL (7), PGL (59), PGL (67), PGL (71), a contradiction. Step 17. If G 1 /N = E 8 (q ), where q is a prime power. Thus q {q 8 + q 7 q 5 q q 3 + q + 1, q 8 q 7 +q 5 q +q 3 q +1, q 8 q 6 +q q +1, q 8 q +1}. Since q 8 +q 7 q 5 q q 3 +q +1 < (q 1)(q 8 + q 7 + q 6 + q 5 + q + q 3 + q + q + 1) = q 9 1 < q 9, it follows that q < q 9 in all cases. Since q 10 E 8 (q ) G G = q(q 1), we get a contradiction. We have thus examined all possibilities of G 1 /N. Now we have just seen if G 1 /N = A 1 (q ), where (q 1) q is a prime power or G 1 /N = A 1 (q ), where (q + 1) q is a prime power, then q = q G =PGL (q). In the other cases we get a contradiction. Since q is odd prime power, (q 1) or (q + 1). Therefore, we have proved that G = PGL (q). This completes the proof of the main theorem. Acknowledgment The author is thankful to the referee for carefully reading the paper for his suggestions remarks. References [1] A. K. Asboei, A new characterization of PGL (p), J. Algebra Appl., 1 no. 7 (013) 5 pages, doi: 10.11/S [] A. K. Asboei A. Iranmanesh, A characterization of linear group L (p), Czechoslovak Math. J., 6 (139) (01), [3] A. K. Asboei, S. S. Amiri, A. Iranmanesh A. Tehranian, A characterization of Symmetric group S r, where r is prime number, Ann. Math. Inform., 0 (01) [] A. K. Asboei, S. S. Amiri, A. Iranmanesh A. Tehranian, A new characterization of sporadic simple groups by NSE order, J. Algebra Appl., 1 no. (013) doi: 10.11/S [5] A. K. Asboei, S. S. Amiri, A. Iranmanesh A. Tehranian, A new characterization of A 7 A 8, An. Stiint. Univ. Ovidius Constanta Ser. Mat., 1 no. 3 (013) [6] A. R. Moghaddamfar W. J. Shi, The number of finite groups whose element orders is given, Beitrage Algebra Geom., 7 no. (006) [7] A. S. Kondratev, On prime graph components of finite simple groups, Math. Sb., 180 no. 6 (1989) [8] A. S. Kondratev V. D. Mazurov, Recognition of alternating groups of prime degree from the orders of their elements, Sib. Math. J., 1 no. (000) [9] C. G. Shao, W. Shi Q. H. Jiang, Characterization of simple K groups, Front. Math. China., 3 (008) [10] D. Gorenstein, Finite Groups, Chelsea Publishing Co., New York, [11] D. Passman, Permutation Groups, W. A. Benjamin, Inc., New York-Amsterdam, [1] G. Frobenius, Verallgemeinerung des sylowschen satze, Berliner sitz., (1895) [13] G. Y. Chen, On structure of Frobenius -Frobenius group, J. Southwest China Normal University, (Chinese), 0 no. 5 (1995) [1] G. Y. Chen, On Thompson s conjecture, J. Algebra, 185 no. 1 (1996) [15] J. G. Thompson, Normal p-complement for finite groups, Math. Z., 7 (1959)

12 8 Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei [16] J. H. Conway, R. T. Curtis, S. P. Norton R. A. Wilson, Atlas of finite groups, Oxford University Press, Eynsham, [17] J. S. Williams, Prime graph components of finite groups, J. Algebra, 69 no. (1981) [18] L. Weisner, On the number of elements of a group which have a power in a given conjugate set, Bull. Amer. Math. Soc., 31 (195) [19] N. Iiyori H. Yamaki, Prime graph components of the simple groups of Lie type over the field of even characteristic, J. Algebra, 155 no. (1993) [0] V. D. Mazurov E. I. Khukhro, Unsolved Problems in group theory, the Kourovka Notebook, 16 ed., Russian Academy of Sciences Siberian Division, Institute of Mathematics, Novosibirsk, 006. [1] W. Shi, A new characterization of the sporadic simple groups in group theory, In Proceeding of the 1987 Singapore Group Theory, Conference, Walter de Gruyter, Berlin, (1989) Alireza Khalili Asboei Department of Mathematics, Farhangian University, Shariati Mazaran, Iran khaliliasbo@yahoo.com