# Composition factors of tensor products of modular representations

In ordinary representation theory over $\mathbb{C}$, all the irreducible modules of a finite group $G$ appear as composition factors of the tensor products $X \otimes \cdots \otimes X$ of a faithful $\mathbb{C}$-representation $X$. This is e.g. Thm. 10.8 in Ch.V of Huppert's Book "Finite Groups I".

This should also be true in positive characteristic, so let $X:G \to \mathrm{GL}(n,K)$ with $\mathrm{char}(K) = p$ be a modular faithful representation. Then all irreducible $KG$-modules appear as composition factors of the tensor products $X \otimes \cdots \otimes X$.

I'm not sure, how to prove this, but I think it shouldn't be difficult.

## 2 Answers

This is true. I believe it may be due to L.G. Kovacs. When $K$ is algebraically closed, the proof is much the same as the complex case, except that one works with the Brauer character $\phi$ afforded by $X.$ For any Brauer irreducible character $\psi,$ let $\theta$ denote the Brauer character of its projective cover. Then some linear combination of powers of $\phi$ vanishes on all non-identity $p$-regular elements, but not on $1_{G}.$ Hence $\langle \theta, \phi^{n} \rangle \neq 0$ for some $n,$ so that the simple module affording $\psi$ is a composition factor of $X \otimes \ldots \otimes X$ ( $n$ factors) The case when $K$ is a finite field follows from the algebraically closed case.

• While the basic facts were established much earlier, there is a short direct proof by Bryant-Kovacs, Tensor products of representations of finite groups. Bull. London Math. Soc. 4 (1972), 133–135 – Jim Humphreys Oct 5 '13 at 19:52
• I think the proof by Bryant and Kovacs is of a rather stronger results (essentially a free summand of a high enough tensor power). There is an earlier ( I think) proof by Kovacs along the character theoretic lines I outlined- this may have appeared in an article by Peter Neumann titled " A letter from L Kovacs" or something like that. This follows the Brauer Blichfeldt pattern, except using Brauer character, if my memory is accurate. Maybe not the first proo though. Maybe I've mixed it up with a variant of Brauer's permutation lemma on Brauer character tables. – Geoff Robinson Oct 5 '13 at 20:43

Questions along this line have appeared on Math Overflow with some frequency, so it's useful to take a look at some of them and their links, for instance here.

As Geoff points out, it's natural here to start with $K$ being algebraically closed; so irreducible means the same thing as absolutely irreducible. In any case, the contrasting methods of Steinberg and Brauer (note his comments) in their two short notes provide lots of information about what is possible in prime characteristic. There is no need to start from scratch here. Both of those articles are available online here and here.

• Thank you for the references. I was only interested in the case, when $K$ is algebraically closed. – KoopaTroopa Oct 5 '13 at 19:53