#### Abstract

Let be a group. Denote by the set of prime divisors of . Let be the graph with vertex set such that two primes and in are joined by an edge if has an element of order . We set to denote the number of connected components of the prime graph . Denote by the set of nonidentity orders of conjugacy classes of elements in . Alavi and Daneshkhah proved that the groups, where with , are characterized by . As a development of these topics, we will prove that if is a finite group with trivial center and with composite, then is isomorphic to .

#### 1. Introduction

In this paper, all groups considered are finite and simple groups are nonabelian simple groups. Let be a finite group and let be its center. For any , suppose that denotes the conjugacy classes in containing and denotes the centralizer of in . We will use to denote the set has a conjugacy class of size . Thompson in 1987 gave the following conjecture with respect to .

*Thompson’s Conjecture* (see [1], Question 12.38). If is a finite simple nonabelian group, is a finite group with trivial center, and , then .

Let denote the set of all prime divisors of . Let be the graph with vertex set such that two primes and in are joined by an edge if has an element of order . We set to denote the number of connected components of the prime graph . A classification of all finite simple groups with disconnected prime graph was obtained in [2, 3]. Based on these results, Thompson’s conjecture was proved valid for many finite simple groups with (see [4, 5]). In Ahanjideh and Xu’s works, the groups , , , , , and are true for Thompson’s conjecture (see [6–10]). Alavi and Daneshkhah proved that the groups with , , and are characterized by (see [11]). So is there a group with connected prime graph for which Thompson’s conjecture would be true? Recently, the groups , , , and were proved valid for this conjecture (see [12–15]). As a development of these topics, we will prove that Thompson’s conjecture is true for the alternating groups of degree with composite.

We will introduce some notations used in the proof of the main theorem. Let be a group and a prime. Then denote by the Sylow -subgroup of . Let and denote the automorphism and outer automorphism groups of , respectively. Let denote the set of element orders of . The other notations are standard (see [16], for instance).

#### 2. Preliminary Results

In this section we will give some preliminary results.

Lemma 1. *Let , , and . Then *(1)*;*(2)* divides ;*(3)*if , then .*

*Proof. *See [12, Lemma 1.2] and [7, Lemma 2.3].

Lemma 2. *If and are finite groups with trivial centers and , then .*

*Proof. *See [12, Lemma 3].

Lemma 3. *Suppose that is a finite group with trivial center and is a prime from such that does not divide for all in . Then a Sylow -subgroup of is elementary abelian.*

*Proof. *See [12, Lemma 4].

Lemma 4. *Let be a normal subgroup of , and . *(1)*If is the image of an element of in , then divides .*(2)*If , then .*(3)*If , then divides .*

*Proof. *See [12, Lemma 5].

Let denote the nonnegative integer such that but .

Lemma 5. *Let be the alternating group of degree , where is a prime. Then the following hold.*(1)*; in particular, .*(2)* for each . Furthermore, , where . In particular, if , then .*

*Proof. *(1) By the definition of Gaussian integer function, we have that

(2) Similarly, as (1), we have that
for an odd prime .

If , .

The proof is completed.

Let be the symmetric group of degree . Assume that the cycle has 1-cycles, 2-cycles, and so on, up to -cycles, where . Then the number of conjugacy classes in is Let be the alternating group of degree .

Lemma 6. *Let . Then, for the size of the conjugacy class of in , one has the following. *(1)*If, for all even , and, for all odd , , then .*(2)*In all other cases, .**In particular, .*

*Proof. *See [17].

Lemma 7. *If is a natural number, then there are at least prime numbers such that . Here*(i)* for ;*(ii)* for ;*(iii)* for ;*(iv)* for ;*(v)* for ;*(vi)* for .**In particular, for every natural number , there exists a prime such that , and, for every natural number , there exists an odd prime number such that .*

*Proof. *See Lemma 1 of [18].

Lemma 8. *Let , and be positive integers such that . Then there exists a prime with the following properties:*(i)* divides ,*(ii)* does not divide for all ,**with the following exceptions: , ; and ; .*

*Proof. *See [19].

Lemma 9. *With the exceptions of the relations and every solution of the equation
**
has exponents ; that is, it comes from a unit of the quadratic field for which the coefficients and are primes.*

Let be a nonabelian simple group and let* O* denote the order of the outer automorphism group of .

Lemma 10. *Let be a nonabelian simple group. Then the orders and their outer automorphism of are as listed in Tables 1, 2, and 3.*

*Proof. *See [22].

#### 3. Main Theorem and Its Proof

In this section, we give the main theorem and its proof.

Theorem 11. *Let be a finite group with trivial center and with composite. Then is isomorphic to .*

*Proof. *We know that are characterized by if (see [12–14]). Then in the following we only consider when .

We divide the proof into the following lemmas.

Lemma 12. *Let . Then the following hold.*(1)*If , then we can write with and conjugacy class sizes of -elements of are
* *for possible with .* *In particular, if is an odd prime divisor of , then conjugacy class sizes of -element of are
* *where and .*(2)*If , then one can write with and conjugacy class sizes of -elements of are
* *for possible with .*(3)*If , then one can write with and conjugacy class sizes of -elements of are
* *In particular, if , then the conjugacy class size of -elements of is
*(4)*The following numbers from are maximality with respect to divisibility.(a) One of the following holds: if ; if ; if and with ;(b).*

*Proof. *From (3) and Lemma 6, we get the desired results.

Lemma 13. *Let be a finite group with trivial center and . Then and .*

*Proof. *Note that . Since , every member from divides the order of and . So by Lemma 12, we have that .

Lemma 14. *Suppose that is a finite group with trivial center and . Then the following hold.*(1)*There exist different primes from such that . In particular, the Sylow -subgroup of is a cyclic group of order where . There does not exist an element of order , , or .*(2)*For all , if is divisible at most by , then the Sylow -subgroup of is of order .*

*Proof. *(1) By Lemma 7, there exist different prime numbers from such that .

From Lemmas 12 and 13, we have that the primes are prime divisors of and do not divide for all . Then by Lemma 3, is elementary abelian. Therefore if , then is an -number.

Let . Consider an element of with
or
by Lemma 12.

Assume that . Let be an element of having order . Then , , and by Lemma 1. Since is abelian, . Hence, . It follows that equals or by Lemma 12.

If equals , then . On the other hand, ; then we have that
or
Obviously, there is no number from such that and .

Therefore equals . In the following, we will consider the following three cases.*Case **1*. if .

Obviously, . Therefore , a contradiction, since , , and the maximality of .*Case **2. * if .

Obviously, . Therefore . Also we get a contradiction as in Case 1.*Case **3. * if and with .

In this case, . It follows that . By Lemma 1, and so , a contradiction.

Assume that . Let . Since is elementary abelian, the numbers and are coprime. Let
Then and . Therefore,
or

On the other hand, the element of is of order . Since the Sylow -subgroup of is elementary abelian, then . It follows that
by Lemma 12.

If , then , a contradiction. Hence . We consider the following three cases.*Case **1. * if .

Obviously, . But , a contradiction, since , , and the maximality of .*Case **2. * if .

Obviously, . But , a contradiction, since , , and the maximality of .*Case **3. * if and with .

In this case, . It follows that since the maximality of . By Lemma 1, and so , a contradiction.

Therefore the Sylow -subgroup of is of order .

Similarly we can prove the other two cases.

There does not exist an element of order , , or .

(2) Without loss of generality, we assume that is divisible at most by .

Assume that . Consider an element of such that
by Lemma 12.

Let . Then there is an element of of order . By Lemma 1 we have that , , and .

If is an -central element, then and is an -number. We consider the following three cases.*Case **1. *.

In this case . Then we have that and so since is an -central element (in fact, if , then there is a conjugacy class size which is a multiple of the number contradicting ). Thus since , a contradiction. *Case **2. *.

We have that . We similarly can rule out this case as “Case 1: ”. *Case **3. * and with .

In this case, or (we only choose the maximality from the -number in ). If the former, then and so since is an -central element (in fact, if , then there is a conjugacy class size which is a multiple of the number contradicting ). Thus since , a contradiction. Then . But there is no member from such that and .

If is a noncentral -element, then we choose an element of order such that . By hypothesis, and, obviously, . Then by Lemma 1.2 of [4], there is an -element such that , , and
On the other hand, we also have . Since , then and so . Therefore . It follows that since the maximality of , and we get , a contradiction.

Let ; then we write . If is elementary abelian, then . Set
Then and . Hence and . Since is an element of order , then
If , then and so . It follows that contradicts the fact that is an -element. Therefore . On the other hand, . Obviously, but . It follows that lies in , which contradicts the maximality of .

Therefore is nonabelian. So we chose an element of order such that . By hypothesis, and, obviously, . Then by Lemma 1.2 of [4], we also have that since the maximality of , and we get , a contradiction.

The lemma is completed.

Lemma 15. *Suppose that is a finite group with trivial center and . Let . Then . In particular, is insoluble.*

*Proof. *Let , and denote by and by the images of an element and a subgroup of in , respectively. Assume that the result is not true; then there is a prime with .

Let with . Then contains a Hall -subgroup of order with . However, Hall -subgroup must be cyclic contradicting Lemma 14.

Let be a Sylow -subgroup of where . If , then is a nontrivial normal subgroup of . Let be an element of order in . So we have that is a divisor of
By coprime action lemma, . In the following, we consider two cases “ and ”.(i) and . In this case, by Lemma 14, we have that the Sylow -subgroup of is of order . Hence there is a Hall -subgroup . Since must be cyclic, then there is an element of order , a contradiction by the proof of Lemma 14.(ii) and . If , , or with and , then the index of in is at most . By Lemma 8, there exists a least divisor of such that divides , and the subgroup must be abelian. It follows that and . If is a nontrivial element of , then the order of is a multiple of . By Lemma 4, lies in the center of a Sylow -subgroup of . This contradicts Lemma 12. Thus .It follows that for .

Therefore . In particular, is insoluble.

Lemma 16. *There is a normal series such that .*

*Proof. *By Lemmas 13 and 14, .

By Lemma 15, we have that , where is a direct product of nonabelian simple groups . Since cannot contain a Hall -subgroup, numbers , , and divide the order of exactly one of these groups that is listed as in Lemma 10, and so we assume that they divide . Since , we let and denote the factor groups and , respectively. If , then a Sylow 2-subgroup of is nontrivial and its center has a nontrivial intersection with . Consider a nontrivial element of such that its image in lies in . Since centralizes , it lies in the center of a Sylow 2-subgroup of and centralizes an element of order , a contradiction. Thus , and is almost simple. Therefore
Obviously, (in fact, if , then , a contradiction, from Lemma 10; if , then there is an element of order with contradicting Lemma 14). In the following, we always assume that . In the following, we consider which is listed as in Tables 1, 2, and 3.*(i) Case **1. * with .

Then with composite and prime. If , then , a contradiction. Therefore is isomorphic to , , , or .

Let be an element of order in . Then is -number since .

Since , then . We have . On the other hand, . It follows that and so there is an element of or of order with and and divide one of the prime divisors of the numbers , , or , which contradicts Lemma 14. Similarly, we can rule out these cases “ and ”.

Therefore .*(ii) Case **2. * is not isomorphic to a sporadic simple group according to Table 3.*(iii) Case **3. * is isomorphic to a simple group of Lie type.

Let be a prime power.(1) with . In this situation, by Lemma 13, and so
It follows that or . If , then is a power of . Since by Lemma 14, this is impossible as . Therefore . It follows that for some as is prime. On the other hand, or . If the former, then and , a contradiction. It follows that . Hence and for some integer . By Lemma 5, , but the equation has no solution in . Furthermore, since has the same order as , we also can rule out.(2) with . Therefore we have that . Since the Sylow -subgroup of is of order , as, otherwise, and thus , a contradiction. It follows that or for some integer . Let . Then with prime, and so . Thus , a contradiction. Let for some integer . Then with prime, and so . It follows that , a contradiction.(3) with . In this situation,
Since the Sylow -subgroup of is of order and , we obtain that or for some integer . If the former, then we have that and , a contradiction. Let for some integer . Then for some . If , then and , a contradiction. It follows that for some and . By Lemma 5, and . If is odd, then . By Lemmas 8 and 9, , , and . Hence or 7. If , we can rule out this case since . If , then and so , a contradiction, since . If is even, then . So or . If the former, then , a contradiction. If the latter, then . It follows that . Let with . Then or and so or . If , then, by Lemma 8, and ; we also can rule out this case as above. So . It follows that . Similarly, as , we can rule out this case.(4). Therefore we have that
It follows that
If , then we can rule out this case since the Sylow -subgroup of is of order . Hence , where . On the other hand, . If , then and , a contradiction. If , then and . By Lemma 5, and so . It is easy to rule out this case by considering the orders of . Similarly, we can exclude that , and .(5). Then we have . It follows that , , or . If , we rule out this case. If , then there exists a prime such that for some integer . Therefore and by Lemma 5. It follows that and so we have a contradiction by considering the order of . Similarly, we also can rule out this case “”.(6). It is easy to see that . It follows that with , with , or . If , then we rule out this case since the Sylow -subgroup of is of order . So with , with , and so there exists a prime such that for some integer . It means that . Therefore . We also can rule out this case by order consideration.(7) with . It follows that . Thus , , or . If , then we rule out this case. If , then there is a prime such that and so by Lemma 5. Hence , a contradiction. Similarly we can rule out this case “”. Similarly .(8), with . We see that . Since the Sylow -subgroup of is of order , then . It follows that or . If , then there exists a prime such that for some integer . If , then , a contradiction. Hence by Lemma 5, and so , a contradiction. If and , then there exists a Frobenius group of with a kernel of order and a complement of order , respectively, and so there is an element of order , which contradicts Lemma 14.(9). We have that . In this case, since has a Sylow -subgroup of order , then , , or . If , then there exists a prime such that and so . By Lemma 5, . It follows that . Order consideration rules out these cases “”. Similarly we can rule out this case “”.(10) with . It is easy to get that
It follows that or . If , then since the Sylow -subgroup of is of order , a contradiction. Hence and so for some integer . It follows that there exists a prime such that and so by Lemma 5. Since the Sylow -subgroup of is of order , then and so , a contradiction.This completes the proof of the lemma.

Lemma 17. *Consider the following.*