decomposition of representations of a product group Suppose $G_i$ are finite groups for $i=1,2$ and G is the direct product of $G_i$. If V is a finite dimensional irreducible representation of $G$, then it is well known that $V$ is a tensor product of $V_i$,$i=1,2$ and each $V_i$ is an irreducible representation of $G_i$.
The question I have is when $V$ is given, is there a canonical way to construct $V_i$ from $V$?
 A: If you choose a representation $W$ of $G_1$ which is  isomorphic to $V_1$, then you can construct $V_2$ as $\mathrm{Hom}_{G_1}(W, V)$. But I don't think there can be a fully choice-free construction of $(V_1, V_2)$.
Here is a hazy argument; if you give me a rigorous definition of choice-free then I may able to do better. Suppose that I find $V_1$, $V_2$ and an isomorphism $V_1 \boxtimes V_2 \to V$. And suppose that you likewise find $V'_1$, $V'_2$ and $V'_1 \boxtimes V'_2 \to V$. If your construction is canonical, you should be able to give canonical isomorphisms $a_1: V_1 \to V'_1$ and  $a_2: V_2 \to V'_2$, making the obvious diagram commute. 
But  the obvious diagram also commutes when $(a_1, a_2)$ is replaced by $(-a_1, -a_2)$. I don't see how you can possibly single out which of $a_1$ and $- a_1$ is better.
A: I agree with David Speyer's answer, and furthermore there is no canonical way to construct $V_i$ from $V$.  This is a subtle and oft-overlooked point in representation theory, in my opinion.  Many texts prove that an irrep of $G_1 \times G_2$ is isomorphic to a tensor product of an irrep of $G_1$ with an irrep of $G_2$.  The typical slick proof relies on character theory -- kind of a cheat, in my view, since it only says something about isomorphism classes.
Here's a categorical explanation of the theorem:  Let $G_1$ and $G_2$ be finite groups, and let $\pi$ be an irrep of $G_1 \times G_2$ on a complex vector space $V$.
Then, for every pair $(\rho_1, W_1)$, $(\rho_2, W_2)$ of representations of $G_1, G_2$, one gets a complex vector space:
$$H_\pi(\rho_1, \rho_2) := Hom_G(\rho_1 \boxtimes \rho_2, \pi).$$
In fact, this extends to a contravariant functor:
$$H_\pi:  Rep_{G_1} \times Rep_{G_2} \rightarrow Vec.$$
Here we use categories of finite-dimensional complex representations and vector spaces.
This is also functorial in $\pi$, yielding a functor:
$$H:  Rep_G \rightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ],$$
where the right side of the arrow denotes the category of functors (for categories enriched in $Vec$).  What this demonstrates is that the canonical thing is to take representations of $G$ to objects of an appropriate functor category related to $Rep_{G_1}$ and $Rep_{G_2}$.  By Yoneda's lemma (for categories enriched in $Vec$), there is an embedding of categories:
$$Rep_{G_1} \times Rep_{G_2} \hookrightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ].$$
It turns out -- and this is where some finiteness is important, and a proof necessarily uses some counting, character theory, or the like -- that for any irrep $\pi$ of $Rep_G$, the functor $H_\pi \in [Rep_{G_1} \times Rep_{G_2}, Vec]$ is representable.  It is not uniquely representable, but it is uniquely representable up to natural isomorphism.
Practically, what this means is that given an irrep $\pi$ of $G$, there exists an isomorphism $\iota: \pi \rightarrow \pi_1 \boxtimes \pi_2$ for some irreps $\pi_1, \pi_2$ of $G_1, G_2$, respectively.  The pair $(\pi_1, \pi_2)$ is not unique, but the triple $(\iota, \pi_1, \pi_2)$ is unique up to unique isomorphism.  This is usually good enough. 
