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.

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.) $\endgroup$ – Frieder Ladisch Jun 8 '15 at 14:24