Each irreducible representation of $G_1 \times G_2$ is isomorphic to a
representation $\rho^1 \otimes \rho^2$, where $\rho^i$ is an irreducible representation of $G_i$ ($i = 1,2$).
Can we extend this result to reducible representations? That is, can we say, for each representation $\pi: G_1 \times G_2 \to GL(V)$, there exist $\rho^1: G_1 \to GL(V_1)$ and $\rho^2: G_2 \to GL(V_2)$ such that $V$ and $\pi$ are isomorphic to $V_1 \otimes V_2$ and $\rho^1 \otimes \rho^2$, respectively?
If this is not true in general, will be a restriction on $G_1$ and $G_2$ to make the statement true?
Best Answer
I assume we're talking about the semisimple case, e.g. finite groups and complex vector spaces.
Any $(G_1\times G_2)$-rep will be a direct sum of irreps; you just described the irreps. Such reps generally won't be "factorizable" as a tensor product. No condition can change this besides triviality. E.g. if $V_1,V_2$ are irreps and $W_1,W_2$ are trivial reps of $G_1,G_2$ (resp),
$$ V\,=\,(V_1\otimes W_2)\,\oplus\,(W_1\otimes V_2) $$
is not equivalent to $U_1\otimes U_2$ for any $G_1$-rep $U_1$ and $G_2$-rep $U_2$.
Proof by contradiction. You could write $U_1,U_2$ themselves as direct sums, and then distribute the tensor product $U_1\otimes U_2$ over the direct sums - counting irreps we see either $U_1$ has two irreps and $U_2$ has one or vice versa, wlog assume the former $U_1=U_1'\oplus U_2''$, and so
$$ V \,\cong\,(U_1'\otimes U_2)\,\oplus\,(U_1''\otimes U_2), $$
which (because multiplicity of irreps is unique) means $U_2\cong W_2\cong V_2$, contra hypothesis.
Here's an advanced perspective.
The $G$-reps, up to isomorphism, equipped with direct sum and tensor, form a semiring $(\mathrm{Rep}(G),\oplus,\otimes)$, like the natural numbers $(\mathbb{N},+,\times)$. If we allow formal differences ("virtual" reps) we get the Burnside ring $\mathrm{Burn}(G)$ (just like how we construct $\mathbb{Z}$ out of $\mathbb{N}$). The irreps form a $\mathbb{Z}$-basis for $\mathrm{Burn}(G)$ as an additive group. The result you cite implies
$$\mathrm{Burn}(G_1\times G_2)\cong\mathrm{Burn}(G_1)\otimes\mathrm{Burn}(G_2) $$
as rings. You (kind of) just asked if every element of the latter is a "simple" tensor.