Are all representations of $G\times H$ induced from representations of $G$ and $H$?

Question

This is a crosspost from math.SE. Suppose $G$ and $H$ are discrete groups. Is it always the case that any finite dimensional complex representation of $G\times H$ is of the form $$ \bigoplus_i V_i \otimes W_i, $$ where $V_i, W_i$ are reps of $G$ and $H$, respectively?

I know this is true when $G$ and $H$ are finite and when the representation of $G\times H$ is completely reducible, but is there a simple counterexample to the general case?

I'm also curious if it is ``usually true," in some sense, that any rep of $G\times H$ has the above form.

Answer 1

Let $G = H = \mathbb{Z}$. Now a $G \times H$ representation is a pair of commuting invertible matrices. Let's try $$\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)\mbox{ and } \left( \begin{array}{cc} -1 & 1 \\ 0 & -1 \\ \end{array} \right).$$ Certainly this rep is indecomposable since it already is after restricting to either group. Why can't it be a tensor product of two representations of $\mathbb{Z}$? One representation would have to be one-dimensional, given by a scalar multiplication. But this is impossible since neither matrix is a scalar.

I think it is true that there is an open set on the representation variety where things decompose into a sum of tensor products as you say but I do not see why exactly.

Answer 2

This is not true, in general. For example, take $G = H = \mathbb Z$. Let $G \times H$ act on $\mathbb C^3$ in such a way that a generator of $G$ carries $e_1$ to $e_2$, and $e_2$ and $e_3$ to $0$, while a generator of $H$ carries $e_1$ to $e_3$, and $e_2$ and $e_3$ to $0$. It is easy to see that this representation is indecomposable, and that it is not of the form above.

[Edit:] as was, correctly, pointed out to me, my maps are not invertible; but you can make them invertible by adding the identity to both.

I had in mind the following. A representation of $G \times H$ corresponds to a $\mathbb C[x^{\pm 1}, y^{\pm 1}]$-module. The representation above corresponds to $\mathbb C[x,y]/((x-1)^2, (y-1)^2, (x-1)(y-1))$.