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 ?

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.
