$\DeclareMathOperator{\Aut}{Aut}$
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 ?