$\DeclareMathOperator{\Aut}{Aut}$
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 ?
Best Answer
$\DeclareMathOperator{\End}{End}$ 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.