A Characterization of PSL(3,q) for q =2 m

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1 Acta Mathematica Sinica, English Series July, 00, Vol.18, No., pp A Characterization of PSL(,q) for q = m A. IRANMANESH S. H. ALAVI B. KHOSRAVI Department of Mathematics, Tarbiat Modarres University, P. O. Box: , Tehran, Iran Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran. iranmana@modares.ac.ir Abstract The order components of a finite group are introduced in [1]. In [9], it is proved that the group PSL(,q), where q is an odd prime power, is uniquely determined by its order components. In this paper, we show that the group PSL(,q), where q = m, is also uniquely determined by its order components. Keywords Finite group, Simple group, Prime graph, Order component MR(000) Subject Classification 0D05, 0D60 1 Introduction Let G be a finite group, π(g) the set of prime numbers dividing G, the order of G. The prime graph Γ(G) of a group G is a graph whose vertex set is π(g), and, two distinct primes p and q are linked by an edge if and only if G contains an element of order pq. Let π i, i =1,,...,t(G) be the connected components of Γ(G). For G even, π 1 will be the connected component containing. Then G can be expressed as a product of some positive integers m i, i =1,,...,t(G) withπ(m i )=π i. The integers m i are called the order components of G. The set of order components of G will be denoted by OC(G). If the order of G is even, we will assume that m 1 is the even order component and m,...,m t(g) will be the odd order components of G. The non-abelain simple groups having at least three prime graph components are obtained by Chen [1, Tables 1,,]. We have exhibited non-abelian simple groups with two order components in Table 1 using [, ]. The following groups are uniquely determined by their order components: Suzuki-Ree groups [4], Sporadic simple groups [5], PSL (q) [1],E 8 (q) [6], G (q) [q 0 (mod )] [7], A p where p and p are primes [8], PSL(,q)whereq is an odd prime power [9] and F 4 (q) whereq is even [17]. In this paper, we prove that PSL(,q)where q = m,m 1are also uniquely determined by their order components. In other words we have: The main theorem Let G be a finite group and M =PSL(,q)whereq = m,m 1and OC(G) =OC(M). Then G = M. Received December 15, 000, Revised August 15, 001, Accepted November 1, 001

2 464 Iranmanesh A. et al. Preliminary Results In this section we bring some preliminary lemmas to be used in the proof of the main theorem. Lemma.1 [, Theorem A] If G is a finite group and its prime graph has more than one component, then G is one of the following groups: (a) Asimplegroup; (b) AFrobenius or-frobenius group; (c) An extension of a π 1 -group by a simple group; (d) An extension of a simple group by a π 1 -solvable group; (e) An extension of a π 1 -group by a simple group by a π 1 -group. Lemma. [, Corollary] If G is a solvable group with at least two prime graph components, then G is either Frobenius group or -Frobenius group and G has exactly two prime graph components one of which consists of the prime dividing the lower Frobenius complement. Lemma. [10, Theorem 1] Let G be a Frobenius group of even order, H, K be Frobenius complement and Frobenius kernel of G, respectively. Then t(g) =, and prime graph components of G are π(h), π(k) and G has one of the following structures: (a) π(k), all Sylow subgroups of H are cyclic; (b) π(h), K is an abelian, H is a solvable group, the Sylow subgroups of odd order of G are cyclic and the -Sylow subgroups of G are cyclic or generalized quaternion groups; (c) π(h), K is an abelian, H is not a solvable group, and there exists H 0 H such that H : H 0, H 0 = Z SL(, 5), ( Z, 5) =1 and the Sylow subgroups of Z are cyclic. Lemma.4 [10, Theorem ] Let G be a -Frobenius group of even order. Then: (a) t(g) =; (b) There exists normal series 1 H K G such that, π 1 = π(g/k) π(h), π(k/h) =π, G/K and K/H are cyclic, G/K Aut(K/H), ( G/K, K/H ) =1and G/K < K/H. Moreover, H is a nilpotent group. Lemma.5 [11, Lemma 8] Let G be a finite group with t(g) and N be a normal subgroup of G. IfN is a π i -group for some prime graph component of G and m 1,m,...m r are some of the order components of G but not a π i -number, then m 1 m m r is a divisor of N 1. Lemma.6 Let M =PSL(,q), whereq = m,m. Then: (a) If p π(m) and p does not divide q, then S p <q,wheres p Syl p (M); (b) If p π 1 (M), p α M and p α 1 0(mod q +q+1 k ), k =(q 1, ), thenp =and α =m, hencep α = q ; (c) If q 1 and q +q 6 (q (q +1),thenq =16. Proof Since M = G = q (q 1) (q +1) (q +q+1) k, k =1 or, it is easy to show that (a) holds. To prove (b) we consider the following two cases: Case 1 m is even. Then q 1(mod)andk =. In this case q +q+1 <p α since p α 1 0 (mod q +q+1 ). Therefore, G = q (q (q +1) q +q+1. Since (q +1,q =1, p α divides

3 A Characterization of PSL(,q) for q = m 465 q or (q or q +1. Ifp α divides q +1,thenq q < 0 and this is impossible because q 16. If p α divides (q,thenrp α =(q 1) for some positive integer r. Sincep α 1 0 (mod q +q+1 ), there exists a positive integer s such that (p α =s(q + q + 1). Therefore, rsp α +sq +=p α, i.e., we must have rs, because if rs, then rsp α +sq +> p α, which is a contradiction. Thus rs =1 or, therefore, r = s =1or,r =ands =1or,r =1 and s =. If r = s =1, then p α =(q + 1), thus (q + 1), which is a contradiction. If r =1 and s =, then p α =6q +, i.e., q 8q =0, and this is a contradiction. If r =and s =1, then p α = q +1, soq 8q 5 =0, which is impossible. Therefore, p α q and hence p =. By assumption, s(q + q +1)=(p α for some positive integer s. It is easy to see that p α >q /4, and therefore, p α = q n 1, n 0, thus sq + sq + s +=q n 1, n 0, i.e., q s +,sos + q. Sincep α q, s(q + q +1) (q, i.e., s q. Therefore, q s + q. Therefore, s + must be one of the numbers q or q or q. Ifs +=q, then q s+1 because sq+s+1 =q n 1, n 0. But q s+ therefore, q, i.e., q =or4,which is a contradiction. If s+ =q, then q s+ because sq +s+ =q n 1, n 0, but q s+, thus q =1, which is a contradiction. Therefore, s +=q. Sincesq + sq + s +=q n 1, n 0, then sq + s +=q n 1, n 0, therefore, s += n 1,thusq = n 1 because s +=q. Therefore, p α = q and α =m. Case m is odd. Then q 1 (mod ) and the odd order component of M is q + q +1. Since M = q (q 1) (q +1)(q +q +1), then p α divides q,(q 1) or q +1. But q +q +1 <p α and q + q +1> (q and q + 1, then we must have p α q, i.e., p =. By assumption t(q + q +1)=p α 1 for some positive integer t, then by a similar method to that of (a), we must have p α = n q, n 0andt +1=q. Therefore, α =m and p α = q. To prove (c), since (q +,q+1)=1, (q 1,q+1)=1 and q + q =(q 1)(q + ), then q+ 6 q 1. Since q 1, (q 1,q+ ) = and therefore, (q +)/6 =andq =16. Lemma.7 Let G be a finite group, M =PSL(,q) where q = m,m 1and OC(G) = OC(M). ThenG is neither a Frobenius group nor a -Frobenius group. Proof If q =or4,m has at least order components. Using Lemmas. and.4, G can not be Frobenius group or -Frobenius group. Let q 8, we have two cases: Case 1 m is even. Then q 1(mod)and G = q (q (q +1)(q + q +1)/. If G is a Frobenius group then by Lemma., OC(G) ={ H, K } where H and K are the Frobenius complement and the Frobenius kernel of G, respectively. Supposing K, then K = q (q (q +1) and H = q +q+1. Let p be a prime which divides q +1. Then S p q +1< q +q+1 where S p is a p-sylow subgroup of K. Since K is nilpotent and (q +1,q(q (q + q + 1)) =1, then S p must be a normal subgroup of G and by Lemma.5 q +q+1 S p 1, which contradicts S p < q +q+1. If H, then H = q (q (q +1) and K = (q + q +1)/. Since H divides K 1, then q (q (q +1) q +q, which is a contradiction. Let G be a -Frobenius group. By Lemma.4 there is a normal series 1 H K G such that K/H = q +q+1 and G/K < K/H. Since K/H = q +q+1 < (q 1) (q+1),there

4 466 Iranmanesh A. et al. exists a prime p such that p (q (q +1)/ andp does not divide G/K, i.e., p H since π 1 = π(g/k) π(h). Therefore, p divides (q 1) or q + 1 and hence S p < (q + q +1)/. But S p must be a normal subgroup of K because H is nilpotent and K = q +q+1 H. Then q +q+1 S p 1, by Lemma.5, contradicting S p < (q + q +1)/. Case m is odd. Then q 1 (mod) and G = q (q (q +1)(q + q +1). IfG is a Frobenius group or a -Frobenius group, a similar treatment to that for the first case leads to a contradiction. Lemma.8 Let G be a finite group. If the order components of G are the same as those of M =PSL(,q) where q = m,m 1, theng has a normal series 1 H K G satisfying the following conditions: (a) H and G/K are π 1 -groups, K/H is a non-abelian simple group with π(k/h) involving π and π,whereπ = π 1 and H is a nilpotent group; (b) The odd order components of M are equal to some of those of K/H, especially, if q =, then t(k/h) and if q =4,thent(K/H) 4. For other cases, t(k/h). Proof (a) This part of the lemma follows from Lemmas.1,.,. and.7 because the prime graph of M has at least two prime graph components. (b) For primes p and q, ifk/h has an element of order pq, theng has an element of this order by (a). Hence by the definition of prime graph components, an odd order component of G must be an odd order component of K/H. Ifq =, then M has three order components. If q =4, then M has four order components and otherwise the number of order components of M is equal to two. Therefore, the proof is completed. In the next section we prove the main theorem. Proof of the Main Theorem By Lemma.8, G has a normal series 1 H K G such that H and G/K are π 1 -groups, and K/H is a non-abelain simple group. If q =, then t(k/h). If q =4, then t(k/h) 4. If q>4, then t(k/h), and also the odd order components of M are some of the odd order components of K/H. Here we consider three cases: q 1(mod),q 1(mod)or q {, 4}. Using Tables 1, we have: Case 1 If q 1 (mod ), then q 16 and m 4 is an even number. By Lemma.8, one of the odd order components of K/H is (q + q +1)/. Step 1 We prove that K/H cannot be an alternating group A n. If K/H = A n, then the odd order components of A n are p or p, i.e., (q +q +1)/ =p or p. If (q +q+1)/ =p, sincep 1 K/H, wehave (q 1)(q+) G, so q +q 6 (q 1) (q+1). Therefore, by Lemma.6(c), q =16, and then p =91. Thus G = , therefore, 11 K/H, which is a contradiction since 11 G. If(q + q +1)/ =p, from Table, p and p are primes. Since p K/H, then (q 1)(q+) G, asaboveq =16 and p =91, which is a contradiction, since p must be a prime number.

5 A Characterization of PSL(,q) for q = m 467 Table 1 The order components of simple groups 1) with t(g) = Group Orcmp1 Orcmp A p, p 5, 6 4 (p )(p )(p p p and p notbothprime A p+1, p 4, 5 4 (p )(p (p +1) p p 1andp +1 notboth prime A p+, p, 4 4 (p (p +1)(p +) p p and p + notboth prime A p 1 (q), (p, q) (, ), (, 4) q p(p 1) Π p 1 i=1 (qi A p(q), q 1 p 1 q p(p+1) (q p+1 Π p 1 i= (qi A p 1 (q) q p(p 1) Π p 1 i=1 (qi ( 1) i ) A p(q), q +1 p +1 q p(p+1) (q p+1 Π p 1 i= (qi ( 1) i ) q p 1 (q 1)(p,q 1) q p 1 q 1 q p +1 (q+1)(p,q+1) q p +1 q+1 (p, q) (, ), (5, ) A () B n(q), n = m 4,q odd q n (q n Π n 1 i=1 (qi q n +1 B p() p ( p +1)Π p 1 i=1 (i p 1 C n(q), n = m q n (q n Π n 1 i=1 (qi q n +1 (,q 1) C p(q), q =, q p (q p +1)Π p 1 i=1 (qi q p 1 (,q 1) D p(q), p 5, q =,, 5 q p(p 1) Π p 1 i=1 (qi q p 1 q 1 1 D p+1 (q), q =, (,q 1) qp(p+1) (q p +1)(q p+1 Π p 1 i=1 (qi q p 1 (,q 1) q n +1 (,q+1) D n(q), n = m 4 q n(n 1) Π n 1 i=1 (qi D n(), n = m +1 5 n(n 1) ( n + 1)( n 1 Π n i=1 (i n 1 +1 D p(), p m +1,p 5 p(p 1) Π p 1 i=1 (i +1 4 D n(), n = m 1 +1 p, m n(n 1) ( n + 1)( n 1 Π n i=1 (i +1 G (q), q ɛ(mod),ɛ= ±1, q> q 6 (q ɛ)(q (q + ɛ) q ɛq +1 D 4 (q) q 1 (q 6 (q (q 4 + q +1) q 4 q +1 F 4 (q), q odd q 4 (q 8 (q 6 (q 4 q 4 q +1 F 4 () E 6 (q) q 6 (q 1 (q 8 (q 6 (q 5 (q (q q 6 +q +1 (,q 1) E 6 (q), q> q 6 (q 1 (q 8 (q 6 (q 5 +1)(q +1)(q q 6 q +1 (,q+1) M J Ru He Mcl Co Co Fi F 5 = HN ) p is an odd prime number. Step We prove that K/H cannot be a simple group of type A n (q )or A n (q )withthe exception of A (q). If K/H = A n (q ), then (n, q )=(, ), (, 4), (1,q ), (p 1,q )or(p, q )andq 1 p 1by Tables 1 and.

6 468 Iranmanesh A. et al. Table The order components of simple groups 1) with t(g) Group Orcmp 1 Orcmp Orcmp Orcmp 4 Orcmp 5 Orcmp 6 A p, p and p 4 (p )(p p p are primes A 1 (q), 4 q +1 q +1 q (q / A 1 (q), 4 q 1 q 1 q (q +1)/ A 1 (q), q q q +1 q 1 A () 8 7 A (4) A 5 () B (q) q q q +1 q + q +1 q 1 q = n+1 > D p() p(p 1) ( p 1 ( p 1 +1)/ ( p +1)/4 p = n +1,n Π p i=1 (i D p+1 () p(p+1) ( p p +1 p+1 +1 p = n 1, n Π p 1 i=1 (i E 7 () F 4 (q) q 4 (q 6 (q 4 q 4 +1 q 4 q +1 q, q> F 4 (q) q 1 (q 4 (q +1) q q q + q q = n+1 > (q +1)(q +q q +1 +q + q +1 G (q), q q 6 (q q + q +1 q q +1 G (q), q = n+1 q (q q q +1 q + q E 7 () E 6 () M M M M J J J HS Sz ON Ly Co F F F 1 = M F = B F = Th ) p is an odd prime number.

7 A Characterization of PSL(,q) for q = m 469 Table The order components of E 8 (q) Group E 8 (q), q 0, 1, 4(mod5) Orcmp 1 q 10 (q 18 (q 14 (q 1 (q 10 (q 8 (q 4 + q +1) Orcmp q 8 + q 7 q 5 q 4 q + q +1 Orcmp q 8 q 7 + q 5 q 4 + q q +1 Orcmp 4 q 8 q 6 + q 4 q +1 Orcmp 5 q 8 q 4 +1 Group E 8 (q), q, (mod5) Orcmp 1 q 10 (q 0 (q 18 (q 14 (q 1 (q 10 (q 8 (q 4 +1) (q 4 + q +1) Orcmp q 8 + q 7 q 5 q 4 q + q +1 Orcmp q 8 q 7 + q 5 q 4 + q q +1 Orcmp 4 q 8 q 4 +1 (1) If n =andq = or 4, then by Table, the odd order components of K/H are,5,7 and 9. Thus (q + q +1)/ must be, 5, 7 and 9, which is impossible. () If (n, q )=(1,q )whereq 1 (mod 4), then the odd order components of K/H are q and (q +1)/ by Table. If (q + q +1)/ =q,then(q +1)/ =(q + q +4)/6, thus q +1 and it is a contradiction because (q +1)/ is an odd order component of K/H. If (q + q +1)/ = (q +1)/, then A 1 (q ) = (q + q + 1)(q +q 4)(q +q /7, since K/H G, thus(q +q / divides q (q (q +1)(q + q +1)/. Therefore, (q +q / divides (q. Since (q = q +q 1 + q 8q+4 then (q +q / divides (q 8q +4)/, i.e., (q + q +1)/ (q 8q +4)/, which is a contradiction. () If (n, q )=(1,q )whereq 1 (mod 4), then the odd order components of K/H are q and (q /, i.e., (q + q +1)/ =q or (q /. If (q + q +1)/ =q,then A 1 (q ) =(q + q +4)(q + q +1)(q + q )/54. We know that K/H G, andbya similar method to that of (), (q + q )/6 (q (q + 1). By Lemma.6(c), we must have q =16, and thus q =91, which is impossible because q must be a power of a prime number. If (q + q +1)/ =(q /, then q = q by Lemma.6(b). Thus (q =, which is a contradiction. (4) If (n, q )=(1,q )where q,thenq +1 and q 1 are odd order components of K/H by Table. Hence (q + q +1)/ =q +1 or q 1. If (q + q +1)/ =q +1 then q =(q + q )/. Since q A 1 (q ), (q + q )/6 (q (q + 1) and by using Lemma.6(c), q =16 and q =90. This is a contradiction because q mustbeapowerofaprime number. If (q + q +1)/ =q 1, then by Lemma.6(b), q = q and thus (q =1, which is a contradiction. (5) If (n, q ) = (p, q )whereq 1 p 1, then the odd order component of K/H is (q p /(q by Table 1. Therefore, (q + q +1)/ =(q p /(q. By Lemma.6(b), q p = q.thusq =q >q,andq p(p+1) >q, which contradicts Lemma.6(a). (6) If (n, q )=(p 1,q q ), then the odd order component of K/H is p 1 (p,q 1)(q 1) by Table 1. Therefore, (q + q +1)/ = q p 1 d(q 1) where d =(p, q, and thus q p = q by Lemma.6(b). Hence (q = d(q. Since p is a prime number, d =1ord = p. If d =1, then q >qand q p(p 1) >q and it contradicts Lemma.6(a). If d = p>, since q p = q,then q p(p 1) >q, which contradicts Lemma.6(a). Now let d = p =. Since q p = q,wehave

8 470 Iranmanesh A. et al. q = q. Therefore, if K/H is a simple group of type A n (q ), then K/H must be isomorphic to M, wherem =PSL(,q), q = m and m is even. By the same method, we can prove that K/H cannot be a simple group of type A n (q ). Step We prove that K/H cannot be a simple group of type B n (q )orc n (q ). Since B n (q )andc n (q ) have the same order components, it is sufficient to discuss one of them, for example C n (q ). If K/H = C n (q ), then by Table 1, (n, q )=( m,q )wherem 1or(n, q )=(p, l), here p being an odd prime number and l =or. (1) If (n, q )=( m,q )wherem 1, then, by Table 1, K/H has an odd order component (q n +1)/(,q. Therefore, (q + q +1)/ =(q n +1)/(,q by Lemma.8, so that q n q + q, but in this case q 16, hence q n >q,andq n >q n. Now if n, then q n >q, which contradicts Lemma.6(a). () If (n, q )=(p, l), where p is an odd prime number and l = or, then an odd order component of K/H is (l p 1)/(,l 1). By Lemma.8 we have (q +q+1)/ =(l p 1)/(,l 1), therefore, l =andl p = q by Lemma.6(b). Hence p =m and thus m =1, which is a contradiction since m 4. Step 4 We prove that K/H cannot be a simple group of type D n (q ). If K/H = D n (q ), then (n, q )=(p, l) wherel =,or5,or(n, q )=(p +1,k), here k = or by Table 1. (1) If (n, q )=(p, l) wherel =or5,or(n, q )=(p +1, ), then K/H has an odd order component ( p / or(5 p /4 by Table 1. Thus (q + q +1)/ =( p / or(5 p /4 by Lemma.8. Thus q or 5 q by Lemma.6(b), which is a contradiction. () If (n, q )=(n, ) where n = p or p +1,thenK/H has an odd order component p 1 by Table 1. Thus (q + q +1)/ = p 1 by Lemma.8. Therefore, p = q by Lemma.6(b), and thus m =1, which is a contradiction. Step 5 We prove that K/H cannot be a simple group of types: D n (q ), D 4 (q ), E 6 (q ), E 6 (q ), E 8 (q ), F 4 (q ), F 4 (q )andg (q ). The proof is similar for each type. Thus we do it for one type, for example, G (q ). If K/H = G (q ), then K/H has odd order component q + q +1orq q +1(orboth of them, when q ) by Tables 1 and. That is, (q + q +1)/ =q + q +1 or q q +1 by Lemma.8. If (q + q +1)/ =q q +1, then q >q,soq 6 >q and it contradicts Lemma.6(a). If (q + q +1)/ =q + q +1,then(q + q +1)/ =(q /(q, thus q = q by Lemma.6(b). Hence q 6 >q, and it contradicts Lemma.6(a). Step 6 We prove that K/H cannot be a simple group of types B (q )whereq = n+1 >, and G (q )whereq = n+1. Again the proof is similar for each type. Thus we choose one type, for example, B (q ). If K/H = B (q )whereq = n+1 >, then K/H have the odd order components q 1, q q +1andq + q + 1 by Table. Therefore, (q + q +1)/ =q 1, q q +1 or q + q + 1 by Lemma.8. If (q + q +1)/ =q 1, then q = q by Lemma.6(b), and thus q >q, which contradicts Lemma.6(a). If (q + q +1)/ =q q +1, then q > q (q 1) 9 >q because q 16, which is a contradiction. If (q + q +1)/ =q + q +1, then (q (q +)= n+1 ( n + 1). Since q 1thenq 1=k for some positive integer k. Thusk(k +1)= n+1 ( n +1),andk = n +1, k +1= n+1, which is a contradiction. Step 7 We prove that K/H cannot be a sporadic simple group, E 7 (), E 7 (), E 6 () or F 4 ().

9 A Characterization of PSL(,q) for q = m 471 If K/H is a sporadic simple group E 6 (), E 7 (), E 7 () or F 4 (), then by using Tables 1 and and Lemma.8, (q + q +1)/ =5, 7, 11, 1, 17, 19,, 9, 1, 7, 41, 4, 47, 59, 61, 71, 7, 17, or 109. But none of these equations have integer solutions. Step 8 We prove that G = M, M =PSL(,q), where q 1andq = m 16. By the above steps, we have K/H = M, i.e., K/H = M = G, thusg = K and H =1. Therefore G = M. Case If q 1 (mod ), then q 8andm isoddwhereq = m. The proof is similar to that of Case 1, and we omit the proof. Case If q =or4,sincepsl(, ) =PSL(, 7), by [1], it remains to prove the main theorem for q =4. Suppose that M =PSL(, 4). By Lemma.8 there exists a normal series 1 H K G where K/H is a non-ablian simple group. Some of the odd order components of K/H are 5,7 and 9. Therefore, t(k/h) 4. By Tables and, K/H can only be one of the following simple groups: A (4), B (q) whereq = n+1 >, E 6 (), E 8 (q) or one of the sporadic simple groups M, J 1, J 4, ON, Ly, F, F 1. Since 5,7 and 9 are the odd order components of K/H, it is easy to see that K/H is neither E 6 () nor one of the above sporadic simple groups. Suppose that K/H = B (q), where q = n+1 >. Since the odd order components of B (q) areq q +1, q 1and q + q +1, we have q q +1=5, q 1=7andq + q + 1 =9 by Lemma.8, which is a contradiction. Since the odd order components of E 8 (q) are congruent to 1 modulo 5, K/H cannot be E 8 (q). Therefore, K/H = M, wherem = A (4) =PSL(, 4). Since M = G then H =1andK = G. Therefore, G = M, wherem =PSL(, 4). The proof of the main theorem is now completed. Remark.1 It is a well-known conjecture of J. G. Thompson that if G is a finite group with Z(G) =1andM is a non-abelian simple group satisfying N(G) =N(M), where N(G)={n G has a conjugacy class of size n}, theng = M. We can give a positive answer to this conjecture by this characterization for the groups under discussion. Corollary. Let G be a finite group with Z(G) =1, M =PSL(,q), whereq = m,m 1 and N(G) =N(M). ThenG = M. Proof By [1, Lemma 1.5], if G and M are two finite groups satisfying the conditions of Corollary., then OC(G) = OC(M). So the main theorem implies this corollary. Shi and Bi in [16] put forward the following conjecture: Conjecture Let G be a group, M a finite simple group. Then G = M if and only if (i) G = M, and, (ii) π e (G) =π e (M), where π e (G) denotes the set of orders of elements in G. This conjecture is valid for groups of alternating type [1], sporadic simple groups [14], and some simple groups of Lie types [15,16]. As a consequence of the main theorem, we prove the validity of this conjecture for the groups under discussion. Corollary. Let G be a finite group and M =PSL(,q), where q = m,m 1. If G = M and π e (G) =π e (M). ThenG = M. Proof By assumption we must have OC(G) = OC(M), then the corollary follows from the main theorem.

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