TITLE: Irreducible representations of direct product of compact groups
QUESTION [7 upvotes]: 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!

REPLY [3 votes]: 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.