A New Characterization of A 11

Transcription

1 International Journal of Algebra, Vol. 8, 2014, no. 6, HIKARI Ltd, A New Characterization of A 11 Yong Yang, Shitian Liu and Yanhua Huang School of Science Sichuan University of Science and Engineering Zigong Sichuan, , P.R. China Copyright c 2014 Yong Yang, Shitian Liu and Yanhua Huang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Let G be a group and ω(g) be the set of element orders of G. Let k ω(g) and s k be the number of elements of order k in G. Let nse(g) ={s k k ω(g)}. The groups A5, A 6, A 7 and A 8 are unique determined by nse(g). But if π(a n ) > 4, then can the alternating A n be characterized by nse only? In this paper, we give an example which is a simple K 5 -group, can be characterized by nse. Namely, we prove that if G is a group, then G = A 11 if and only if nse(g)=nse(a 11 ). Mathematics Subject Classification: 20D05, 20D06, 20D20 Keywords: Element order, Alternating group, Thompson s problem, Number of elements of the same order 1 Introduction In 1987, J. G. Thompson posed a very interesting problem related to algebraic number fields as follows (see [20]). Thompson s Problem. Let T (G) ={(n, s n ) n ω(g) and sn nse(g)}, where s n is the number of elements with order n. Suppose that T (G) =T (H). If G is a finite solvable group, is it true that H is also necessarily solvable?

2 254 Yong Yang, Shitian Liu and Yanhua Huang It is easy to see that if G and H are of the same order type, then nse(g) = nse(h), G = H. Theorem 1.1 [18, 16, 10, 3] Let G be a group and H be a simple K i -group, A 12 or A 13. Then G = H and nse(g)=nse(h) if and only if G = H. Not all groups can be determined by nse(g) and G. Let A, B be two finite groups, G := A B means the semidirect of A, B and A G. For example. In 1987, J. G. Thompson gave an example as followings. Let G 1 =(C 2 C 2 C 2 C 2 ) A 7, G 2 = L 3 (4) C 2, where both G 1 and G 2 are maximal subgroups of M 23. Then nse(g 1 )=nse(g 2 )={1, 435, 2240, 6300, 8064, 6720, 5040, 57600}, but G 1 G 2. Comparing the sizes of elements of same order but disregarding the actual orders of elements in T (G) of the Thompson s Problem, in other words, it remains only nse(g), whether can it characterize finite simple groups? Theorem 1.2 [6, 19, 22, 12, 7, 8, 11, 9] Let G be a group and H be Some projective special linear groups, U 3 (5), U 3 (7) or L 5 (2). Then nse(g)=nse(h) if and only if G = H. A finite group G is called a simple K n -group, if G is a simple group with π(g) = n. Recently some groups with that π(g) = 3, 4 are characterized by nse only, but there is no new group with that π(g) = 5 which is characterized by nse. So in this paper, it is shown that the groups A 11 also can be characterized by nse only (A 11 is a simple K 5 -group). We introduce some notations. Let a.b denote the products of an integer a by an integer b. Let G be a group and r a prime. Then we denote the number of the Sylow r-subgroup P r of G by n r (G) orn r. Let A n be the alternating group of degree n. Let ω(g) be the set of element orders of G. Let k ω(g) and s k be the number of elements of order k in G. Let nse(g) ={s k k ω(g)}. The other notations are standard (see [1]). 2 Some preliminary results Lemma 2.1 [2] Let G be a finite group and m be a positive integer dividing G. IfL m (G) ={g G g m =1}, then m L m (G). Lemma 2.2 [19] Let G be a group containing more than two elements. If the maximal number s of elements of the same order in G is finite, then G is finite and G s(s 2 1).

3 A new characterization of A Lemma 2.3 [14] Let G be a finite group and p π(g) be odd. Suppose that P is a Sylow p-subgroup of G and n = p s m with (p, m) =1. If P is not cyclic and s>1, then the number of elements of order n is always a multiple of p s. To prove G = A n, we need the structure of simple K n -groups with n = 11, 12. Lemma 2.4 [21] Let G be a simple K 4 -group. Then G is isomorphic to one of the following groups: (1) A 7, A 8, A 9 or A 10. (2) M 11, M 12 or J 2. (3) One of the following: (a) L 2 (r), where r is a prime and r 2 1=2 a 3 b v c with a 1, b 1, c 1, and v is a prime greater than 3. (b) L 2 (2 m ), where 2 m 1=u, 2 m +1 = 3t b with m 2, u, t are primes, t>3, b 1. (c) L 2 (3 m ), where 3 m +1 = 4t, 3 m 1=2u c or 3 m +1 = 4t b, 3 m 1=2u, with m 2, u, t are odd primes, b 1, c 1. (4) One of the following 28 simple groups: L 2 (16), L 2 (25), L 2 (49), L 2 (81), L 3 (4), L 3 (5), L 3 (7), L 3 (8), L 3 (17), L 4 (3), S 4 (4), S 4 (5), S 4 (7), S 4 (9), S 6 (2), O + 8 (2), G 2 (3), U 3 (4), U 3 (5), U 3 (7), U 3 (8), U 3 (9), U 4 (3), U 5 (2), Sz(8), Sz(32), 2 D 4 (2) or 2 F 4 (2). Lemma 2.5 [5] Each simple K 5 -group is isomorphic to one of the following simple groups: (1) L 2 (q) with π(q 2 1) =4. (2) L 3 (q) with π((q 2 1)(q 3 1)) =4. (3) U 3 (q) with q satisfies π((q 2 1)(q 3 + 1)) =4. (4) O 5 (q) with π(q 4 1) =4. (5) Sz(2 2m+1 ) with π((2 2m+1 1)(2 4m+2 + 1)) =4. (6) R(q) where q is an odd power of 3 and π(q 2 1) =3and π(q 2 q+1) = 1.

4 256 Yong Yang, Shitian Liu and Yanhua Huang (7) The following 30 simple groups: A 11, A 12, M 22, J 3, HS, He, McL, L 4 (4), L 4 (5), L 4 (7), L 5 (2), L 5 (3), L 6 (2), O 7 (3), O 9 (2), PSp 6 (3), PSp 8 (2), U 4 (4), U 4 (5), U 4 (7), U 4 (9), U 5 (3), U 6 (2), O + 8 (3), O 8 (2), 3 D 4 (3), G 2 (4), G 2 (5), G 2 (7) or G 2 (9). Lemma 2.6 Let G be a simple K n -group such that 3 4 G , where n =4, 5. Then G is isomorphic to A 11 or M 22. Proof. We will prove the Lemma by the following two steps. Step 1. G is a simple K 4 -group. From Lemma 2.4(1)(2)(4), order consideration rules out these cases. Therefore we consider Lemma 2.4(3). Case a. G = L 2 (r). We can assume that r =3, 5, 7, 11. Let r = 3, then 3 2 1=2 3, a contradiction. Let r = 5. Then 5 2 1=2 3.3, a contradiction. Let r = 7. Then 7 2 1=2 4.3, a contradiction. Let r = 11. Then = and G = L 2 (11). But 3 4 L 2 (11) since L 2 (11) = Case b. G = L 2 (2 m ). We can assume that u =3, 5, 7, 11. Let u = 3. Then 2 m 1 = 3 and so m = 2. So we have 5 = 3t b, the equation has no solution in N. Let u = 5. But the equation 2 m 1 = 5 has no solution in N. Let u = 7. Then 2 m 1 = 7 and so m = 3. But from the equation =3t b, we have t = 3 and b = 1, a contradiction. Let r = 11. Then the equation 2 m 1 = 11 has no solution in N. Case c. G = L 2 (3 m ). If 3 m +1=4t, 3 m 1=2u c, then we consider that t =3, 5, 7, 11. Let t =3, 5, 7, 11. Then the equation 3 m +1=4.t has no solution in N.

5 A new characterization of A If 3 m +1=4t b,3 m 1=2u, then we assume that u =3, 5, 7, 11. Let u =3, 5, 7, 11. Then the equation 3 m 1=2.t has no solution in N. Step 2. G is a simple K 5 -group. In view of Lemma 2.5(7), G = A 11 or M 22. Let G = L 2 (q). If q =2, 3, then π(g) = 1 which contradicts π(q 2 1) =4. If q =4, 9, 5, 7, then π(g) = 2 which contradicts π(q 2 1) =4. If q =8, 16, 32, 128, 27, 81, 25, then π(g) = 3 which contradicts π(q 2 1) =4. If q = 64, then 13 L 2 (64), a contradiction. Similarly we can rule out the remaining cases. This completes the proof. Lemma 2.7 [13, Theorem 9.3.1] Let G be a finite solvable group and G = mn, where m = p α 1 1 pαr r, (m, n) =1. Let π = {p 1,,p r } and h m be the number of Hall π-subgroups of G. Then h m = q β 1 1 qs βs satisfies the following conditions for all i {1, 2,,s}: (1) q β i i 1 (mod p j ) for some p j. (2) The order of some chief factor of G is divided by q β i i. 3 Main theorem and its proof Let G be a group such that nse(g)=nse(l 2 (r)), and s n be the number of elements of order n. By Lemma 2.2 we have G is finite. We note that s n = kφ(n), where k is the number of cyclic subgroups of order n. Also we note that if n>2, then φ(n) is even. If m ω(g), then by Lemma 2.1 and the above discussion, we have { φ(m) s m m d m s (1) d Theorem 3.1 Let G be a group. Then G = A 11 if and only if nse(g)=nse(a 11 )= {1, 18315, , , , , , , , , , , , , , }.

6 258 Yong Yang, Shitian Liu and Yanhua Huang Proof. If G = A 11, then from [1], nse(g)=nse(a 11 ). So we assume that nse(g)=nse(a 11 ). By (1), π(g) {2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 89, 101, , , }. Since > 1 is the only odd number, then s 2 = and 2 π(g). If 2.13 ω(g), then by Lemma 2.1, s 2 + s 13 + s 2.13,a contradiction since s 2.13 nse(g). Therefore 2.13 ω(g). It follows that the Sylow 13-subgroup of G acts fixed point freely on the set of elements of order 2 and P 13 s 2, a contradiction. Hence 13 π(g). Similarly we can prove that the primes 17, 23, 29, 41, 89, 101, , , do not belong to π(g). If 2.31 ω(g), then by Lemma 2.3 of [15], s 2.31 = s 31.t for some integer t and s 2.31 = s 31. By Lemma 2.1, s 2 + s 31 + s 31 (= ), a contradiction. Then we consider that the Sylow 31-subgroup of G acts fixed point freely on the set of elements of order 2 and P 31 s 2, a contradiction. Hence 31 π(g). Therefore we have that π(g) {2, 3, 5, 7, 11, 37}. If3, 5, 7, 11, 37 π(g), then s 3 = , s 5 = , s 7 = or , s 11 = and s 37 = If 2 a ω(g), then since φ(2 a ) s 2 a,0 a 9. By Lemma 2.1, P 2 1+s 2 + s s 2 i with i =2, 3,..., 9 and P If 3 a ω(g), then 1 a 5. Let exp(p 3 ) = 3. Then by Lemma 2.1, P 3 1+s 3 and P Let exp(p 3 )=3 2. Then P 3 1+s 3 + s 3 2 and P (when s 3 2 = ). Let exp(p 3 )=3 3. Then P 3 1+s 3 + s s 3 3 and P (when s 3 2 = , s 3 3 = or s 3 2 = , s 3 3 = ). Let exp(p 3 )=3 4. Then P 3 1+s 3 + s s s 3 4 s 3 2 = , s 3 3 = , s 3 4 = ). and P (when Let exp(p 3 )=3 5. Then P 3 1+s 3 + s s s s 3 5 and P (when s 3 2 = , s 3 3 = , s 3 4 = , s 3 5 = ). Therefore P If 2.3 ω(g), then by Lemma 2.3 of [15], s 2.3 = s 3.t for some integer t, then s 2.3 = s 3 since s 2.3 nse(g). Similarly s = or ; s = or But by Lemma 2.1, s 2 + s s s 3 + s s s 2 3.3, a contradiction by easy computation. Hence ω(g).

7 A new characterization of A If ω(g), then by Lemma 2.3 of [15], s = s 3 2.t for some integer t, then s = s 3 2 where s 3 2 {237600, , , , , , , , , , , }. Let s 3 2 {237600, }. If s 3 2 = , then s = , , If s 3 2 = , then s = , , Let s 3 2 {457380, , , , , , , , , }. Then s = s 3 2. By Lemma 2.1, s 2 + s 3 + s s s 2.3 2, a contradiction by easy computer s calculations. So we have ω(g). If 5 a ω(g), then 1 a 3. Let exp(p 5 ) = 5. Then P 5 1+s 5 and P Let exp(p 5 )=5 2. Then P 5 1+s 5 + s 5 2 and P Let exp(p 5 )=5 3. Then P 5 1+s 5 + s s 5 3 and P Therefore P If 2.5 ω(g), then by Lemma 2.3 of [15], s 2.5 = s 5.t for some integer t and s 2.5 = s 5. If ω(g), then s =4.s 2 2.t for some integer t and s = (in this case s 2 2 = ). If ω(g), then s = (in this case s 2 3 = ). By Lemma 2.1, s 2 + s s s 5 + s s s 2 3.5, a contradiction. Hence ω(g). If ω(g), then s = s 5 2.t for some integer t. Then s = But by Lemma 2.1, s 2 + s 5 + s s s 2.5 2, a contradiction by computer computation. Therefore ω(g). If 3.5 ω(g), then by (1), s 3.5 {237600, , , , , , , , , }. If ω(g), then by Lemma 2.3 of [15], s =4.s 3 2.t for some integer t and s = (in this case s 3 2 = ). If ω(g), then s =4.s 3 3.t for some integer t, but the equation has no solution in N. Therefore ω(g). If ω(g), then by Lemma 2.1, s 2 + s 3 + s 5 + s s s s 2.3.5, a contradiction. Hence ω(g). If 7 a ω(g). Then 1 a 3. Let s 7 =

8 260 Yong Yang, Shitian Liu and Yanhua Huang Let exp(p 7 ) = 7. Then by Lemma 2.1, P 7 1+s 7 and P Let exp(p 7 )=7 2. Then P 7 1+s 7 + s 7 2 and P Let exp(p 7 )=7 3. Then s 7 + s s 7 3, a contradiction since s 7 3 nse(g). Let s 7 = Let exp(p 7 ) = 7. Then by Lemma 2.1, P 7 1+s 7 and P 7 7. Let exp(p 7 )=7 2. Then P 7 1+s 7 + s 7 2 and P (when s 7 2 = ). Let exp(p 7 )=7 3. Then s 7 + s s 7 3 and P Therefore P If 2.7 ω(g), then s 2.7 = s 7.t for some integer t and s 2.7 = , , If ω(g), then s =6.s 2 2 for some integer t, but the equation has no solution since s nse(g). So ω(g). If ω(g), then s = s 7 2.t for some integer t and s = s 7 2 = , , But by Lemma 2.1, s 2 + s 7 + s s s 2.7 2, a contradiction. So ω(g). If 3.7 ω(g), then s 3.7 = (in this case s 7 = ). Similarly we have that ω(g); ω(g); 5.7 ω(g). If ω(g), then by Lemma 2.1, s 2 +s 3 +s 7 +s 2.7 +s 3.7 +s 2.3.7, a contradiction. Hence ω(g). If 11 π(g). Then 1 a 2. Let exp(p 11 ) = 11. Then P 11 1+s 11 and P Let exp(p 11 )=11 2. Then P 11 1+s 11 + s 11 2 and P Therefore P If 2.11 ω(g), then by Lemma 2.3 of [15], s 2.11 = s 11. But by Lemma 2.1, s 2 + s 11 + s 2.11 (= ), a contradiction. Hence 2.11 ω(g). Similarly 3.11, 5.11, 7.11 ω(g). If 37 a π(g), then a = 1 and so P To remove the prime 37, we show that the prime 3 belongs to π(g). Assume that 3 π(g). If 5, 7, 11, 37 π(g), then G is a 2-group. Since ω(g) = 10 and nse(g) = 16, then we can rule out this case. If 37 π(g), then since P 37 = 37, n 37 = s 37 /φ(37) = and 3 π(g), a contradiction.

9 A new characterization of A Let 11 π(g). Let exp(p 11 ) = 11. Then P 11 = 11. Since n 11 = s 11 /φ(11) = , then 3 π(g), a contradiction. Let exp(p 11 )=11 2. Then P 11 =11 2. Since s 11 2 = or , and n 11 = s 11 2/φ(11 2 )= or , then 3 π(g), a contradiction. Let 7 π(g). Let s 7 = * Let exp(p 7 ) = 7. Then P If P 7 = 7, then n 7 = s 7 /φ(7) = , 3 π(g), a contradiction. If P 7 =7 2, then since π(g) {2, 3, 5, 7, 11, 37} and the above argument, we can assume that π(g) {2, 3, 5, 7}. Therefore k k k k k k k k k k k k k k 14 =2 a.5 c.7 2 where k 1,..., k 14, a, and c are non-negative integers and 0 14 k i 1. It is easy to see that the i=1 equation has no solution. * Let exp(p 7 ) = 7 2. Then P Since n 7 = s 7 2/φ(7 2 ) = , 3, 11 π(g), a contradiction. Let s 7 = Let exp(p 7 ) = 7. Then P 7 7. Since n 7 = s 7 /φ(7 2 )= or , 3, 11 π(g), a contradiction. Let exp(p 7 )=7 2. Then P If P 7 =7 2, then 3 π(g) since s 7 2 { , , }, a contradiction. If P 7 =7 3, then we can rule out this case as the case s 7 = , exp(p 7 )=7 and P 7 =7 2 Let exp(p 7 )=7 3. Then P Since n 7 = s 7 3/φ(7 3 )= , 3, 11 π(g), a contradiction. Let 5 π(g). Let exp(p 5 ) = 5. Then P If P 5 = 5, then n 5 = , 3 π(g), a contradiction. If P 5 =5 2, then k 1 +

10 262 Yong Yang, Shitian Liu and Yanhua Huang k k k k k k k k k k k k k 14 =2 a.5 2 where k 1,..., k 14 and a are non-negative integers. Since nse(g) = 16 and ω(g) = 15, then the equation has no solution. Let exp(p 5 )=5 2. Then P 5 =5 2. Since n 5 = s 5 2/φ(5 2 ), 3 π(g) as s 5 2 {237600, , , , , , , , , }, a contradiction. Let exp(p 5 ) = 5 3. Since s 5 3 = , , then 3 π(g), a contradiction. Therefore 3 π(g). If 3.37 ω(g), then by Lemma 2.3 of [15], s 3.37 =2.s 37.t for some integer t. But the equation has no solution since s 3.37 nse(g). Therefore 3.37 ω(g), it follows that the Sylow 37-subgroup of G acts fixed point freely on the set of elements of order 3. Hence P 37 s 3, a contradiction. Thus 37 π(g). Therefore π(g) {2, 3, 5, 7, 11}. In the following, we consider the following cases: {2, 3}, {2, 3, 5}, {2, 3, 7}, {2, 3, 11}, {2, 3, 5, 7}, {2, 3, 5, 11}, {2, 3, 7, 11} and {2, 3, 5, 7, 11}, Case a. π(g) ={2, 3}. In this case, k k k k k k k k k k k k k k 14 =2 a.3 b where k 1,..., k 14, a and b are non-negative integers. Since nse(g) = 16 and ω(g) = 17, then the equation has no solution. Case b. π(g) ={2, 3, 5}. In this case, k k k k k k k k k k k k k k 14 =2 a.3 b.5 c where k 1,..., k 14, a, b and c are non-negative integers, and 0 14 k i 10. Since G i= , then the equation has no solution. Similarly we can rule out these cases: {2, 3, 7}, {2, 3, 11} Case c. π(g) ={2, 3, 5, 7}. In this case, k k k k k k k k k k k k k k 14 =2 a.3 b.5 c.7 d where k 1,..., k 14, a, b, c and d are non-negative integers, and 0 14 k i 14. Since G i=1

11 A new characterization of A , then (a, b, c, d)=(8, 4, 2, 2), (9, 4, 2, 2), (7, 5, 2, 2), (5, 6, 2, 2), (6, 5, 2, 2), (3, 7, 2, 2),(4, 7, 2, 2), (9, 2, 3, 2), (7, 3, 3, 2), (8, 3, 3, 2), (6, 4, 3, 2), (7, 4, 3, 2), (4, 5, 3, 2), (5, 5, 3, 2), (3, 6, 3, 2), (1, 7, 3, 2), (2, 7, 3, 2), (9, 7, 1, 1), (9, 5, 2, 1), (8, 6, 2, 1), (9, 6, 2, 1), (6, 7, 2, 1), (7, 7, 2, 1), (4, 7, 3, 1), (5, 7, 3, 1), (5, 6, 6, 3, 1), (6, 6, 3, 1), (7, 5, 3, 1), (8, 5, 3, 1), (9, 4, 3, 1), (9, 3, 1, 3), (8, 4, 1, 3), (6, 5, 1, 3), (7, 5, 3, 1), (3, 6, 1, 3), (4, 6, 3, 1), (2, 7, 3, 1), (3, 7, 1, 3), (1, 7, 2, 3), (2, 6, 2, 3), (3, 6, 2, 3), (4, 5, 2, 3), (5, 5, 2, 3), (5, 4, 2, 3), (6, 4, 2, 3), (7, 3, 2, 3), (8, 3, 2, 3), (9, 2, 2, 3). Let G = Then by the programme of [6], there is no group H such that G = = H and nse(g)=nse(h). Similarly we can rule out the other cases. Similarly we can rule out these cases: {2, 3, 5, 11}, {2, 3, 7, 11}. Case d. π(g) ={2, 3, 5, 7, 11}. Since 2.11 ω(g), then we consider that the Sylow 2-subgroup of G acts fixed point freely on the set of elements of order 11 and P 2 s 11. Hence P Similarly we can prove that P 3 3 4, P 5 5 2, P 7 7 and P Therefore we can assume that G =2 a.3 b.5 c Since G =2 a.3 b.5 c.7.11, then G = or G = Assume that G is soluble. Since s 11 = and P 11 = 11, then n 11 = s 11 /φ(11) = By Lemma 2.7, 5 1 (mod 11), a contradiction. Hence G is insoluble. Therefore there is a normal series 1 K L G such that L/K is a simple K i -group with i =3, 4, 5. Let L/K be a simple K 3 -group. Then by [4], L/K = A 5, A 6, L 2 (7), L 2 (8), U 3 (3) or U 4 (2). It is easy to prove that L/K is not a simple K 3 -group. For instance, assume that L/K = U 4 (2). Let G = G/K and L = L/K. Then U 4 (2) L = LC G (L)/C G (L) G/C G (L) =N G (L)/C G (L) Aut(L) Set M = {xk xk C G (L)}, then G/M = G/C G (L) and so U 4 (2) G/M Aut(U 4 (2)). Therefore G/M = U 4 (2), G/M = SU 4 (2).

12 264 Yong Yang, Shitian Liu and Yanhua Huang If G/M = U 4 (2) or G/M = U 4 (2). Then M = or and M is normal in G. It follows that there is a Frobenius group with a kernel of order 11 and a complement of order 2 and there exists an element of order 3.11, a contradiction. Therefore L/K is a simple K i -group with i =4, 5. Then by Lemma 2.6, L/K = A 11 or M 22. Let L/K = A 11. Then G/M = A 11 or S 11. If G/M = A 11, then M = 2 and M = Z(G). It follows that there is an element of order 2.11, a contradiction. If G/M = S 11, then M = 1. But nse(g) nse(s 11 ). So we rule out this case. Let L/K = M 22. Then G/M = M 22 or 2.M 22 and M =2.3 2 or 3 2. If the former, there exists a Frobenius group of a kernel of order 3 and a complement of order 2. It following that there is an element of order 3.11, a contradiction. If the latter, we also can rule out. Therefore G = = A 11. By assumption, nse(g)=nse(a 11 ), then by [17], G = A 11. This completes the proof. Acknowledgments The first author is supported by Education Involution(Grant No: JC-0909 and JG-1102). The second author is supported by the Department of Education of Sichuan Province (Grant No: 12ZB085, 12ZB291 and 13ZA0119) and by the Opening Project of Sichuan Province University Key Laborstory of Bridge Non-destruction Detecting and Engineering Computing (Grant No: 2013QYJ02). The authors are very grateful for the helpful suggestions of the referee. References [1] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, Atlas of finite groups, Oxford University Press, Eynsham, 1985, Maximal subgroups and ordinary characters for simple groups, With computational assistance from J. G. Thackray. [2] G. Frobenius, Verallgemeinerung des sylowschen satze, Berliner Sitz (1895), [3] S. Guo, S. Liu, and W. Shi, A new characterization of alternating group A 13, Far East J. Math. Sci. 62 (2012), no. 1,

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