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

Acta Mathematica Sinica, English Series July, 00, Vol.18, No., pp. 46 47 A Characterization of PSL(,q) for q = m A. IRANMANESH S. H. ALAVI B. KHOSRAVI Department of Mathematics, Tarbiat Modarres University, P. O. Box: 14115 17, Tehran, Iran Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran. E-mail: 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

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

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

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 = 1 5 1 17, 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.

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 () 6 4 5 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 () 11 5 1 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 1 6 5 11 J 7 5 7 Ru 14 5 7 1 9 He 10 5 7 17 Mcl 7 6 5 7 11 Co 1 1 9 5 4 7 11 1 Co 10 7 5 7 11 Fi 17 9 5 7 11 1 F 5 = HN 14 6 5 6 7 11 19 1) 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.

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) 6 5 7 9 A 5 () 15 6 5 7 11 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 () 6 11 5 7 7 17 11 1 17 19 1 4 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 +1 6 5 7 E 7 () 11 1 19 7 41 757 109 61 7 547 E 6 () 6 9 5 7 11 1 17 19 M 11 4 5 11 M 7 5 7 11 M 7 5 7 11 M 4 10 5 7 11 J 1 5 7 11 19 J 7 5 5 17 19 J 4 1 5 7 11 9 1 7 4 HS 9 5 7 11 Sz 1 7 5 7 11 1 ON 9 4 5 7 11 19 1 Ly 8 7 5 6 7 11 1 7 67 Co 18 6 5 7 11 F 18 1 5 7 11 1 17 F 4 1 16 5 7 11 1 17 9 F 1 = M 46 0 5 9 7 6 11 1 41 59 71 17 19 9 1 47 F = B 41 1 5 6 7 11 1 1 47 17 19 F = Th 15 10 5 7 1 19 1 1) p is an odd prime number.

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

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 ().

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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