Is there any method of producing representations of the direct product of two groups $G\times H$ which are not trivial, i.e. not the direct sum of a representation of $G$ and a representation of $H$? I guess that this will depend on the groups, but I am interested in any general comments or methods that can be made.

The motivation is studying group algebra $\mathbb{K}G-\mathbb{K}H$ bimodules, where the first problem is finding interesting examples.

EDIT: On looking at the question `Irreducible representations of a semidirect product', I see that the tensor product of irreps of $G$ and $H$ will do. (I should have noticed that!) That question suggests that these are all irreps of the direct product, and that all the irreps of the direct product are obtained in this manner - is there an obvious reason why this is true? (I hope I read the other question correctly, the last comment might only have referred to a particular example).

PS I am obviously not an expert in representation theory!!

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    $\begingroup$ Your terminology is unusual: direct product, or direct sum, or simply product are common. $\endgroup$ – P Vanchinathan Oct 24 '16 at 13:26
  • $\begingroup$ OK, I will change that - thx. $\endgroup$ – Edwin Beggs Oct 24 '16 at 13:29
  • $\begingroup$ Consider the following construction: $\phi: G\to GL(V), \psi: H\to GL(V)$, $\rho(gh)= \phi(g)\psi(h)$, $(g,h)\in G\times H$. Then $\rho$ is typically not a direct sum of two representations. $\endgroup$ – Moishe Kohan Oct 24 '16 at 13:52
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    $\begingroup$ The problem is that $\rho(gh)\neq \rho(hg)$ in general, but $gh=hg$ in $G\times H$. Essentially the whole problem is how to get the actions of $G$ and $H$ to commute. $\endgroup$ – Edwin Beggs Oct 24 '16 at 14:15
  • $\begingroup$ Are you assuming $G,H$ are finite, $\Bbb K$ is algebraically closed and $|G|,|H|$ are invertible in $\Bbb K$? (So that we have Maschke's theorem and Schur's lemma.) $\endgroup$ – arctic tern Oct 24 '16 at 15:24

I'll assume $G,H$ are finite groups and the base field is $\mathbb{C}$. The Schur orthogonality relations give a way to determine if a representation is irreducible: compute the norm of its character and see if it equals one. We can use this to prove

Claim. Suppose $V$ is a $G$-rep and $W$ an $H$-rep. Then $V\otimes W$ is a $G\times H$-irrep if and only if ($V$ is a $G$-irrep and $W$ is an $H$-irrep).

Proof. It should be clear that if either of $V$ or $W$ is reducible then $V\otimes W$ will be a reducible representation of $G\times H$. Conversely, if $V$ and $W$ are irreps we may compute

$$|\chi_{V\otimes W}|^2=\frac{1}{|G\times H|}\sum_{(g,h)} |\chi_{V\otimes W}(g,h)|^2=\frac{1}{|G|}\left(\sum_g|\chi_V(g)|^2\right)\frac{1}{|H|}\left(\sum_h|\chi_W(h)|^2\right)=1.$$

(Of course you may augment this for fields besides $\mathbb{C}$ as long as Schur's still applies.)

In fact, this classifies all irreducible representations of $G\times H$:

Claim. If $U$ is a $G\times H$-irrep then $U\cong V\otimes W$ for some $G$-irrep $V$ and $H$-irrep $W$.

Idea. Pick $u\in U$ so that $V=\mathrm{span}\,Gu$ is a $G$-irrep of $U$, then let $W=\mathrm{span}\,Hu$. There is a well-defined map $V\otimes W\to U$ given by $gu\otimes hu\mapsto (g,h)u$. It is clearly $G\times H$-equivariant, and surjective since $U$ is a $G\times H$-irrep (so every nonzero $u\in U$ is a cyclic generator as a simple module over the group algebra). Not sure of a good way to check injectivity.

Since irreps of $G$ and $H$ already give irreps of $G\times H$, it suffices to check that the map

$$\mathrm{Irr}(G)\times\mathrm{Irr}(H)\to\mathrm{Irr}(G\times H) $$

is a bijection. Injectivity can be determined from Schur's orthogonality relations again. To see surjectivity, we can show both sets have the same number of elements. From the Artin-Wedderburn decomposition we know the number of irreps of $G\times H$ equals the dimension of the center of its group algebra, but then notice

$$\begin{array}{ll} \dim Z(\mathbb{C}[G\times H]) & =\dim Z(\mathbb{C}[G]\otimes\mathbb{C}[H]) \\ & =\dim \big(Z(\mathbb{C}[G])\otimes Z(\mathbb{C}[H])\big) \\ & =\big(\dim Z(\mathbb{C}[G]\big)\cdot\big(\dim Z(\mathbb{C}[H])\big). \end{array}$$


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