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.
Best Answer
$$ \mathrm{End}_{kG}(U \otimes_k V) = \left(\mathrm{End}_k(U \otimes_k V)^N\right)^{G/N} = \mathrm{End}_k(V)^{G/N} = k. $$ 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 $V$.