# subrepresentations of the product of nonisomorphic irreducible representations

Let G be a finite group, and let V and W be two irreducible representations of G over a field k.

How do I show that if V and W are not isomorphic, then the G-subrepresentations of the product V x W are precisely {0}, V x {0}, {0} x W, and V x W itself?

I'm trying to apply Schur's lemma.

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Let a $G$-submodule of $V \times W$ be $M$. Consider the projection of $V \times W$ to $V$ and $W$ respectively. Then it's easy to see that the case where the image is $0$ and $0$ corresponds to $M = 0$, $0$ and $W$ corresponds to $0 \times W$, $V$ and $0$ corresponds to $V \times 0$.
The last case is trickier. Suppose that $M$ projects onto $V$ and $W$. Consider $V \times 0 \cap M$. It's either $V \times 0$ or $0 \times 0$, since $V$ is irreducible. If $M$ contains both $V \times 0$ and $0 \times W$, we are done since they generate $V \times W$. Otherwise WLOG assume that $V \times 0 \cap M = 0$, and we hope to derive a contradiction.
For any $(v_1,w), (v_2,w) \in M$, we must have $v_1 = v_2$ since otherwise their difference gives us a nontrivial element of $V \times 0$. Each $w \in W$ must show up as the second coordinate of some element in $M$ since $M$ projects onto $W$. In fact there is a unique element $(v,w)$ in $M$ with second coordinate w as explained in the answer. The map $w \to (v,w) \to v$ is then a $G$-linear map $W \to V \times W \to V$, since the $W \to M$ is (by forementioned uniqueness) and $M \to V$ is.This map is nonzero (otherwise $M$ projects to $0$ in $V$). Thus it must be an isomorphism, contradiction.
each $w \in W$ must show up as the second coordinate of some element in $M$ since $M$ projects onto $W$. In fact there is a unique element $(v,w)$ in $M$ with second coordinate $w$ as explained in the answer. The map $w \to (v,w) \to v$ is the map written in the answer, which must be $G$-linear since the $W \to M$ is (by forementioned uniqueness) and $M \to V$ is. –  Soarer Sep 11 '11 at 7:56