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

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.

• You mean the $V_{i}$ and $W_{i}$ are irreducibe reps? – Geoff Robinson Jun 11 '13 at 18:40
• I'm not assuming $V_i$ and $W_i$ are irreducible. – Eric O. Korman Jun 11 '13 at 19:44

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.
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.
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 $\mathbb C[x,y]/((x-1)^2, (y-1)^2, (x-1)(y-1))$.