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$\DeclareMathOperator{\Aut}{Aut}$

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 ?

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$\DeclareMathOperator{\End}{End}$ 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 $g: \mathbb{C}G_1 \to \End_\mathbb{C}(V_1)$ 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.

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  • $\begingroup$ Thanks for your answer. Unfortunately in the course I've taken we haven't dealt with group algebras (only Lie algebras of Lie groups) nor simple modules. (In fact the main audience of the course is 2nd year physics students who have never even heard of the concept of module). Is there a way to prove it using another argument ? $\endgroup$ – Daniel Zimmer Jun 20 '17 at 20:44
  • $\begingroup$ Well the group algebra is just formal linear combinations of group elements, so I suppose you could just go through everything I did in terms of group elements. Of course that's probably not a satisfying answer, as it's still essentially using the group algebra. Beyond that, I think the group algebra is genuinely unavoidable here. There is no good analog of character theory for arbitrary infinite groups, so ultimately you need to use linear algebra and as soon as you do any linear algebra you are working with the group algebra. Of course if someone else has a way I'd love to see it. $\endgroup$ – Nate Jun 20 '17 at 21:55

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