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Suppose I have two unitary representations $U_V, U_W$ of a group $G$ on finite-dimensional vector spaces $V$ and $W$. I know that the tensor product representation $U_V\otimes U_W$ need not be irreducible, even if the original representations are. But I can decompose it into a direct sum of irreducible representations:

$$U_V\otimes U_W = \oplus_{k}U_k$$

Now, suppose I want to do things the other way round. I have some favourite irreducible representation $U_X$ on a vector space $X$. Can I always find vector spaces $V, W$ and non-trivial representations $U_V, U_W$ such that $U_X$ appears as an irreducible subrepresentation of $U_V\otimes U_W$? If so, how?

If not true in general, are there specific groups or special types of groups (e.g., Lie groups) where it holds?

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It looks as though you want $U_V$ and $U_W$ to be irreducible, and that you're just concerned with finite dimensional representations?

Suppose that there is a non-trivial irreducible representation $U_Y$ not isomorphic to $U_X$ and let $U_Y^*$ be its dual. Then the trivial representation is a subrepresentation of $U_Y^*\otimes U_Y$, and so $U_X$ is a subrepresentation of $U_X\otimes U_Y^*\otimes U_Y$, and therefore of $U_Z\otimes U_Y$ for some irreducible constituent $U_Z$ of $U_X\otimes U_Y^*$. Since we're assuming that $U_X\not\cong U_Y$, $U_Z$ can't be trivial.

This deals with any group with at least two non-isomorphic non-trivial irreducible representations.

If there's only one non-trivial irreducible representation $U_Y$, then it must be self dual, and the trivial representation is a constituent of $U_Y\otimes U_Y$. If moreover $U_Y$ is not one-dimensional, then $G$ can't act trivially on $U_Y\otimes U_Y$, and so $U_Y$ itself must be a constituent of $U_Y\otimes U_Y$.

So the only cases where your condition doesn't hold are when there are no non-trivial finite-dimensional irreducible representations, or when there is only one, that is one-dimensional.

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