Irreducible representations of direct product of compact groups I am trying to prove that for $G$ and $H$ two compact groups, it's true that any irreducible $G \times H$-representation $U$ is isomorphic to a tensor product of an irreducible $G$-representation $V$ and an irreducible $H$-representation $W$, ie. that $U \simeq V \otimes W$. 
After thinking a lot about this, I found the following argument in some online lecture notes (Lemma 22.6 in https://math.berkeley.edu/~teleman/math/RepThry.pdf): 
"From the properties of the tensor product of matrices, it follows
that the character of $V \otimes W$ at the element $g \times h$ is 
$\chi_{V \otimes W} (g \times h) = \chi_V (g) \cdot \chi_W (h)$. Now, a conjugacy class in $G \times H$ is a Cartesian product of conjugacy classes in $G$ and $H$, and character theory ensures that the $\chi_V$ and $\chi_W$ form a Hilbert space basis of the $L^2$ class functions on the two groups. It follows that the $\chi_V (g) \cdot \chi_W (h)$ form a Hilbert space basis of the class functions on $G\times H$, so this is a complete list of irreducible characters. $\blacksquare$"
I understand the claim about the conjugacy classes of the direct product and everything else, except how the conclusion that $\chi_V \cdot \chi_W$ are a full set of irreducible characters for the product group follows from this. 
I just don't get why a class function $f : G \times H \rightarrow \mathbb{C}$ can be written as a linear combination of the various $\chi_V \cdot \chi_W$, as $V$ and $W$ vary through all irreps of $G$ and $H$. I tried to convince myself of this by looking at, say,
\begin{align*}
f_G : G \times \lbrace e \rbrace &\longrightarrow \mathbb{C}, \text{ and}\\
f_H : \lbrace e \rbrace \times H &\longrightarrow \mathbb{C},
\end{align*}
and use that since $f_G$ and $f_H$ are class functions, they can each be written as a linear combination of $\lbrace \chi_V \rbrace_V$ and $\lbrace \chi_W \rbrace_W$ respectively. But I can't seem to connect this fact to the conclusion I'm trying to reach. 
Any tips on why the specific argument works or how the result might be proved in a different way are most welcome! 
 A: Take $U$ a finite dimensional representation of $G\times H$. Consider only the $G$ action. Then every element from $H$ is an intertwining operator. Decompose $U$ into $G$-isotypic components. Then $H$ invariates each of them. Note that for every irreducible representation of $G$ the intertwining operators are scalers. Therefore,  each $G$ isotypic of type $\rho$ of $U$ can be written as $\rho\otimes \eta$, where $\eta$ is a representation of $H$. Therefore, each finite dimensional (continuous) representation of $G\times H$ can be written as a direct sum of $\rho_1\otimes \rho_2$, where $\rho_1$ is an irrep of $G$ and $\rho_2$ is an irrep of $H$. From here it is easy also to show that $\rho_1\otimes \rho_2$ are irreducible. 
This argument avoids most of the analysis. It only uses Schur's lemma on intertwining operators. One needs to argue that $\eta$ above is continuous, if we started with a continuous representation of $G\times H$, which is straightforward, looking at matrix coefficients. 
A: The basic idea is that $L^2(G)\otimes L^2(H) \cong L^2(G\times H)$ by the canonical isomorphism $A\otimes B \mapsto A(g)B(h)$. Then the statement follows.
A: This result requires (in general) that the underlying field is algebraically closed. The problem is that the intertwining operators are not (in general) scalars. 
A simple counterexample over the field $\mathbb{R}$ is to consider the natural 2-dimensional irreducible representation of the cyclic group $C_{12}$. However, one can write $C_{12}\simeq C_3\times C_4$, but it's easy to see that this representation is not (isomorphic to) the tensor product of irreducible reps of $C_3$ and $C_4$. 
Presumably (I've not checked the details) the OP's statement is true if either of the groups $G$ or $H$ has the property that every irreducible rep of that group is absolutely irreducible (over the field in question).  For example, if $G=S_n$, the symmetric group. Would anyone like to confirm?
