CHARACTERIZATION OF PSL(3,Q) BY NSE

CHARACTERIZATION OF PSL(,Q) BY NSE S. ASGARY an N. AHANJIDEH Communicate by Alexanru Buium Let G be a group an π e(g) be the set of element orers of G. Suppose that k π e(g) an m k is the number of elements of orer k in G. Set nse(g) := {m k : k π e(g)}. Let M = P SL(, q), where q is a prime power an r = (q + q + 1)/(, q 1) is a prime number an G be a finite group such that r G, r G an G = M. In this paper, we prove that G = M if an only if nse(g) = nse(m). AMS 010 Subject Classification: 0D05, 0D0. Key wors: set of the number of elements of the same orer, prime graph, the classification theorem of finite simple groups. 1. INTRODUCTION If n is an integer, then we enote by π(n) the set of all prime ivisors of n. If G is a finite group, then π( G ) is enote by π(g). We enote by π e (G) the set of element orers of G. Set m k = m k (G) := {g G : the orer of g is k} an nse(g) := {m k (G) : k π e (G)}. Throughout this paper, we enote by φ the Euler totient function. If G is a finite group an r is a prime, then we enote by S r (G) a Sylow r-subgroup of G, by Syl r (G) the set of Sylow r-subgroups of G an n r (G) is the number of Sylow r-subgroups of G. cl G (x) enotes the size of the conjugacy class of G containing x. Let n be a positive integer an p be a prime number. Then n p enotes the p-part of n. The prime graph GK(G) of a finite group G is a graph whose vertex set is π(g) an two istinct primes p an q are joine by an ege if an only if G contains an element of orer pq (we write p q). Let t(g) be the number of connecte components of GK(G) an let π 1, π,..., π t(g) be the connecte components of GK(G). If π(g), then we always suppose that π 1 (G). G can be expresse as a prouct of co-prime positive integers OC i, i = 1,,..., t(g), where π(oc i ) = π i. These OC i s are calle the orer components of G an the set of orer components of G will be enote MATH. REPORTS 19(69), 4 (017), 45 48

46 S. Asgary an N. Ahanjieh by OC(G). Also if π(g), we call OC,..., OC t(g) the o orer components of G. The sets of orer components of finite simple groups with isconnecte prime graph can be obtaine using [10] an [17]. Let M t (G) := {g G : g t = 1}. Then G an H are of the same orer type if an only if M t (G) = M t (H), t = 1,,.... In 1987, J.G. Thompson put forwar the following problem: Thompsons Problem. Let T (G) = {(k, m k ) : k π e (G), m k nse(g)}, where m k is the number of elements of G of orer k. Suppose that H is a group with T (G) = T (H). If G is solvable, then is it true that H is also necessarily solvable? It is easy to see that if T (G) = T (H), then nse(g) = nse(h) an G = H. We say that the group G is characterizable by nse (an the orer) if every group H with nse(g) = nse(h) (an G = H ) is isomorphic to G. Note that not all groups can be characterizable by nse. For instance, let G = Z 4 Z 4 an H = Z Q 8. Then G = H an nse(g) = nse(h), while G H. In [1], it is shown that the simple groups P SL(, q), where q 1, are characterizable by nse. In [9] an [15], it is prove that P SL(, p), where p is a prime, is characterizable by nse. Also, in [8], [] an [], the authors respectively showe that for the prime number p, P GL(, p), the alternating groups A n, where n {p, p + 1, p + } an, C n (), D n () an D n+1 (), where n + 1 = p, are characterizable by nse uner some extra conitions. Throughout this paper, let q be a prime power such that q +q+1 (,q 1) is a prime, namely r an M = P SL(, q). In this paper, we are going to stuy the characterization of M by nse. In fact, we prove the following theorem: Main Theorem. Let G be a finite group such that r G, r G an G = M. Then G = M if an only if nse(g) = nse(m).. PRELIMINARIES Definition.1 ([5]). Let a be a natural number an r be a prime such that (a, r) = 1. If n is the smallest natural number such that r (a n 1), then r is name a Zsigmony prime of a n 1. Lemma.1 ([5]). Let a an n be natural numbers, then there exists a Zsigmony prime of a n 1, unless (a, n) = (, 1), (a, n) = (, 6) or n = an a = s 1 for some natural number s. Remark.1. If l is a Zsigmony prime of a n 1, then Fermat s little theorem shows that n l 1. Put Z n (a) = {l : l is a Zsigmony prime of a n 1}. If r Z n (a) an r a m 1, then we can see at once that n m.

Characterization of PSL(,Q) by nse 47 Lemma. ([4]). Let G be a Frobenius group of even orer with kernel K an complement H. Then t(g) =, the prime graph components of G are π(h) an π(k) an the following assertions hol: (1) K is nilpotent; () K 1 (mo H ). Lemma.. If x is an element of M {1}, then either cl M (x) r = M r or cl M (x) r < M r an cl M (x) = M (q 1) GL(1, q ) Proof. It follows from [1, Corollary.8]. = GL(, q) GL(1, q ). Lemma.4 ([6]). Let t be a positive integer iviing G. Then t M t (G). From Lemma.4, it may be conclue that: Corollary.1. For a finite group G: (i) if n G, then n s n m s ; (ii) if n π e (G), then m n = φ(n)k, where k is the number of cyclic subgroups of orer n in G. In particular, φ(n) m n. (iii) if R Syl r (G) is cyclic of prime orer r, then m r = n r (G)(r 1); (iv) if P Syl p (G) is cyclic of prime orer p an r π(g) {p}, then m rp = n p (G)(p 1)(r 1)k, where k is the number of cyclic subgroups of orer r in C G (P ). Lemma.5 ([11]). If n 6 is a natural number, then there are at least s(n) prime numbers p i such that (n + 1)/ < p i < n. Here s(n) = 1, for 6 n 1; s(n) =, for 14 n 17; s(n) =, for 18 n 7; s(n) = 4, for 8 n 41; s(n) = 5, for 4 n 47; s(n) = 6, for n 48.. MAIN RESULTS In this section, let G be a finite group such that r G, r G an nse(g) = nse(m). In the following, we are going to bring some useful lemmas which will be use uring the proof of the main theorem:

48 S. Asgary an N. Ahanjieh 4 Lemma.1. m r (M) = 1 (r 1)q (q 1)(q 1). Proof. By [7], N M (S r (M)) =. n r (M) = q 1 (q 1)(,q 1) an hence, M N M (S r (M)) = 1 q (q 1)(q 1). Now, since S r (M) is cyclic, Corollary.1(iii) implies that m r (M) = n r (M)(r 1) = 1 (r 1)q (q 1)(q 1), as esire. Lemma.. (i) For u π e (M), either r m u (M) or u = r; (ii) For every u π e (G), r m u (G) if an only if m u (G) = m r (M); (iii) m r (G) = m r (M); (iv) m (G) = m (M). Proof. (i) Obviously, m u (M) = cl M (x k ), where x k s are selecte O(x k )=u from istinct conjugacy classes of M. Thus Lemma. completes the proof of (i). (ii) Since m u (G) nse(g) = nse(m), (i) completes the proof. (iii) By Corollary.1(i), we have r (1 + m r (G)) an hence, r m r (G). Thus m r (G) = m r (M), by (ii). (iv) By Corollary.1(ii), for every u π e (M), φ(u) m u (M). Thus if u >, then m u is even. On the other han, (1 + m (M)) an hence, m (M) is o. Applying the same reasoning shows that the only o number in nse(g) is m (G) an hence, m (G) = m (M), as wante. Lemma.. For every s π(g) {r}, sr π e (G). Proof. Suppose on the contrary, sr π e (G). Since r G, we euce that S r (G) is cyclic an hence, Corollary.1(iv) forces m rs (G) = (r 1)(s 1)n r (G)k, for some natural number k. Thus m rs (G) = m r (G)(s 1)k an hence, one of the following hols: (i) r m rs (G). Then r (s 1)k an hence, m rs (G) > M, along with Lemmas.1 an.(iii). Thus m rs (G) nse(m), which is a contraiction. (ii) r m rs (G). Then Lemma.(ii) shows that m rs (G) = m r (G) an hence, by Corollary.1(iv), s =. On the other han, Corollary.1 (i) an Lemma.(iv) show that r (1 + m (G) + m r (G) + m r (G)) = 1 + m (M) + m r (G). Now, since by Corollary.1 (i) an Lemma.(i), r (1 + m r (G)) an r m (M), we euce that r m r (G), which is a contraiction. Corollary.1. r is an o orer component of G.

5 Characterization of PSL(,Q) by nse 49 Proof. It follows from Lemma.. Lemma.4. (i) n r (G) = n r (M) = 1 q (q 1)(q 1); (ii) 1 M G an G 1 r(r 1)q (q 1)(q 1). Proof. Since S r (G) is cyclic, Corollary.1(iii) forces m r (G) = φ(r)n r (G). Thus by Lemma.(iii), φ(r)n r (M) = m r (M) = m r (G) = φ(r)n r (G) an hence (i) follows. Now let s π(g) {r}. By Lemma., S s (G) acts fixe point freely on the set of elements of orer r in G an hence, G s m r (G) = φ(r)n r (G). Also, G r = r an n r (G) = 1 q (q 1)(q 1). Thus M / G an G 1 r(r 1)q (q 1)(q 1). Proof of the main theorem. If G = M, then it is obvious that nse(g) = nse(m). Now we assume that nse(g) = nse(m). In the following, we show that G has a normal series 1 H K G such that K/H is a non-abelian simple group. Let x G be an element of orer r. Since r is the maximal prime ivisor of G an an o orer component of G, C G (x) =< x >. Set H = O r (G), the largest normal r -subgroup of G. Since < x > acts on H fixe point freely, H is a nilpotent group. Suppose that K be a normal subgroup of G such that K/H is a minimal normal subgroup of G/H. Then K/H is a irect prouct of copies of same simple group. Since r K/H an r K/H, K/H is a simple group. On the other han, since < x > is a Sylow r-subgroup of K, G = N G (< x >)K by the Frattini argument an so G/K ivies r 1. Now, it s not too har to prove that K/H r. Therefore, G has a normal series 1 H K G such that K/H is a non-abelian simple group. In the following, assume that = (, q 1) an q = p α, where p is a prime an α is a positive integer. Also, for convenience let f(q) = 1 r(r 1)q (q 1)(q 1). We are going to continue the prove of the main theorem in the following steps: Step 1. K/H is not a Sporaic simple group. Proof. Suppose that K/H is a Sporaic simple group. Thus r = q +q+1 {5, 7, 11, 1, 17, 19,, 9, 1, 7, 41, 4, 47, 59, 67, 71}. If q +q+1 {5, 11}, then since q is a prime power, we get a contraiction. Assume that q +q+1 = 7 an = 1. Thus q =. But P SL(, ) =..7 an K/H {M, J 1, J, HS}, so 5 K/H, which is a contraiction. If =, then q = 4 an P SL(, 4) = 6..5.7. Also, K/H {M, J 1, J, HS}. Now, similar to the above we get a contraiction. The same argument rules out the other possibilities of r. Step. K/H cannot be an alternating group A m, where m 5.

40 S. Asgary an N. Ahanjieh 6 Proof. If K/H = A m, then since q +q+1 = r π(k/h), r m. Also, since q is a prime power, r 7. Thus by Lemma.5, there exists a prime number u π(a m ) π(g) such that (r + 1)/ < u < r. Lemma.4(ii) forces u 1 r(r 1)q (q 1)(q 1). We can check at once that u q, u q 1 an u r 1. Thus u Z (q). It follows that u = r, where r = 7 an q = 4. So M = P SL(, 4) = 6..5.7. Since A m ivies G, we get m {7, 8}. Note that H ivies G / K/H an S 7 (G) acts fixe point freely on H an hence, 7 H 1. Therefore, consiering the orers of G an K/H shows that either H = 8 an m = 7 or H = 1. If H = 8 an m = 7, then we can assume that H is a -elementary abelian group an hence, A 7 GL(, ). Thus A 7 ivies GL(, ), which is a contraiction. If H = 1, then G = A 7, S 7, A 8 or S 8, which in two former cases m 7 (G) = m 7 (A 7 ) = 70 5760 = m 7 (M), which is a contraiction. If G = A 8, then 144 nse(g) nse(m) an if G = S 8, then 76 nse(g) nse(m), contraicting our assumptions. Step. K/H = P SL(, q). Proof. By Steps 1 an, an the classification theorem of finite simple groups, K/H is a simple group of Lie type such that t(k/h) an r OC(K/H). Thus K/H is isomorphic to one of the following groups: Case 1. Let t(k/h) =. Then OC (K/H) = r = q +q+1. Thus we have: 1.1. If K/H = C n (q ), where n = u, then q n +1 (,q 1) = q +q+1. If (, q 1) =, then q n +1 = r, so q n = (q +q+1) an hence, (q, r) = (q, q) = 1. Also, since p is o, we have p q 1 or p q 1 an hence, (p α ) n = K/H p G p 1 (,p ) (q 1) p (q 1) p ( q +q+1 ) p < ( (q +q+1) ) < (p α ) n. Therefore, n <, which is a contraiction. If (, q 1) = 1 an = 1, then we have q n = q(q + 1), which is impossible. If (, q 1) = 1 an =, then q = α an q + 1 = r. Thus α + 1 = q +q+1 an hence, α = (q 1)(q+). Since q 1, q + an hence, α, which is a contraiction. The same reasoning completes the proof in the case when either K/H = B n (q ) or K/H = D n (q ), where n = u 4. 1.. If K/H = C s () or B s (), where s is prime, then s 1 = q +q+1. So s = (q + q + + ) an hence, either (, q) = 1 an s+1 > q + q + + or q = s =. In the former case, since s = K/H G = (q 1) (q 1) < ( q +q+1 ) < (s+1), s < (s + 1) an hence, s =. So q(q + 1) {1, 8}. Obviously, q(q + 1) 8. If q(q + 1) = 1, then q =, which is a contraiction. If q = s =, then r = 1 an M = P SL(, ) = 4..1, so 5 G. On the other han, 5 C () = B (), which is a contraiction.

7 Characterization of PSL(,Q) by nse 41 1.. If K/H = C s (), where s is prime, then s 1 = q +q+1 an hence, s = q +q+1+. Now, if q {, 4}, then q +q+1 = 7 an hence, s =. In these cases, P SL(, ) =..7 an P SL(, 4) = 6..5.7. So 9 G. On the other han, 9 C (), which is a contraiction. If q {, 4}, then (, q) = 1 an hence, s = K/H G ( q 1 s. So s < s, which implies that s =. This forces q +q+1 ) (q 1) ( q +q+1 ) < ( q +q+1+ ) = =, which is impossible. 1.4. If K/H = D s (q ), where s 5 is prime an q =,, 5, then q s 1 Thus q s() (q i 1) q 1 = r. 1 (r 1)q (q 1)(q 1). On the other han, r 5 = (q s 1) 5 < q 5s an q s().q s() (q 1) 5 < q s() (q i 1) 1 (r 1)q (q 1)(q 1) < r 5 s() s()+, which implies that q < q 5s an hence, s < 5, which is a contraiction. 1.5. If K/H = D n (), where 9 n = m + 1 an n is not prime, then n 1 +1 = q +q+1. Thus (, q) = 1 an n 1 = (q + q + ) an hence, n > q + q +. Since n(n 1) = K/H G = (q 1) (q 1) < (q + q + ) < n, we obtain n 1 <, which is impossible. 1.6. If K/H = D n (), where n = m + 1 5, then n 1 + 1 = q +q+1. Thus n 1 = q +q+1 an hence, (, q) = 1. Therefore, n(n 1) = K/H G < ( q +q+1 ) = (n 1), so n <, which is impossible. 1.7. If K/H = D s+1 (q ), where s is an o prime an q =,, then q s 1 (,q 1) = r. Thus q s(s+1) (q s + 1)(q s+1 1) (q i 1) 1 (r 1)q (q 1)(q 1). Also, r 5 = (q s 1) 5 < q 5s an q s(s+1) q s(s+1)/ < q s(s+1) (q s + 1)(q s+1 1) (q i (,q 1) 5 1) 1 (r 1)q (q 1)(q 1) < r 5, which implies that q s(s+1)/ < q 5s an hence, s <, which is a contraiction. 1.8. If K/H = D s (), where 5 < s m + 1 an s is an o prime, then s +1 4 = r an s() ( i 1) 1 (r 1)q (q 1)(q 1). Thus r 5 = (s +1) 5 104 < 5s an s() < s() ( i 1) 1 (r 1)q (q 1)(q 1) < r 5, which implies that s(s 1) < 5s an hence, s 1 < 5, which is a contraiction.

4 S. Asgary an N. Ahanjieh 8 1.9. If K/H = G (q ), where < q ɛ (mo ) an ɛ = ±1, then q ɛq + 1 = q +q+1. Thus we can check at once that K/H > f(q), which is a contraiction. If K/H = F 4 (q ), where q is o, then similar to the above, we get a contraiction. 1.10. If K/H = F 4 (), then K/H = 11 5 1 an q +q+1 = 1. If = 1, then q = an P SL(, ) = 4..1, so 5 G, which is a contraiction. If =, then q(q + 1) = 8, which is impossible. 1.11. If K/H = P SU(4, ), then K/H = 6 4 5. Thus q +q+1 = 5, which is impossible. 1.1. If K/H = P SL(s, q ), where (s, q ) (, ), (, 4) an s is an o prime, then r = other han, r 5 = q s s() 1 (s,q 1)(q 1) an q (q s 1) 5 (s,q 1) 5 (q 1) 5 (q i 1) 1 (r 1)q (q 1)(q 1). On the < q 5s an q s(s 1) s < q s() (q i 1) 1 (r 1)q (q 1)(q 1) < r 5, which implies that s(s 1) s < 5s. Hence s =, 5. If s = 5, then q +q+1 = q 4 +q +q +q +1 (5,q 1) an hence, (q, q) = 1. Also, q 10 (q 1)(q 1)(q 1)(q 4 1 1) (r 1)q (q 1)(q 1). But q 10 1 (r 1)q (q 1)(q 1), which is a contraiction. If s =, then q 1 (,q 1)(q 1) = q +q+1 an hence, q +q +1 (,q 1) = q +q+1. Now, we ivie the proof into the following subcases: (i) Suppose that (q, q) 1 an q q. (1) Let (, q 1) = 1 an = 1. Thus q + q + 1 = q + q + 1, so q = q an K/H = P SL(, q). () If (, q 1) = an =, then q +q +1 = q +q+1 an similar to the above, K/H = P SL(, q). () If (, q 1) = 1 an =, then q + q + 1 = q +q+1 an hence, q(q + 1) = q + q +, which implies that q q + q +. Since q q, q, thus q = an hence, q (q + 1) = 4, which is a contraiction. (4) If (, q 1) = an = 1, then q +q +1 = q + q + 1 an hence, q(q + 1) = (q 1)(q +), which implies that q q + q. Thus q ( ) an hence, q =. It follows that 18 = q +q. Hence, q = 4. But (s, q ) (, 4) by assumption, which is a contraiction. The same argument completes the proof when q q. (ii) Assume that (q, q) = 1. (1) If (, q 1) = 1 an = 1, then q + q + 1 = q + q + 1 an hence, q (q + 1) = q(q + 1). Since (q, q) = 1, we get a contraiction.

9 Characterization of PSL(,Q) by nse 4 () If (, q 1) = an =, then q +q +1 = q +q+1 an similar to the above, we get a contraiction. () If (, q 1) = an = 1, then q +q +1 = q +q +1 an hence, q +q = q(q + 1). Thus (q 1)(q +) = q(q + 1), which implies that k(k + 1) = q(q + 1). Consiering the ifferent possibilities of q leas us to get a contraiction. (4) If (, q 1) = 1 an =, then similar to the above, we get a contraiction. 1.1. If K/H = P SL(s + 1, q ), where (q 1) (s + 1) an s is an o prime, then r = q s 1 q 1 s(s+1) an q (q s+1 1) (q i 1) 1 (r 1)q (q 1)(q 1). On the other han, r 5 = (q s 1) 5 < q 5s an q s(s+1) s < q s(s+1) (q 1) 5 (q s+1 1) (q i 1) 1 (r 1)q (q 1)(q 1) < r 5 < q 5s, which implies that s + 1 < 6. Hence s =, so q + q + 1 = q +q+1. Since (q 1) (s + 1), q {,, 5}, which implies that K/H = P SL(4, ), K/H = P SL(4, ), K/H = P SL(4, 5). If K/H = P SL(4, ), then since P SL(4, ) = A 8, Step leas us to get a contraiction. If K/H = P SL(4, ), then q +q+1 = 1 an hence q =. So K/H G, which is impossible. The same reasoning rules out the case when K/H = P SL(4, 5). 1.14. If K/H = E 6 (q ), then r = q 6 +q +1 (,q 1) an q 6 (q 1 1)(q 8 1)(q 6 1)(q 5 1)(q 1) 1 (r 1)q (q 1)(q 1). On the other han, r 5 = (q 6 +q +1) 5 < q 45 (,q 1) 5 an q 6 (q 1 1)(q 8 1)(q 6 1)(q 5 1)(q 1) 1 (r 1)q (q 1)(q 1) < r 5 < q 45, which is a contraiction. The same reasoning rules out the case when K/H = E 6 (q ), where q >. 1.15. If K/H = D 4 (q ), then r = q 4 q + 1 an q 1 (q 4 + q + 1)(q 6 1)(q 1) 1 (r 1)q (q 1)(q 1). But q 1 (q 4 + q + 1)(q 6 1)(q 1) > 1 (r 1)q (q 1)(q 1), which is a contraiction. 1.16. If K/H = P SU(s+1, q ), where (s, q ) (, ), (5, ), (q +1) (s+1) an s is an o prime, then r = q s s(s+1) +1 q +1 an q (q s+1 1) (q i ( 1) i ) 1 (r 1)q (q 1)(q 1). Moreover, r 5 = ( q s +1 q +1 )5 = (q q s +...+1) 5 < q 5() an q s(s+1).q s() +s < q s(s+1) (q s+1 1) (q i ( 1) i ) 1 (r 1)q (q 1)(q 1) < r 5, which implies that s + s < 5(s 1). Thus s <, which is a contraiction. 1.17. If K/H = P SU(s, q ), where s is an o prime, then r = q s +1 (q +1)(s,q +1) an

44 S. Asgary an N. Ahanjieh 10 q s() (q i ( 1) i ) 1 (r 1)q (q 1)(q 1). Also, r 5 = q 5() an q s() q s() < q s() (q s +1) 5 (q +1) 5 (s,q +1) 5 < (q i ( 1) i ) 1 (r 1)q (q 1)(q 1) < r 5. This implies that s < 5. Thus s = an hence, r = q q +1 (,q +1). This shows that q = q 1, q(q + 1) = k(k + 1) or q (q 1) = k(k + 1). If q = q + 1, then we can see that K/H G, which is a contraiction. Consiering the ifferent possibilities of k in two latter cases shows that (q, q ) = (4, ), so G 5 = 5. Since 5 Aut(P SU(, )), so H 5 = 5. But S 7 (G) acts fixe point freely on S 5 (H) an hence, 7 5 1, which is a contraiction. Case. Let t(k/h) =. Then r {OC (K/H), OC (K/H)}:.1. If K/H = P SL(, q ), where 4 q, then the o orer components of K/H are q +1 an q 1. If q +1 = r, then q = r 1 = (q +q+1) 1 an hence, either q = q(q + 1) or q = (q 1)(q+), which are impossible. So let q 1 = r. Thus q = α > 4, q = r + 1 = (q +q+1) + 1 an n r (K/H) = q (q +1) n r (G) = 1 q (q 1)(q 1). If = 1, then q(q +1) = ( α 1 1) an hence, either q = or (q (q + 1), K/H ) 6. If q =, then q = 8, so n r (K/H) n r (G), which is a contraiction. In the latter case, q (q + 1) (q 1), which is impossible. Now let =. Then we can see at once that (q + )(q 1) = 6( α 1 1) an either (q, q ) = 1 or q = 4 an q = 8. Thus if (q, q ) = 1, then the above statements show that r + 1 = q 8 q + 1, which is impossible. If q = 4 an q = 8, then H 5 = 5 an hence, 7 5 1, which is a contraiction... If K/H = P SL(, q ), where 4 q 1, then q = r or (q + 1)/ = r. If q = r, then n r (K/H) = (q + 1) n r (G) = 1 q (q 1)(q 1) an either q + 1 = q + q + or q + 1 = (q +q+4). So we can see at once that q + 1, which is a contraiction. If (q +1) = r, then n r (K/H) = 1 q (q 1) n r (G) = 1 q (q 1)(q 1) an q = q + q + 1 or q = (q +q 1). This forces q = 5 an hence, r =, which is impossible. The same reasoning completes the proof when K/H = P SL(, q ) an 4 q + 1... If K/H = P SU(6, ) or K/H = P SL(, ), then K/H = 15 6 5 7 11 or K/H = 7. It follows that q +q+1 {7, 11}. We can check that q +q+1 11 an hence q +q+1 = 7. So q {, 4}. Thus P SU(6, ) G an hence, K/H = P SU(6, ). Also, if K/H = P SL(, ) an q = 4, then as mentione in the previous cases 7 5 1, which is a contraiction. Therefore, K/H = P SL(, ) = M, as esire..4. If K/H = D s (), where s = t + 1 5, then s +1 4 = q +q+1 or +1 =

11 Characterization of PSL(,Q) by nse 45 q +q+1. If s +1 4 = r, then s() ( i 1) On the other han, r 5 = (s +1) 5 104 < 5s an s() s < s() 1 (r 1)q (q 1)(q 1). ( i 1) 1 (r 1)q (q 1)(q 1) < r 5. This implies that s() < 6s an hence, s < 4, which is a contraiction. The same reasoning rules out the other possibility..5. If K/H = D s+1 (), where s = n 1 an n, then s + 1 = q +q+1 or s+1 + 1 = q +q+1. If s + 1 = r, then s = q(q + 1) or s = (q 1)(q+), which is impossible. The same reasoning rules out the other possibilities..6. If K/H = G (q ), where q 0 (mo ), then q ± q + 1 = q +q+1. We know that K/H G an G f(q), so K/H f(q). Since K/H = q 6 (q 1)(q 6 1) an either q (q ±1) = q(q +1) or q (q ±1) = 1 (q 1)(q +), we can check at once that q 6 G p, which is a contraiction..7. If K/H = G (q ), where q = t+1 >, then q q + 1 = q +q+1 or q + q + 1 = q +q+1. Let (, q) = 1. If q q + 1 = q +q+1, then q > q +q+1. Thus ( t+1 ) = K/H G < ( q +q+1 ) < ( t+1 ), which is a contraiction. Now let q + q + 1 = q +q+1. If = 1, then t+1 ( t + 1) = q(q + 1). Now, since (, q) = 1, q an hence, q ( t + 1) an (q + 1) = t+1, which is impossible. If =, then q + q + 1 = q +q+1 an hence, t+ ( t + 1) = (q 1)(q + ). Thus either t+1 (q 1) an (q + ) ( t + 1) or t+1 (q + ) an (q 1) ( t + 1). This forces (q 1) = t+1 an (q + ) = ( t + 1). This guarantees that G t+. On the other han, (t+1) = K/H G t+, which is a contraiction. Now assume that (, q) 1. So = 1 an hence, q ± q + 1 = q + q + 1. This forces q = t+1 an q + 1 = t ± 1, which is impossible..8. If K/H = F 4 (q ), where q is even, then q 4 + 1 = q +q+1 or q 4 q + 1 = q +q+1. Thus r 6 = (q 4 + 1) 6 < (q 5 ) 6 = q 0 an q 4 (q 1 1)(q 8 1)(q 6 1)(q 1) 1 r(r 1)q (q 1)(q 1). Thus q 6 1 r(r 1)q (q 1)(q 1) < r 6 < q 0, which is a contraiction..9. If K/H = E 7 (), then r {7, 17}. Therefore, either r = 7 an q = 8 or r = 17 an q = 19. So either M = P SL(, 8) = 9 7 7 or M = P SL(, 19) = 4 4 5 19 17. On the other han, 1 E 7 (), so K/H G, which is a contraiction..10. If K/H = E 7 (), then r {757, 109}. One can check at once that (q +q+1) 109. If q +q+1 = 757, then = 1 an q = 7. On the other han, P SL(, 7) = 4 9 7 1 757 an 5 E 7 (), which is a contraiction.

46 S. Asgary an N. Ahanjieh 1.11. If K/H = F 4 (q ), where q = t+1, then r = q ± q + q ± q + 1. In both cases, one can check at once that K/H > G, which is a contraiction. Case. Let t(k/h) {4, 5}. Then as follows: r {OC (K/H), OC (K/H), OC 4 (K/H), OC 5 (K/H)},.1. If K/H = E 6 (), then q +q+1 {1, 17, 19}. Obviously, q +q+1 17. If q +q+1 = 19, then = an q = 7. Thus M = P SL(, 7) = 5 7 19. On the other han 11 E 6 (), which is a contraiction. The same reasoning rules out the case when r = 1 an q =... If K/H = P SL(, 4), then q +q+1 {5, 7, 9}. It is easy to check that q +q+1 {5, 9}. If q +q+1 = 7, then either q = or q = 4. In the former case, K/H G, which is a contraiction. So q = 4 an hence, K/H = P SL(, 4) = M, as esire... If K/H = B (q ), where q = t+1 an t 1, then r {q 1, q ± q + 1}. Let q 1 = r an = 1. Thus ( t 1) = q(q + 1). If q =, then q + 1 = an hence, t = 1 an M = P SL(, ). Therefore, 5 G an 5 K/H, which is a contraiction. This forces q t 1 or q t + 1 an hence, q(q + 1) ( t + 1)( t 1 + 1). Therefore, t = an q = 5. Thus K/H G, which is a contraiction. If q 1 = r an =, then we can see that (. t 1 1) = q(q + 1). If q =, then q + 1 = 5 an t = 1, which is impossible as escribe above. Thus q + 1 = an hence, q 1 =. Also, r 1 =. So (t+1) K/H G 5, which is a contraiction. Now assume that q + q + 1 = r. If =, then q +q = t+1 ( t + 1) an hence, (q 1)(q + ) =. t+1 ( t + 1). Since q 1, q 1 = k for some positive integer k. Thus k(k + 1) = t+1 ( t + 1) an hence, k(k + 1) = t+1 ( t +1 ). Now, if t+1 k, then k+1 t +1 an if t+1 k+1, then k t +1, which are impossible. If = 1, then q +q +1 = q + q +1 an hence, q(q +1) = t+1 ( t +1), which is impossible. The same reasoning rules out the case when q q + 1 = r..4. If K/H = E 8 (q ), then r { q 10 +q 5 +1 q q +1 = q 8 q 7 + q 5 q 4 + q q + 1, q 10 q 5 +1 = q 8 + q 7 q 5 q 4 q + q + 1, q 10 +1 = q 8 q 6 + q q +1 q +1 q 4 q + 1, q 8 q 4 + 1}. Thus r < q 9. On the other han, r 5 < q 45 an G 1 (r 1)q (q 1)(q 1) < r 5. Since q 10 K/H an K/H G, we get a contraiction. The above cases show that K/H = P SL(, q).

1 Characterization of PSL(,Q) by nse 47 Step 4. G = M. Proof. By the previous argument an step, G has a normal series 1 H K G such that K/H = M = P SL(, q). We claim that H = 1. Suppose on the contrary, H 1, so H. Let t π(h). By Frattini s argument N G (S t (H))H = G, so N G (S t (H))/N H (S t (H)) = G/H. Since r G/H, r N G (S t (H)) an hence, N G (S t (H)) contains an element x of orer r. Also, by Lemma., rt π e (G), thus < x > acts fixe point freely on S t (H) {1}, so r S t (H) 1. On the other han, S t (H) H an since G = G/K K/H H an K/H = P SL(, q), H r 1. This forces r r 1, which is impossible. Therefore H = 1 an hence, K = M. We know that G AutM. If q = p m, then since by Lemma., GK(G) is isconnecte, by [14], G/K =< ϕ > < θ >, where ϕ is the fiel automorphism of orer u an θ is the graph automorphism of orer. Since G = M, the orer of ( ϕ is 1 an ) hence, G = K or G = K < θ >. If G = K < θ >, then m K < θ > > m (K), but nse(g) contains exactly one o number m (K), which is a contraiction. Therefore G = K = M. REFERENCES [1] N. Ahanjieh, Thompson s conjecture for some finite simple groups. J. Algebra 44 (011), 05 8. [] N. Ahanjieh an B. Asaian, NSE characterization of some alternating groups. J. Algebra Appl 14() (015). [] S. Asgary an N. Ahanjieh, NSE characterization of some finite simple groups. To appear in Sci. Ann. Comput. Sci., 015. [4] G.Y. Chen, On Frobenius an -Frobenius group. J. South China Normal Univ. Natur. Sci. E. 0(5) (1995), 485 487. [5] W. Feit, On large Zsigmony primes. Proc. Amer. Math. Soc. 10(1) (1988), 9 6. [6] G. Frobenius, Verallgemeinerung es Sylowschen Satze. Berliner Sitz (1985), 981 99. [7] B. Huppert, Enliche Gruppen. Springer-Verlag, 1967. [8] A. Khalili Asboei, A new characterization of P GL(, p). J. Algebra Appl. 1(7) (01). [9] A. Khalili Asboei an A. Iranmanesh, A characterization of the linear groups L (p). Czechoslovak Math. J. 64(19) (014), 459 464. [10] A.S. Konratev, Prime graph components of finite simple groups. Math. USSR-Sb. 67(1) (1990), 5 47. [11] A.S. Konratev an V.D. Mazurov, Recognition of Alternating groups of prime egree from their element orers. Sib. Math. J. 41() (000), 94 0. [1] M. Khatami, B. Khosravi an Z. Akhlaghi, A new characterization for some linear groups. Monatsh. Math. 16 (011), 9 50. [1] P. Kleiman an M. Liebeck, The subgroup structure of the finite classical groups. Lonon Math. Soc. Lecture Note Ser., 1990. [14] M.S. Lucio, Prime graph components of finite almost simple groups, Ren. Semin. Mat. Univ. Paova 10 (1999), 1.

48 S. Asgary an N. Ahanjieh 14 [15] C. Shao an Q. Jiang, A new characterization of P SL(, p) by nse. J. Algebra Appl. 1(4) (014). [16] R. Shen, C.G. Shao, Q.H. Jiang, W.J. Shi an V. Mazurov, A new characterization of A 5. Monatsh. Math. 160 (010), 7 41. [17] J.S. Williams, Prime graph components of finite groups. J. Algebra 69 (1981), 487 51. Receive 19 January 016 Shahrekor University Department of Mathematics Shahrekor, Iran soleyman.asgary@stu.sku.ac.ir Shahrekor University Department of Mathematics Shahrekor, Iran ahanjieh.nea@sci.sku.ac.ir