International Journal of Group Theory ISSN (print): 51-7650, ISSN (on-line): 51-7669 Vol. 5 No. 1 (016), pp. 17-8. c 016 University of Isfahan www.theoryofgroups.ir www.ui.ac.ir 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 01. 17
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 1989. 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)}.
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
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,
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 3. 3.3, 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
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.
Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei 3 3.1. 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. 3.3. 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. 5.1. 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.
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. 6.1. 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. 7.1. 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. 7.3. 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. 8.1. 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 + 1 5. Then q = r 1 + 1 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.
Int. J. Group Theory 5 no. 1 (016) 17-8 A. Khalili Asboei 5 8.3. 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 1 +1 1 3r(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. 8.5. 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) 1 +1 +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. 1 +1 1 +1) 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. 9.1. 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. 9.3. 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 () = 63 3 11 5 7 3 11 13 17 19 31 3 73 17 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) = 3 363 5 73 11 133 17 19 37 1 61 73 57 757 1093 q = 757 or 1093. By Lemma., G =PGL (757) or PGL (1093), a contradiction.
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 () = 36 3 9 5 7 11 13 17 19 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 () = 11 3 3 5 13 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. 1.1. 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),
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/S0199881350000. [] A. K. Asboei A. Iranmanesh, A characterization of linear group L (p), Czechoslovak Math. J., 6 (139) (01), 59-6. [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) 13-3. [] 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/S0199881501587. [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) 3-50. [6] A. R. Moghaddamfar W. J. Shi, The number of finite groups whose element orders is given, Beitrage Algebra Geom., 7 no. (006) 63 78. [7] A. S. Kondratev, On prime graph components of finite simple groups, Math. Sb., 180 no. 6 (1989) 787 797. [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) 9-30. [9] C. G. Shao, W. Shi Q. H. Jiang, Characterization of simple K groups, Front. Math. China., 3 (008) 355-370. [10] D. Gorenstein, Finite Groups, Chelsea Publishing Co., New York, 1980. [11] D. Passman, Permutation Groups, W. A. Benjamin, Inc., New York-Amsterdam, 1968. [1] G. Frobenius, Verallgemeinerung des sylowschen satze, Berliner sitz., (1895) 981-993. [13] G. Y. Chen, On structure of Frobenius -Frobenius group, J. Southwest China Normal University, (Chinese), 0 no. 5 (1995) 85 87. [1] G. Y. Chen, On Thompson s conjecture, J. Algebra, 185 no. 1 (1996) 18 193. [15] J. G. Thompson, Normal p-complement for finite groups, Math. Z., 7 (1959) 33 35.
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, 1985. [17] J. S. Williams, Prime graph components of finite groups, J. Algebra, 69 no. (1981) 87 513. [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) 9 96. [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) 335 33. [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) 531-50. Alireza Khalili Asboei Department of Mathematics, Farhangian University, Shariati Mazaran, Iran Email: khaliliasbo@yahoo.com