Is the tensor product of irreducible representations of different groups irreducible? $\DeclareMathOperator{\Aut}{Aut}$
Let $G_1$ and $G_2$ be two groups, provided with two irreducible linear representations 
$$R_1 : G_1 \to \Aut(V_1) \text{ and } R_2 : G_2 \to \Aut(V_2),$$
$V_1$ and $V_2$ being two finite-dimensional vector spaces over $\mathbb{C}$.
If $G_1$ and $G_2$ are finite, one can show that the tensor product representation
$$R_{\otimes} = R_1\otimes R_2 : G_1\times G_2 \to \Aut(V_1\otimes V_2),$$
defined for any couple $(g_1,g_2)$ by $R_{\otimes}(g_1,g_2)=R_1(g_1)\otimes R_2(g_2)$ 
is again irreducible, using the Schur orthogonality relations. Indeed one has 
$$\begin{alignat}{2}|\chi_{\otimes}|^2 & = \frac{1}{|G_1\times G_2|} \sum\limits_{(g_1,g_2)} |\chi_\otimes (g_1,g_2)|^2 = \frac{1}{|G_1||G_2|} \sum\limits_{g_1,g_2} |\chi_1(g_1)|^2 |\chi_2(g_2)|^2 \\ & = \left(\frac{1}{|G_1|} \sum\limits_{g_1} |\chi_1(g_1)|^2\right) \left(\frac{1}{|G_2|} \sum\limits_{g_2} |\chi_2(g_2)|^2\right) = 1. \end{alignat}$$
The question is, is the result still true when $G_1$ and $G_2$ are not assumed to be finite and, if so, how can we prove it ?
 A: $\DeclareMathOperator{\End}{End}$
So long as $V_1$ and $V_2$ are still finite dimensional this will still hold. Moreover it holds for finite dimensional simple modules over algebras not just groups, but I'll stick to the group case.
First note that the maps $f:\mathbb{C}G_1 \to \End_\mathbb{C}(V_1)$ and $g: \mathbb{C}G_2 \to \End_\mathbb{C}(V_2)$ are surjective maps of algebras. This is a consequence / simplest case of the Jacobson density theorem.
Now the map you care about is $f\otimes g: \mathbb{C}G_1 \otimes \mathbb{C}G_2 \to \End_\mathbb{C}(V_1 \otimes V_2) \cong \End_\mathbb{C}(V_1)\otimes \End_\mathbb{C}(V_2)$ and a tensor product of two surjective maps into finite dimensional vector spaces is again surjective. So $V_1 \otimes V_2$ is a simple $\mathbb{C}G_1 \otimes \mathbb{C}G_2 \cong \mathbb{C}[G_1\times G_2]$ module, as desired.
A: Assume $G_1$ and $G_2$ are finite groups.
It suffices to check that the only $G_1\times G_2$-homomorphisms $V_1\otimes V_2\to V_1\otimes V_2$ are scalar multiples (the converse of Schur's lemma).
They correspond to $G_1$-homomorphisms $V_1\to\hom_{G_2}(V_2,V_1\otimes V_2)$. However, I claim
$$\begin{align*}\varphi\colon V_1&\to \hom_{G_2}(V_2,V_1\otimes V_2)\\
v_1&\mapsto (v_2\mapsto v_1\otimes v_2)\end{align*}$$is an isomorphism of $G_1$-representations. It is easily checked that $\varphi$ respects the $G_1$-action. To check $\varphi$ is an isomorphism, we count dimensions. If $\dim V_1=n_1$, then $V_1\otimes V_2\cong V_2^{\oplus n_1}$ as $G_2$-representations, so by Schur's lemma
$$\hom_{G_2}(V_2,V_1\otimes V_2)\cong\hom_{G_2}(V_2,V_2^{\oplus n_1})\cong\hom_{G_2}(V_2,V_2)^{\oplus n_2}\cong k^{\oplus n_1}.$$
Thus, both sides have the same dimension $n_1$. $\varphi$ is injective, so it must be an isomorphism (of vector spaces).
We conclude (again by Schur's lemma) that the space of $G_1$-homomorphisms $V_1\to\hom_{G_2}(V_2,V_1\otimes V_2)$ must be $1$-dimensional (consisting of scalar multiples of $\varphi$). Thus, the space of $G_1\times G_2$-homomorphisms $V_1\otimes V_2\to V_1\otimes V_2$ must also be $1$-dimensional, as desired.
