If I know how to decompose a vector space in irreducible representations of two groups, can I understand the decomposition as a rep of their product? Let $V$ be a (complex, finite-dimensional) vector space. Suppose that two (finite) groups $A$ and $B$ act on $V$. Furthermore, suppose that these actions commute, so that the direct product $A \times B$ also acts on $V$.
As a representation of $A$, we can decompose $V$ as the direct sum $$ V = \bigoplus_i V_i ,$$ where each $V_i$ is an irreducible representation. And we can do the same by considering $V$ as a representation of $B$, say
$$V = \bigoplus_j V_j' .$$
Also, it is known that every irreducible representation of a direct product of groups $A \times B$ is of the form $V_i \otimes V_j'$, where $V_i$ is irreducible for $A$ and $V_j'$ is irreducible for $B$.
My question is : If I know the two decompositions of $V$ as both a representation of $A$ and of $B$, can I understand its decomposition $$ V = \bigoplus_{i,j} V_i \otimes V_j' $$ into irreducible representations of $A \times B$ in a simple way, or is it a hard problem in general?
 A: It depends on how well you understand the action. The other answers say "no," but I could just as well say "yes."
If $\lambda$ is an irreducible representation of $A$ and $V_\lambda$ is the $\lambda$-isotypic summand of $V$, that is, the sum of all subrepresentations isomorphic to $\lambda$; and if $\mu$ is a an irreducible representation of $B$ and $V_\mu$ is the $\mu$-isotypic summand of $V$, then the $\lambda\otimes \mu$-isotypic summand of $V$ is $V_\lambda\cap V_\mu$.
Of course, there is only one answer. This is exactly saying that you need to understand how $B$ acts on $V_\lambda$. But since it lies in $V$, that is understanding how $B$ acts on $V$ and how $V_\lambda$ lies in $V$. If you understand $V_\lambda$ and $V_\mu$ as concrete subspaces of $V$, rather than multiplicities, then you do. Specifically, you need to understand them enough to compute their intersection.
A: You have to understand how $B$ acts on
$\operatorname{Hom}_A(V_i, V)$ for an irreducible representation $V_i$ of $A$. This depends on your context.
If you can do that then the evaluation map induces an isomorphism
$$\bigoplus_i V_i \otimes \operatorname{Hom}_A(V_i, V) \to V.$$
A: To be totally clear: no, the decomposition as a representation of $A$ and the decomposition as a representation of $B$ separately don't determine the decomposition as a representation of $A \times B$, because this is not enough information by itself to determine which irreducibles of $A$ pair with which irreducibles of $B$ in general.
The smallest counterexample is $A = B = C_2$ acting on a $2$-dimensional vector space $V$ such that, as a representation of either $A$ or $B$, $V$ decomposes as a direct sum of the trivial representation $1$ and the sign representation $-1$. This means that $V$ could be either $1 \otimes 1 + (-1) \otimes (-1)$ or $1 \otimes (-1) + (-1) \otimes 1$ (the $+$ here is a direct sum but I find writing direct sums and tensor products together annoying to read) and you can't tell which. You can construct a similar counterexample out of any pair of groups $A, B$ which both have non-isomorphic irreducibles of the same dimension.
What you can do instead is the following. If you understand the action of $A$, then you get a canonical decomposition of $V$ as a direct sum
$$V \cong \bigoplus_i V_i \otimes \text{Hom}_A(V_i, V)$$
where $V_i$ are the irreps of $A$ and $\text{Hom}_A(V_i, V)$ is the multiplicity space of $V_i$. If instead of considering the multiplicity space we just write a direct sum of a bunch of copies of $V_i$ that decomposition is not canonical and not unique. The point of doing this canonically is that if the action of $B$ commutes with the action of $A$ then the action of $B$ naturally descends to a family of actions on each multiplicity space $\text{Hom}_A(V_i, V)$. Once you decompose each of these actions into irreps of $B$ then you've understood the entire action of $A \times B$; this is how you prove that result you cite about direct products. Of course you could also start with $B$ instead.
