# Analogy between product of conjugacy classes and irreps: is there analog of Thompson conjecture ?

The Thompson conjecture: in a finite simple non-abelian group, there exists a conjugacy class such that every element of the group can be expressed as a product of two elements from that conjugacy class.

The conjecture is still open despite much progress (see below). There are plenty analogies between conjugacy classes and irreducible representations of finite simple groups (see below), so it is natural to ask what is known on the following questions:

Question: is it true that in any finite simple group there exists an irreducible representation $\chi$ such that $\chi\otimes \chi$ contains all irreducible representations ? (It might be one should consider $\chi \otimes \bar \chi$.) What about modular representations ? If is not true - how strongly is it violated ?

Background 1: Azad, Fisman 1987 An analogy between products of two conjugacy classes and products of two irreducible characters in finite groups mentions the following nice results:

• A finite group B is isomorphic to the first Janko group J, if and only if C*C = G for EVERY nontrivial conjugacy class C of G. The analogous theorem in terms of characters is that a finite group G is isomorphic to J, if and only if $Irr(\chi^2) = Irr(G)$ for EVERY nontrivial irreducible character of G, where $Irr(\chi^2)$ is the set of all the irreducible constituents of $\chi^2$ [ACH].

• If $C$ and $D$ are non-trivial conjugacy classes of a finite group G such that either $CD = mC + nD$ or $CD = mC^{-1} + nD$ where m,n are non-negative integers, then G is NOT a non-abelian simple group. And similar result on irreps.

The paper contains references for further analogies between products of conjugacy classes and irreps.

Background 2. There is lots of research devoted to Thompson conjecture. It is strictly stronger than Ore conjecture (see MO-Humphreys for history), which says that every element in finite simple groups is commutator. Now it is proved.

Significant progress made by Gordeev, Ellers 1998 On the Conjectures of J. Thompson and O. Ore.

Ore conjecture finally proved in [LOST]: M. W. Liebeck, E. A. O’Brien, A. Shalev, P. H. Tiep – The Ore conjecture, J. Eur. Math. Soc. 12 (2010), 939–1008.

Bourbaki Seminar Mars 2013 by G. Malle gives overview of that works.

See also MO, MO, products of conjugacy classes are related to quantum cohomologies of curves - see MO.

• For finite simple groups of Lie type it is in fact the case that every (ordinary) irreducible character occurs in the square of the Steinberg character, except for some special unitary groups. This was proved by Heide, Saxl, Tiep, and Zalesski - arxiv.org/abs/1209.1768. Aug 28, 2017 at 17:31
• Actually the answer to your question is no for those special unitary groups. Aug 28, 2017 at 17:33
• @JayTaylor Thank you very much ! I would suggest convert comments to an answer, may adding some details Aug 30, 2017 at 8:33
• @Jay Taylor The answer to Alexander's question is indeed "no", but this is not due to this property of the Setinberg representation proved in arxiv.org/abs/1209.1768. This is due to the more precise fact proved in loc. cit.: for finite simple groups G, there exists an irrep $\pi$ which does not occur in the square of the regular representation by conjugation. So Heide, Saxi, Tiep and Zalesski prove the in fact stronger fact. It is false that for all $G$ and all irrep $\pi$, there exists an irrep $\chi$ such that $\pi$ occurs in $\chi\otimes {\bar \chi}$. Aug 30, 2017 at 9:28
• @PaulBroussous Thanks for clarifying. My comment was a reaction to reading this statement at the end of their article. I'll develop my comments into a proper answer. Aug 30, 2017 at 14:05

In the following article

Heide, Gerhard; Saxl, Jan; Tiep, Pham Huu; Zalesski, Alexandre E. Conjugacy action, induced representations and the Steinberg square for simple groups of Lie type. Proc. Lond. Math. Soc. (3) 106 (2013), no. 4, 908–930.

Heide, Saxl, Tiep, and Zalesski show that if $G$ is a finite simple group of Lie type then every irreducible character is a constituent of $\mathrm{St}^2$ unless $G = \mathrm{SU}_n(q)$ where $n$ is coprime to $2(q+1)$, see Theorem 1.2. Here $\mathrm{St}$ denotes the Steinberg character of $G$.

In this exceptional case, namely $G = \mathrm{SU}_n(q)$ with $n$ coprime to $2(q+1)$, they show that there exists no irreducible character $\chi$ of $G$ such that either $\chi^2$ or $\chi\overline{\chi}$ contains all irreducible characters of $G$, see Lemma 5.3.

It's possibly worthwhile noting that the corresponding statement for the symmetric group is an open problem. If $n$ is a triangular number then Saxl has given a specific character of $\mathfrak{S}_n$ whose square conjecturally contains all irreducible characters; it's labelled by a staircase partition.

Remark: One can find a preprint version of the above article here.

As Taylor's answer shows, the simple group PSU(3,3) is a counterexample. Interestingly, the conjecture fails rather dramatically for this group since the irreducible character of degree 6 is not a constituent of $\chi\overline\chi$ for ANY irreducible character $\chi$ of $G$. This is a Magma computer observation, and I have no idea if a similar phenomenon occurs more generally.