SOME PROBLEMS ABOUT PRODUCTS OF CONJUGACY CLASSES IN FINITE GROUPS

. We summarize several results about non-simplicity, solvability and normal structure of ﬁnite groups related to the number of conjugacy classes appearing in the product or the power of conjugacy classes. We also collect some problems that have only been partially solved.

trivial class and another class.Finally, in Section 5, we will outline the advances on the case in which the power of a class is a union of two classes being one of them the inverse of the other.Throughout this survey, we will approach to some characterizations in terms of irreducible characters.

Products of conjugacy classes: Arad and Herzog's conjecture
In [2], Z. Arad and M. Herzog conjectured one of the most significant problems concerning the product of two conjugacy classes.Problem 2.1 (Arad and Herzog's conjecture).In a non-abelian simple group the product of two conjugacy classes is not a conjugacy class.
This was first checked for several families of simple groups such as for alternating groups, Suzuki groups, PSL 2 (q), non-abelian simple groups of order less than one million, and 15 of the 26 sporadic simple groups (see [2]).Since then, several authors have shown the conjecture to hold for certain groups.The following characterization in terms of irreducible characters can be used to check the conjecture, for example, for all the sporadic simple groups and for each group whose character table is known (for instance by using the Atlas [1] or GAP [9]).
We will denote by x G the conjugacy class of each element x ∈ G. Theorem 2.2.[13,Preliminaries] Let G be a group and let a, b, c ∈ G be non-trivial elements of G.
The following conditions are equivalent: 1) for all χ ∈ Irr(G).By using Theorem 2.2, J. Moori and H. P. Tong-Viet showed in [13] that the conjecture is true for the following families of simple groups: PSL 3 (q), PSU 3 (q), where q is a prime power, 2 G 2 (q) with q = 3 2m+1 , PSp 4 (q) where q is a prime-power, PSp 2n (3) with n ≥ 2, and PSU n (2), with (n, 3) = 1 and n ≥ 4.More recently, R. Guralnick, G. Malle and P.H. Tiep validated Arad and Herzog's conjecture in several more cases [12].They also proved that if G is a finite simple group of Lie type and A and B are non-trivial conjugacy classes, either both semisimple or both unipotent, then AB is not a conjugacy class.In addition, they obtained a strong version of Arad and Herzog's conjecture for simple algebraic groups, and in particular, they show that almost always the product of two conjugacy classes in a simple algebraic group consists of infinitely many conjugacy classes.A weaker variation of Arad and Herzog's conjecture is posed in [7].Let K 1 , . . ., K n be conjugacy classes with n ≥ 2 such that the product Since the product of conjugacy classes is G-invariant, for each i there exists a set of Let K be a class.When considering products of classes, a particular case is KK −1 and the simplest situation is KK −1 = 1 ∪ D where D is a class.Under this hypothesis we obtain that the group is not simple by means of the Classification of the Finite Simple Groups (CFSG) and we conjecture that the subgroup ⟨K⟩ is solvable.Unfortunately, we have only been able to get such solvability in some particular cases.
Theorem 2.5.[5, Theorem C] Let K be a conjugacy class of a finite group G and suppose that and ⟨K⟩/⟨D⟩ is cyclic.In addition, (1) If |D| = |K| − 1, then ⟨K⟩ is metabelian.More precisely, ⟨D⟩ is p-elementary abelian for some prime p.
(2) If |D| = |K|, then ⟨K⟩ is solvable with derived length at most 3. (3 We remark that KK −1 = 1 ∪ D forces D to be real, but it does not necessarily imply that K is real too.Moreover, under the assumption of Theorem 2.5, if K is real, then it can be easily proved that D is real too, and hence, KK −1 = K 2 = 1 ∪ D. This case will be addressed in Section 4.
The next natural and simplest case when considering the product of a class and its inverse has the Also in this case it can be checked that the group is not simple.To prove this as well as to obtain the non-simplicity in Theorem 2.5, the following characterization in terms of characters is used.www.SID.ir In particular, if Under the hypotheses of Theorem 2.5 the group G need not be solvable.

Power of class which is a class
In [11], Guralnick and Navarro confirmed Arad and Hergoz's conjecture for the particular case of the square of a conjugacy class.They demonstrated that when the square of a conjugacy class of G is again a class, then G is not a non-abelian simple group.However, the result goes much further.The following assertions are equivalent: (1) K 2 is a conjugacy class of G. (2 (3) χ(x) = 0 or |χ(x)| = χ(1) for all χ ∈ Irr(G), and In this case, [x, G] is solvable, and ⟨K⟩ = ⟨x⟩[x, G] too.Furthermore, if the order of x is a power of a prime p, then [x, G] has normal p-complement.
Observe that K 2 = K can never happen since, by Theorem 3.1, this implies that (3) If x is a p-element for some prime p, then N has normal p-complement.
With regard to Problem 2.3, the assertion is proved to be true for the particular case of the power of a conjugacy class.We find a normal subgroup of a group when there is a conjugacy class satisfying that some of its powers is again a conjugacy class, and obtain an equivalent property in terms of the irreducible characters of the group.This result is in fact a generalization of Theorem 3.1 and its proof is similar.
Notice that under the hypotheses of the above theorem, G is not a non-abelian simple group.This happens because we know that K = xN and if N = 1, then x is central, and if N = G, then K = G, so N is always a non-trivial proper subgroup of G.
The following corollaries are obtained as a consequence of the above theorem.
, where π(t) denotes the set of primes dividing the number t.
Corollary 3.6.[7, Corollary 2.8] Let G be a finite group and let π be a set of primes.Suppose that for each conjugacy class K of π-elements of G there exists n ∈ N such that K n is a conjugacy class.
The authors also obtain the solvability of the subgroup generated by a class when one of its powers is a class by employing similar arguments to those of Guralnick and Navarro, which appeal to the CFSG.Theorem 3.3 joint with Theorem 3.2 are the main ingredients to prove the next theorem.

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The authors make emphasis in the cases where the CFSG is not necessary to obtain the solvability in Theorem 3.7.This occurs for classes of elements of prime order, classes of 2-elements and real classes.
Theorem 3.8.[7,Theorem 2.11] Let K be a conjugacy class of an element x ∈ G. Suppose that there exists n ∈ N with n ≥ 2 satisfying that K n is a conjugacy class.Then: (1) If o(x) is a prime p, then ⟨K⟩ is solvable.
(3) If K n = D where D is a real conjugacy class, then ⟨K⟩ is solvable.Also, D 3 = D and D is a class of a 2-element.
In the particular case of Theorem 3.8(3), if in addition n = 2 a for some a ∈ N, then more information can be given.Observe that if a real conjugacy class K satisfies that there exists n ∈ N such that K n = D where D is a conjugacy class, then D is also a real class.However, if a class K satisfies that there exists n ∈ N such that K n = D with D real, then K need not be real.A trivial example of this situation occurs in Z 4 .

Power of a class which is a union of the trivial class and another class
As we have explained in Section 2, if K and D are conjugacy classes of a group G such that KK −1 = 1 ∪ D, then G is not simple (see [6] for more details on this problem).A natural question is what happens if the power of a class is union of the trivial class and another one, so we establish the following problem.In particular, for all χ ∈ Irr(G).).It can be checked by using the character tables (included in GAP) that for any of these groups, any two non-trivial conjugacy classes of it and n < 6, there exists an irreducible character that does not satisfy the equation of Theorem 5.6.

Problem 2 . 3 .
[7, Conjecture 1.1]In a non-abelian simple group the product of n conjugacy classes, with n a fixed natural number and n ≥ 2, is not a conjugacy class.

Theorem 3 . 1 .
[11, Theorem 2.2]  Let G be a finite group and K = x G the conjugacy class of x in G.

Corollary 3 . 4 . [ 7 , 1 . 3 . 5 . [ 7 ,
Corollary 2.4] Let K = x G with x ∈ G such that K n is a conjugacy class for some n ∈ N with n ≥ 2. Then |K r | = |K| for all r ∈ N and K o(x)+1 = K and K o(x)−1 = K −1 .Moreover, K m is a conjugacy class for all m ∈ N such that (m, o(x)) = Corollary Corollary 2.5 and Theorem 2.12]

Proposition 3 . 9 .
(See [7, Theorem 2.12]) Let G be a finite group and let K be a conjugacy class of an element x ∈ G.If K n = D where D is a real conjugacy class with n = 2 a for some a ∈ N, then |K| is odd and o(x) = 2 a+1 .

Problem 4 . 1 .Theorem 5 . 6 .
Let G be a finite group and let K be a conjugacy class.If K n = 1 ∪ D for some n ∈ N and n ≥ 2, and some conjugacy class D, then ⟨K⟩ is solvable.In particular, G is not a non-abelian simple group.It can be easily proved that if K n = 1 ∪ D for some n ∈ N, then KK −1 = 1 ∪ D, so by Theorem 2.5 the group is not simple.However, the converse is not true.Let G = SL(2, 3) and let K be one of the two conjugacy classes of elements of order 6, which satisfies |K| = 4.It follows thatKK −1 = 1 ∪ Dwhere D is the unique conjugacy class of size 6.However, there is no n ∈ N with K n = 1 ∪ D. Problem 4.1 is proved in[4] for the particular case n = 2 without using the CFSG, and the structure of ⟨K⟩ and ⟨D⟩ are completely determined.Observe that if K n = 1 ∪ D for some n ∈ N and K is real,then K 2 = 1 ∪ D.DOI: http://dx.doi.org/10.22108/ijgt.2019.111448.1480www.SID.irLet G be a finite group and let K be a conjugacy class of an element x ∈ G. Then K n = D ∪ D −1 where D is a conjugacy class if and only if there exist positive integers m 1 and m 2 such that χ(x) n |K| n = χ(1) n−1 |D|(m 1 χ(x n ) + m 2 χ(x −n )) for all χ ∈ Irr(G) and |K| n = (m 1 + m 2 )|D| where m 1 = |K| n |G| ∑ χ∈Irr(G) χ(x) n χ(x n ) χ n−1 (1) and m 2 = |K| n |G| ∑ χ∈Irr(G) χ(x) n χ(x n ) χ n−1 (1) .

Remark 5 . 7 .
Recall that the smallest integer m satisfying C m = G for each non-trivial conjugacy class C of G is called the covering number of G.The covering number of each sporadic simple group is at most 6 ([14] and[2]

1480 www.SID.ir conjecture
we can write DOI: http://dx.doi.org/10.22108/ijgt.2019.111448.wasproved, then we would automatically have Problem 2.3 solved.Let K 1 , . . ., K n be conjugacy classes of G and write The typical non-solvable situation is a group of type Z.S.2, where |Z| = 3, Z is in the center of Z.S, and in addition, S is a non-solvable group acted by an automorphism of order 2 such that the non-trivial elements of Z are conjugate by this automorphism.