On powers of conjugacy classes in finite groups

Conjecture ([4, Conjecture 2])

Let 𝐾 and 𝐷 be conjugacy classes of a finite group 𝐺 and suppose that K n = D ∪ D - 1 for some integer n ≥ 2 . Then ⟨ K ⟩ is a (normal) solvable group of 𝐺.

Suppose that 𝐺 is a finite group and 𝐾 is a conjugacy class of 𝐺 such that K 3 = D ∪ D - 1 for some conjugacy class 𝐷 of 𝐺. Then ⟨ K ⟩ is solvable.

Theorem 2.1 ([1, Theorem 1.11])

Let 𝐺 be a group with finite subsets 𝐴 and 𝐵. If the product A ⁢ B is a normal subset of 𝐺, then there exists a normal subgroup 𝑁 of 𝐺 such that

1. A ⁢ B ⁢ N = A ⁢ B ;

2. | A ⁢ B | ≥ | A ⁢ N | + | B ⁢ N | - | N | .

Theorem 2.2 ([5, Theorem A (a)])

Suppose K = x G is a conjugacy class of an element 𝑥 of a group 𝐺 and 𝐾 is finite. Let 𝑁 denote the normal subgroup [ x , G ] . If | K 2 | = μ ⁢ | K | with μ < 3 2 , then K r = x r ⁢ N for all r ≥ 2 .

Theorem 2.3 ([3, Theorem A])

Let 𝐺 be a finite group, and let 𝑁 be a normal subgroup of 𝐺. Let K = x G be the conjugacy class of an element x ∈ G . Suppose that x ⁢ N = K ∪ K - 1 . Then 𝑁 is solvable, and as a consequence so is ⟨ K ⟩ .

Lemma 2.4 ([4, Lemma 2])

Let 𝐺 be a finite group and 𝐾, 𝐿 and 𝐷 non-trivial conjugacy classes of 𝐺 such that K ⁢ L = D with | D | = | K | . Then ⟨ L ⟩ is solvable.

Proof of Theorem A

We argue by induction on the order of 𝐺. First, we prove that we can assume that 𝐺 has no non-trivial normal solvable subgroups. Otherwise, if 𝐿 is a non-trivial normal solvable subgroup of 𝐺 and G ¯ denotes the factor group G / L , we have K ¯ 3 = D ¯ ∪ D ¯ - 1 (note this is still true if D ¯ = 1 ¯ ), and by induction, ⟨ K ¯ ⟩ is solvable, so the result follows as 𝐿 is solvable. Furthermore, if D = D - 1 , then the result follows from [4, Theorem A]. Also, by [4, Theorem C], either | D | = | K | / 2 or | D | = | K | , and in the first case, the result follows too. Therefore, for the rest of the proof, we will also assume D ≠ D - 1 and | D | = | K | .

We take A = K and B = K 2 in Theorem 2.1 and obtain that there exists a normal subgroup 𝑁 of 𝐺 satisfying K 3 ⁢ N = K 3 , that is, D ∪ D - 1 = N ⁢ ( D ∪ D - 1 ) and

Let us consider the set K ⁢ N , which is the union of (left) cosets of 𝑁. If K ⁢ N is a single coset, say x ⁢ N , with x ∈ K , then

so by applying Theorem 2.3 (there is no loss in assuming x 3 ∈ D ), we get the solvability of ⟨ D ⟩ , contradicting our first assumption. Therefore, we can assume that K ⁢ N consists of at least 2 cosets of 𝑁, and hence | K ⁢ N | ≥ 2 ⁢ | N | . Also, since | K 2 ⁢ N | = | K ⁢ ( K ⁢ N ) | ≥ | K ⁢ N | , from equation (3.1), we obtain

that is,

Next, we show that it can be assumed that | K 2 | ≥ 3 2 ⁢ | K | . Indeed, if | K 2 | < 3 2 ⁢ | K | , by applying Theorem 2.2, we have K 3 = x 3 ⁢ [ x , G ] with x ∈ K , and then Theorem 2.3 applies again to obtain the solvability of ⟨ D ⟩ , which again contradicts our first assumption.

Now, we apply the fact that | K 2 | ≥ 3 2 ⁢ | K | to equation (3.1), and then

so in particular,

Next, we prove that 𝑁 cannot be trivial. Indeed, if N = 1 , then | K | ≤ 2 , and so, for x ∈ K , it certainly follows that Z ⁢ ( C G ⁢ ( x ) ) is a non-trivial normal solvable subgroup of 𝐺, a contradiction. In addition, if | N | = 1 2 ⁢ | K | , then equation (3.3) forces that | K ⁢ N | = | K | , and hence K ⁢ N = K . Now, if we take 𝐶 to be any non-trivial conjugacy class of 𝐺 contained in 𝑁, we necessarily have K ⁢ C = K , and then, by applying Lemma 2.4, we conclude that ⟨ C ⟩ is solvable, a contradiction too. Henceforth, we will assume that

We know that D ∪ D - 1 = N ⁢ ( D ∪ D - 1 ) . Since N ⁢ D is a union of conjugacy classes of 𝐺, there exist exactly three possibilities: N ⁢ D = D , N ⁢ D = D - 1 or N ⁢ D = D ∪ D - 1 . In the first two cases, using Lemma 2.4 and reasoning as above, we can deduce that 𝑁 is (non-trivial) solvable, which is a contradiction. Therefore, we will assume that N ⁢ D = D ∪ D - 1 . Now, since the set D ⁢ N is the union of, say 𝑡, cosets of 𝑁, by using equation (3.2), we have

Moreover, by taking into account that 2 ⁢ | K | = t ⁢ | N | and applying equation (3.4), we obtain 2 ⁢ | K | > t 2 ⁢ | K | , which means that t ≤ 3 . However, since t ≥ 3 by equation (3.5), we conclude that t = 3 , and hence

The rest of the proof consists of getting a contradiction. Let us consider again the set K ⁢ N , which can be written as the union of s ≥ 2 cosets of 𝑁. If s ≥ 3 , since | K 2 ⁢ N | = | K ⁢ ( K ⁢ N ) | ≥ | K ⁢ N | ≥ 3 ⁢ | N | , then equations (3.1) and (3.6) give

a contradiction. Therefore s = 2 , and this means that | K ⁢ N | = 2 ⁢ | N | . Write, for instance, K ⁢ N = x ⁢ N ∪ y ⁢ N , with x , y ∈ K . On the other hand, we have

Furthermore, | K 2 ⁢ N | = | ( K ⁢ N ) 2 | is multiple of | N | for K 2 ⁢ N being the union of cosets of 𝑁, so by the above inequality, there are exactly two possibilities for | K 2 ⁢ N | : either | K 2 ⁢ N | = 2 ⁢ | N | or 3 ⁢ | N | . We distinguish both possibilities. Suppose first that | K 2 ⁢ N | = 2 ⁢ | N | . Then, in particular, | K 2 ⁢ N | = | K ⁢ N | , and this yields the following equalities:

These equalities imply that either x 2 ⁢ N = y ⁢ x ⁢ N and x ⁢ y ⁢ N = y 2 ⁢ N , which clearly leads to x ⁢ N = y ⁢ N , a contradiction; or x 2 ⁢ N = y 2 ⁢ N and x ⁢ y ⁢ N = y ⁢ x ⁢ N . However, in this case, we have

This implies that | K 3 | = 2 ⁢ | N | , contradicting the fact that | K 3 | = 2 ⁢ | K | by equation (3.6). Finally, suppose that | K 2 ⁢ N | = 3 ⁢ | N | . Then equation (3.1) gives

which again contradicts equation (3.6). ∎

We give an easy example of a group satisfying the conditions of Theorem A with | K | = | D | and D ≠ D - 1 . Observe that these are exactly the assumptions that we have at the beginning of the proof of Theorem A. Let

be the semidihedral group of order 16 and Z = ⟨ z ⟩ a cyclic group of order 3. If we take G = H × Z and K = ( a ⁢ z ) G = { a ⁢ z , a 3 ⁢ z } , then we have K 3 = D ∪ D - 1 with D = { a , a 3 } .