Let $k$ be an algebraically closed field of characteristic 0, let $G$ be any group and $N\unlhd G$ a normal subgroup. Let $U$ be a finite-dimensional and irreducible $kG$-module, such that $U$ is also an irreducible $kN$-module. Moreover, let $V$ be a finite-dimensional irreducible $k(G/N)$-module (so it is also a $kG$-module, on which $N$ acts trivially).

Then the tensor product $U\otimes_k V$ is a finite-dimensional $kG$-module. In fact, by a theorem of Chevalley, it is semi-simple.

Q: Under which conditions is $U\otimes_k V$ an irreducible $kG$-module?

My hope is kinda that it always is, but I know just enough about representation theory to know that I do not know enough to make guesses ;-), and that infinite groups behave quite differently than finite groups.

Thus, if the module is not always irreducible, then I'd like to learn about (a) counterexamples, and (b) conditions on $G$ and/or $N$ that make it true. For example, I know that the answer is affirmative if $G$ is finite (see e.g. Corollary 6.17 in Isaacs book "Character Theory of Finite Groups").

Motivation: I would like to prove a certain module of this kind to be irreducible for a research project -- I think I do have an ad-hoc proof, but it is rather ugly and technical, deeply exploiting the structure of my group, slinging around with concrete bases and vectors, etc. -- and it feels like there should be some more elegant and fundamental approach than my caveman style solution. Unfortunately, I do not know much about representations of infinite groups.

For what it's worth, in my setup, $G$ is infinite but $N$ is actually finite; and $G/N$ is a Coxeter group, so $G$ is finitely presented.

  • $\begingroup$ Can you clarify your remark about the case when $G$ is finite? I don't see immediately how to apply the discussion surrounding Clifford's Theorem in the book by Isaacs. (Note too that it's probably safest here to start with an algebraically closed field of characteristic 0, to postpone questions about absolute irreducibility.) $\endgroup$ Jun 8, 2015 at 14:18
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    $\begingroup$ Maybe you are assuming that $k$ is algebraically closed? Because otherwise, there are counterexamples when $G$ is finite, e.g. $G= C_5 \rtimes C_4$, $N=C_5$, $k= \mathbb{Q}$, $U$ the unique faithful irreducible $\mathbb{Q}G$-module and $V=\mathbb{Q}[i]$. Here $U$ is irreducible, but not absolutely irreducible as $\mathbb{Q}N$-module, and we have $U\otimes_{\mathbb{Q}} V \cong U \oplus U$. Notice that Corollary 6.17 in Isaacs book assumes $k$ is algebraically closed. (So I would agree with @JimHumphreys here.) $\endgroup$ Jun 8, 2015 at 14:24
  • $\begingroup$ Yes, sorry, I meant algebraically closed (in fact, I am mostly interested in $k=\mathbb{C}$). I'll correct my question. $\endgroup$
    – Max Horn
    Jun 8, 2015 at 14:25
  • $\begingroup$ @JimHumphreys: I thought it was a pretty much direct application of Corollary 6.17 on page 85? Perhaps I am making a silly mistake here? $\endgroup$
    – Max Horn
    Jun 8, 2015 at 14:29
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    $\begingroup$ @JimHumphreys: Isaacs deals with characters instead of modules, which is o.k. in characteristic $0$. The module $U$ of the question corresponds to $\chi\in \operatorname{Irr}(G)$ in Isaacs' Cor. 6.17, and the module $V$ of the question to the $\beta\in \operatorname{Irr}(G/N)$ in Isaacs. Then $U\otimes V$ has character $\beta \chi$, and Cor. 6.17 tells us that these are irreducible. So I would agree with Max that this is a direct application of Cor. 6.17. $\endgroup$ Jun 8, 2015 at 14:42

1 Answer 1


$$ \mathrm{End}_{kG}(U \otimes_k V) = \left(\mathrm{End}_k(U \otimes_k V)^N\right)^{G/N} = \mathrm{End}_k(V)^{G/N} = k. $$ The second equality holds because $\mathrm{End}_{kN}(U) = k$ (thanks to the assumption that $k$ is algebraically closed).

EDIT: Using Chevalley's theorem is probably overkill. Let $k$ be an arbitrary commutative field, assume that $U$ is irreducible and that $\mathrm{End}_{kN}(U)=k$. Assume that $V$ is irreducible. Then $U \otimes_k V$ is irreducible: clearly it is a semi-simple $kN$-module, and the set of sub-$kN$-modules $W \subset U \otimes_k V$ is in bijection with sub-$k$-vector spaces of $V$, by $W \mapsto (W \otimes_k U^*)^N \subset \mathrm{End}_{kN}(U) \otimes_k V = V$. Moreover, $W$ is a $kG$-submodule of $U \otimes_k V$ if and only if $(W \otimes_k U^*)^N$ if a sub-$k(G/N)$-module of $V$.

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    $\begingroup$ Here is further justification of the second equality. It is well known that $$ \mathrm{End}_k(U \otimes V) \cong (U \otimes V)^* \otimes (U \otimes V) \cong (U^* \otimes U) \otimes (V^* \otimes V) \cong (\mathrm{End}_k U) \otimes (\mathrm{End}_k V) .$$ Since $N$ acts trivially on $\mathrm{End}_k V$, this implies $$ \bigl( \mathrm{End}_k(U \otimes V) \bigr)^N \cong (\mathrm{End}_k U)^N \otimes (\mathrm{End}_k V) .$$ Since $(\mathrm{End}_kU)^N = k$, this is isomorphic to $\mathrm{End}_k V$. $\endgroup$ Jun 8, 2015 at 16:40
  • $\begingroup$ @user74720 Thanks, perfect -- as I hoped, a beautiful solution :). I'd like to use that, (of course giving credit) -- any real name I can put to you, user74720? (Feel free to contact me by email, if you'd rather not say here) $\endgroup$
    – Max Horn
    Jun 9, 2015 at 6:38
  • $\begingroup$ @DaveWitteMorris Thanks for the elaboration $\endgroup$
    – Max Horn
    Jun 9, 2015 at 6:39
  • $\begingroup$ @MaxHorn I'm absolutely fine with you using that without credit. $\endgroup$
    – user74720
    Jun 9, 2015 at 12:45

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