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Let $G$ be a finite group and $M$ a finitely generated $\mathbb{Z}G$-module. Then $M \otimes k$ is a representation of $G$ for any field $k$. I am interested in the number of ways we can turn $M \otimes k$ into a (non associative) $kG$-algebra. Equivalently, I would like to compute $\operatorname{Hom}_{kG}(M\otimes M \otimes k, M \otimes k)$.

Suppose that we know that $M\otimes \mathbb{C}$ is irreducible and that $M \otimes M \otimes \mathbb{C}$ contains a unique composition factor isomorphic to $M \otimes \mathbb{C}$. Then we know, by Schur's lemma, that $\operatorname{Hom}_{\mathbb{C}G}(M\otimes M \otimes \mathbb{C}, M \otimes \mathbb{C}) \cong \mathbb{C}$. What can we say about $\operatorname{Hom}_{kG}(M\otimes M \otimes k, M \otimes k)$?

I have the feeling that $\operatorname{Hom}_{kG}(M\otimes M \otimes k, M \otimes k) \cong k$ if $k$ is algebraically closed and $\operatorname{char}(k) \nmid \left| G \right|$. Is this correct and if so, why? In the other cases, I think that a good knowledge of modular representation theory might give a sufficient answer but I am not sure where I should start looking in the literature.

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  • $\begingroup$ Is it correct that $M \otimes k$ means $M \otimes_{\mathbb Z} k$, but $M \otimes M \otimes k$ means $M \otimes_{\mathbb ZG} M \otimes_{\mathbb Z} k$? $\endgroup$
    – LSpice
    Commented Mar 28, 2018 at 13:53
  • $\begingroup$ @LSpice Yes, that is correct. By $M \otimes M \otimes k$, I really mean the tensor product of the representation $M \otimes_\mathbb{Z} k$ with itself. It could also have written $M \otimes_{\mathbb{Z}G} M \otimes_{\mathbb{Z}G} kG$. $\endgroup$
    – Michiel Van Couwenberghe
    Commented Mar 28, 2018 at 14:34
  • $\begingroup$ Do you want to assume $M$ is $\mathbb{Z}$-free? Otherwise it could have a large summand annihilated by $\text{char}(k)$ and $M\otimes k$ would have little to do with $M\otimes\mathbb{C}$. $\endgroup$
    – Jeremy Rickard
    Commented Mar 28, 2018 at 14:36
  • $\begingroup$ @JeremyRickard I am not really sure. Probably, if that is the only way that I can say something useful. How can we check whether $M$ is $\mathbb{Z}$-free? Or can you always find a $\mathbb{Z}$-free $N$ such that $N \otimes \mathbb{C} \cong M \otimes \mathbb{C}$? $\endgroup$
    – Michiel Van Couwenberghe
    Commented Mar 28, 2018 at 14:49
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    $\begingroup$ @MichielVanCouwenberghe, you can take $N$ to be the quotient of $M$ by its torsion submodule. $\endgroup$
    – LSpice
    Commented Mar 28, 2018 at 15:02

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