Research Article
A Characterization of Alternating Group A28 by Conjugate Class Sizes
School of Science, Sichuan University of Science and Engineering, Zigong Sichuan, 643000, People�s Republic of China
All groups in this study considered are finite and simple groups mean simple non-abelian groups. Denote the alternating and symmetric groups of degree n by An and Sn, respectively. Set Aut(G) denotes the automorphism group of a group G. Let ω(G) denote the set of element order of G. Denote the set of nonidentity orders of conjugate classes of elements in G by N(G). The other notations are standard (Conway et al., 1985).
With respect to N(G), Thompson gave the following known conjecture. Thompsons Conjecture (Mazurov and Khukhro, 2010). If L is a finite simple non-Abelian group, G is a finite group with trivial center and N(G) = N(L), then GL.
For a finite group G, we set π(G) = π(|G|). Let, GK(G) be a graph with vertex set π(G) such that two primes p and q in π(G) are joined by an edge if G has an element of order pq. We set s(G) denote the number of connected components of the prime graph GK(G). A classification of all finite simple groups with disconnected prime graph was obtained in Kondratev (1989) and Williams (1981) study. Based on these results, Thompsons conjecture was proved valid for all finite simple groups with s(G)≥2 (Guiyun, 1996; Chen, 1999). So whether there is a group with connected prime graph for which Thompsons conjecture would be true? Recently, the groups A10, A16 and A22 were proved valid for this conjecture (Vasilev, 2009; Gorshkov, 2012; Xu, 2013). Whether is there a group Ap+5 characterized by N(G) in connection to Thompsons Conjecture? In this study, we give an example for Ap+5, namely, it will be proved that if G is a finite group with trivial center and N(G) = N(A28), then G is isomorphic to A28.
MATERIAL AND METHODS
Some preliminary results are given in this section.
Lemma 1: Let x, y∈G, (|x|), |y|) = 1 and xy = yx. Then:
• | CG(xy) = CG(x)∩CG(y) |
• | |xG| divides |(xy)G| |
• | If |xG| = |(xy)G|, then CG(x)≤CG(y) |
Proof: See Lemma 1.2 of Vasilev (2009) and Lemma 2.3 of Ahanjideh and Ahanjideh (2013).
Lemma 2: If P and H are finite groups with trivial centers and N(P) = N(H), then π(P) = π(H).
Proof: See Lemma 3 of Vasilev (2009).
Lemma 3: Suppose that G is a finite group with trivial center and p is a prime from π(G) such that p2 does not divide |xG| for all x in G. Then a Sylow p-subgroup of G is elementary abelian.
Proof: See Lemma 4 of Vasilev (2009).
Lemma 4: Let, K be a normal subgroup of G and = G/K:
• | If is the image of an element x of G in Then divides |xG| |
• | If (|x|, |K|) = 1, then = xK/K |
• | If y∈K, then |yk| divides |yG| |
Proof: See Lemma 5 of Vasilev (2009).
Lemma 5: Let L = A28. Then the following hold:
• | |L| = 224.313.56.74.112.132.17.19.23 |
• | The following numbers from N(L) are maximality with respect to divisibility: |
224.313.55.74.112.132.17.19, 224.313.56.74.112.132.17.23,
224.313.56.74.11.132.19.23, 224.313.56.74.112.17.19.23,
224.313.55.74.112.132.17.19.23, 222.312.56.74.112.132.17.19.23,
• | 23'-numbers in N(L)\{1} are: |
224.313.55.74.112.132.17.19, 26.34.5.7.13, 33.52.7.13, 23.32.7.13
19'-numbers in N(L)\{1, 23'_number} are:
224.311.56.74.112.132.17.23, 222.313.55.74.112.132.17.23
224.39.56.74.112.132.17.23, 222.312.56.74.112.132.17.23
29.33.53.72.11.13.23, 26.35.52.72.11.13.23
23.3.53.72.11.13.23, 26.34.53.72.11.13.23, 34.52.72.11.13.23
25.34.52.72.11.13.23, 22.35.52.72.11.13.23, 27.34.5.72.11.13.23
27.34.52.74.11.13.23, 24.35.52.7.11.13.23, 26.34.52.7.11.13.23
17'-numbers in N(L)\{1,19'-number, 23'-number} are:
• | Any number from N(L)\{1} is divisible by 13 or 11. |
Proof: James and Kerber (1981).
In this section, the main theorem and its proof is given.
Theorem: Let, G be a finite group with trivial center and N(G) = N(A28). Then G is isomorphic to A28.
Proof: By Lemma 2, we have that π(G) = {2, 3, 5, 7, 11, 13, 17, 19, 23}.
The desired result were obtained by first showing that there is no elements of order 17.19, 17.23 and 19.23 sec proving that if K is a maximal normal subgroup of G, then K is a {2,3}-group, in particular, G is insoluble and third getting that G is isomorphic to A28.
Step 1: Let p∈{17, 19, 23}. Then the Sylow p-subgroup S of G is of order p. There is no elements of order 17.19, 17.23 and 19.23.
Let, p∈π(G). Since p2 does not divide |xG| for all x in G, then by Lemma 3, the Sylow p-subgroup of G is elementary abelian. In particular, if |x| = p, then |xG| is a p'-numbers.
Assume that p = 23 and |S|≥232. Then there is an element x of G such that |xG| = 224.313.56.74.112.17.19.23 by Lemma 5.
Set 23|x|. Let y be an element of CG(x) having order 23. Then CG(xy) = CG(x)∩CG(y) and so |xG| |(xy)G| and |yG| |(xy)G|. Since |yG| is a 23'-number, then |yG| is equal to 224.313.55.74.112.132.17.19, 26.34.5.7.13, 33.52.7.13, 23.32.7.13.
In these four cases, 224.313.55.74.112.132.17.19.23 ||(xy)G| which contradicts Lemma 5.
Set 23 ||x|. Then we write |x| = 23 m. Since S is elementary abelian, then the numbers 23 and m are coprime. Let u = x23 and v = xm. Then x = uv and CG(x) = CG(u)∩CG(v). Therefore |vG|||xG| = 224.313.56.74.112.17.19.23. On the other hand, |v| = 23 and so by Lemma 5, |vG| is equal to 224.313.55.74.112.132.17.19, 26.34.5.7.13 or 23.32.7.13. It follows that 13||xG|, a contradiction.
Assume that p = 19 and |S|≥192. Then exists an element x of G such that |xG| = 224.313.55.74.112.17.19 by Lemma 5.
Set 19 |x|. Let y be an element of CG(x) having order 19. Since S is elementary abelian and |y| = 19, then |yG| is a 19'-number. Therefore |yG| equals to:
26.34.5.7.13, 33.52.7.13, 23.32.7.13, 224.311.56.74.112.132.17.23
222.313.55.74.112.132.17.23, 224.39.56.74.112.132.17.23
222.313.55.74.112.132.17.23, 29.33.53.74.11.13.23,
26.35.52.72.11.13.23, 23.3.53.72.11.13.23, 26.34.53.72.11.13.23,
34.52.72.11.13.23
24.33.52.7.11.13.23, 26.34.52.7.11.13.23
Therefore 13||xG|, a contradiction.
Set 19||x|. Then let |x| = 19 m. Since S is elementary abelian, then the two numbers 19 and m are coprime. Let u = x19 and v = xm. Then x = uv, CG(x) = CG(u)∩CG(v) and |v| = 19. By Lemma 5, |vG| equals to:
26.34.5.7.13, 33.52.7.13, 23.32.7.13, 224.311.56.74.112.132.17.23,
222.313.55.74.112.132.17.23, 224.39.56.74.112.132.17.23,
222.312.56.74.112.132.17.23, 29.33.53.72.11.13.23,
26.35.52.72.11.13.23, 28.3.53.72.11.13.23, 26.34.53.72.11.13.23,
34.52.72.11.13.23, 25.34.52.72.11.13.23, 22.35.52.72.11.13.23,
27.34.5.72.11.13.23, 27.34.52.74.11.13.23, 24.33.52.7.11.13.23 or
26.34.52.7.13.23.
It follows that 13||vG||xG| = 224.313.56.74.112.17.19.23, also a contradiction.
Assume that p = 17 and |S|≥172. Then by Lemma 5, there exists an element x of G such that |xG| = 224.313.56.74.112.17.19.23.
Set 17 |x|. Let y be an element of CG(x) having order 17. Since S is elementary abelian and |y| = 17, then |yG| is 17'-number. By Lemma 5, |yG| equals to:
26.34.5.7.13, 33.52.7.13, 23.32.7.13, 29.33.53.72.11.13.23
26.35.52.72.11.13.23, 23.3.53.72.11.13.23, 26.34.53.72.11.13.23
34.52.72.11.13.23, 25.34.52.72.11.13.23, 22.35.52.72.11.13.23
27.34.5.72.11.13.23, 27.34.52.11.13.23, 24.33.52.7.11.13.23
26.34.52.7.11.13.23, 223.313.56.74.112.132.19.23
224.312.56.73.112.132.23, 222.312.56.74.112.132.19.23
221.311.56.74.112.132.19.23, 224.39.56.74.112.132.19.23
219.312.56.74.112.132.19.23, 220.313.56.74.112.132.19.23
223.39.56.74.112.132.19.23, 221.313.55.74.112.132.19.23
219.312.56.74.112.132.19.23, 217.311.56.74.112.132.19.23
224.311.55.74.112.132.19.23, 222.311.56.74.112.132.19.23
222.312.56.73.112.132.19.23, 219.311.56.74.112.132.19.23
219.312.55.74.112.132.19.23, 221.310.55.74.112.132.19.23
221.311.54.74.112.132.19.23, 218.311.56.73.112.132.19.23
218.310.55.73.112.132.19.23, 217.39.55.73.11.132.19.23
222.313.55.74.112.132.19.23, 223.311.56.74.112.132.19.23
222.313.56.73.112.132.19.23, 218.311.56.74.112.132.19.23
224.311.55.74.112.132.19.23, 224.312.56.74.112.132.19.23
28.35.53.72.11.13.19.23, 27.37.52.72.11.13.19.23
27.37.53.7.11.13.19.23, 23.35.53.72.11.13.19.23
29.35.53.72.11.13.19.23, 25.36.53.72.11.13.19.23
25.35.53.72.11.13.19.23, 25.33.53.72.11.13.19.23
We also have 13||vG|| |xG|, a contradiction.
Set 17||x|. Then let |x| = 17 m. Since S is elementary abelian, then the numbers 17 and m are coprime. Let u = x17 and v = xm. Then |v| = 17, x = uv and CG(x) =CG(u)∩CG(v). Therefore |uG| |xG| and |vG| |xG|. On the other hand, |vG| is 17'-number and so |vG| is equal to:
26.34.5.7.13, 33.52.7.13, 23.32.7.13, 29.33.53.72.11.13.23
26.35.52.72.11.13.23, 23.3.53.72.11.13.23, 26.34.53.72.11.13.23
34.52.72.11.13.23, 25.34.52.72.11.13.23, 22.35.52.72.11.13.23
27.34.5.72.11.13.23, 27.34.52.11.13.23, 24.33.52.7.11.13.23
222.312.56.74.112.132.19.23, 221.311.56.74.112.132.19.23
224.39.56.74.112.132.19.23, 219.312.56.74.112.132.19.23
220.313.56.74.112.132.19.23, 223.39.56.74.112.132.19.23
221.313.55.74.112.132.19.23, 219.312.56.74.112.132.19.23
217.311.56.74.112.132.19.23, 224.311.55.74.112.132.19.23
222.311.56.74.112.132.19.23, 222.312.56.73.112.132.19.23
219.311.56.74.112.132.19.23, 219.312.55.74.112.132.19.23
221.310.55.74.112.132.19.23, 221.311.54.74.112.132.19.23
218.311.56.73.112.132.19.23, 218.310.55.73.112.132.19.23
217.39.55.73.11.132.19.23, 222.313.55.74.112.132.19.23
223.311.56.74.112.132.19.23, 222.313.56.73.112.132.19.23
218.311.56.74.112.132.19.23, 224.311.55.74.112.132.19.23
224.312.56.74.112.132.19.23, 28.35.55.74.11.13.19.23
27.37.52.72.11.13.19.23, 27.37.53.7.11.13.19.23
23.35.53.72.11.13.19.23, 29.35.53.72.11.13.19.23
25.36.53.72.11.13.19.23, 25.35.53.72.11.13.19.23
25.35.53.72.11.13.19.23, 29.34.53.7.11.13.19.23
26.34.53.72.11.13.19.23, 28.38.5.72.11.13.19.23
It follows that 13||vG|| |xG|, a contradiction.
Therefore the Sylow p-subgroup is of order p for p∈{17, 19, 23}.
There is no element of order 17.19, 19.23 and 17.23.
Step 2: Let, π = {2, 3, 5}. Then Oπʹ, π (G) = Oπ (G). In particular, G is insoluble.
Let, K = Oπ (G), = G/K and denote by x* and by H* the images of an element x and a subgroup H of G in , respectively. If the result is invalid, then there exists p∈π (L)\π such that Op≠1.
Let Op≠ 1 for p∈ {17,1923}. Then contains a Hall {p, q}-subgroup of order p.q for p∈{17,1923}\{p}. This subgroup must be cyclic and so there is an element of order p.q which contradicts Step 1.
Let P be a Sylow 13-subgroup of . If O13 ≠ 1, then A = Z (O13 ) is a nontrivial normal subgroup of . Let x* be an element of order 23 in . Then is a divisor of 224.313.55.74.112.132.17.19, 26.34.5.7.13, 33.52.7.13, or 23.32.7.13. It is easy to get A = CA (x*)×[A, x*]. Hence the index of CA (x*) is at most 132. Obviously n = 11 is the least number with that 23|13n-1 and so [A, x*] (x*) is abelian. It means that [A, x*] = 1 and A = CA (x*). Let z* be a nontrivial element of Z (p)∩A, then 23 (z*) |. By Lemma 4, the preimage z in G lies in the center of the Sylow 13-subgroup, contradicting Step 1. So O13 = 1.
Let P be a Sylow 11-subgroup of . If O13≠ 1, then A = Z(O13 ) is a nontrivial normal subgroup of . Let x* be an element of order 23 in . Then is a divisor of 224.313.55.74.112.132.17.19,26.34.5.7.13, 33.52.7.13 or 23.32.7.13. It is easy to get A = CA (x*)×[A,x*]. Hence the index of CA(x*) is at most 132. Obviously n = 11 is the least number with that 23|13n and so [A,x*](x*) is abelian. It means that [A,x*] = 1 and A = CA (x*). Let z* be a nontrivial element of Z(P)∩A, then 23. By Lemma 4, the preimage z in G lies in the center of the Sylow 11-subgroup, contradicting Step 1. So O13 = 1.
Let P be a Sylow 7-subgroup of . If O11≠ 1, then A = Z(O11) is a nontrivial normal subgroup of . Let x* be an element of order 23 in . Then is a divisor of 224.313.55.74.112.132.17.19, 26.34.5.7.13, 33.52.7.13 or 22.32.7.13. It is easy to get A = CA(x*)×[A,x*]. Hence the index of CA(x*) is at most 112. Obviously n = 11 is the least number with that 23|11n-1 and so [A,x*](x*) is abelian. It means that [A,x*] = 1and A = CA(x*). Let z* be a nontrivial element of Z(P)∩A, then 23||. By Lemma 4, the preimage z in G lies in the center of the Sylow 7-subgroup, contradicting Step 1. So O11 = 1.
Therefore Oπ',π(G) = Oπ(G). In particular, G is insoluble.
Step 3: G is isomorphic to A28.
By Lemma 2, π(G) = {2,3,5,7,11,13,17,19,23} and Oπ(G) is a maximal normal soluble subgroup of G with π = {2,3,5}.
From Step 2, G is insoluble and so there is a normal series K≤H≤G with that H/K is a direct product of nonabelian simple groups isomorphic to the groups listed as listed in (Zavarnitsine, 2009), namely, H/K = S1×S2 ×Sk. We know that G can not contain a Hall {17, 19, 23}-subgroup, the numbers 17, 19 and 23 divide the order of exactly one of these groups and so assume that 17, 19, 23||S1|. Therefore, S1≤ . If k>1, then there is a Sylow 7-subgroup P7 of /S1. Let M = H/K and Z = Z(P7). Then Z∩M/S1 is nontrivial. Consider an element x of S2× ×Sk such that its image in lies in Z. Since x centralizes S1, then x∈Z and so centralizes an element of order 23, a contradiction.
So, we have H/K = S1. We know that H/K≤ ≤Aut(H/K). Since 17, 19, 23||G|, then H/K is isomorphic to An with n = 23, 24, 25, 26, 27, 28. Thus An≤ ≤ Aut(An).
If H/K = An with n = 23, 24, 25, then since K is a {2, 3, 5}-group, 13||G|, a contradiction.
Let H/K = A26 or S26. Then there exists an element x* of order 23 in H/K such that |(x*)H/K| = 222 .310 .55.73.112.132.17.19 and |CH/K(x*)| = 3.23. Let x be an element of order 23 in H corresponding to x*, by Lemmas 4 and 5, |xH| |xG| 224.313.55.74.112.132.17.19 and |CG(x)| = |5.23|≤|CH(x)|. Therefore by Lemma 4:
a contradiction.
Let, H/K = A27 or S27. Then there exists an element x* of order 23 in H/K such that |(x*)H/K| = 220.313.56.112.132.17.19 and |CH/K(x*)| = 23.23. Let x be an element of order 23 in H corresponding to x*, by Lemmas 4 and 5, |XG| = 224.313.55.74.112.132.17.19 and |CG(x)| = 5.23≤|CH(x)|. Therefore by Lemma 4:
It follows that 5||K∩CH(x)|, in particular x centralizes an element of order 23 in H/K, a contradiction.
Let H/K ≅ S28. Then there is an element x* of order 23 such that |(x*) | = 225.313.55.73.112.132.17.19, a contradiction.
Hence H/K is isomorphic to A28. Now we prove K = 1. We refine the normal series 1<K<G into the chief ones. Let K be a nontrivial p-group with p∈{2, 3, 5}.
Let p = 2. Then G≅K.A28 or K×A28. If the former, there is an element of order 23 such that 225.313.55.74.112.132.17.19, a contradiction. If the latter, then by Lemma 2.7 of Jiang et al. (2011) also there is an element of order 23 such that 225.313.55.74.112.132.17.19, a contradiction.
For p = 3, 5, we also can similarly rule out as the case "p = 2". Therefore K = 1 and G≅A28.
This completes the proof.
Using the properties of N(A28), we proved that the group A28 is characterizable by the set of its conjugate classes sizes.
As it was proved that the group A10 can be characterized by its order and two special conjugacy classes sizes. Then obviously, we also have the following result.
Corollary 7. Let G be a finite group with trivial center. Assume that N(G) = N(A28) and |G| = |A28|. Then G≅A28.
The project was supported by the Opening Project of Sichuan Province University Key Laborstory of Bridge Non-destruction Detecting and Engineering Computing (Grant No. 2013QYJ02 and 2014QYJ04) and by the Scientific Research Project of Sichuan University of Science and Engineering (Grant No. 2014RC02). The authors are very grateful for the helpful suggestions of the referee.