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It's an easy task to prove with character theory that if $V_1$ and $V_2$ are irreducible representations of $G_1$ and $G_2$ respectively, then $V_1 \otimes V_2$ is an irreducible representation of $G_1 \times G_2$.

However, I'm having some trouble proving the converse, that is, every irreducible representation of $G_1 \times G_2$ is the tensor product of irreducible reps of $G_1$ and $G_2$. Any suggestions?

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I assume that $G_1,G_2$ are finite groups and you look at finite-dimensional representations over some algebraically closed field?

If $V_i$ are the irreducible representations of $G_1$ and $W_j$ are the ones of $G_2,$ we have $$|G_1| = \sum_i \dim(V_i)^2,~ |G_2| = \sum_j \dim(W_j)^2,$$ hence $$|G_1 \times G_2| = \sum_{i,j} \dim(V_i \boxtimes W_j)^2.$$ This shows that the $V_i \boxtimes W_j$ are the only irreducible representations of $G_1 \times G_2$ (since their squared dimensions already add up to $|G_1 \times G_2|$).

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Let the two groups be $G$ and $H$, and consider an irrep $V$ of $G\times H$.

Let $U$ be an irrep of $G$ which is a direct summand of the restriction $V\mathord\downarrow_G$ of $V$ to the subgroup $G=G\times\{1\}\subseteq G\times H$. The vector space $\hom_G(U,V\mathord\downarrow_G)$ of morphisms of $G$-reps, which is then nonzero, has a natural action of $H$: if $f\in\hom_G(U,V\mathord\downarrow_G)$ and $h\in H$, then we can define $h\cdot f$ putting $(h\cdot g)(u)=(1,h)\cdot f(u)$ for all $u\in U$. Consider now the linear map $$\phi_U:U\otimes\hom_G(U,V\mathord\downarrow_G)\to V$$ such that $\phi_U(u\otimes f)=f(u)$ for all $u\in U$. It is easy to see that $\phi_U$ is a map of $G\times H$-modules, and that it is nonzero. It follows, since $V$ is irreducible, that $\phi_U$ is surjective. Moreover, one can see at once that $$\dim\hom_G(U,V\mathord\downarrow_G)\leq\frac{\dim V}{\dim U},$$ and this implies that $\phi$ is in fact injective.

If $\hom_G(U,V\mathord\downarrow_G)$ were a reducible $H$-rep, then $U\otimes\hom_G(U,V\mathord\downarrow_G)$ would be a reducible $G\times H$-rep, and that is absurd because $V$ is an irrep.

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  • $\begingroup$ Why is it the case that irreducibility of V implies the pairing is surjective? $\endgroup$ – Paul T Oct 1 '19 at 4:54

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