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 $V$?



decomposition of representations of a product group
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. 


