An irreducible representation of a finite group $G$ is a representation that cannot be expressed as the direct sum of $G$-representations. Irreducible representations are useful because of Maschke's theorem, which allows us to decompose arbitrary finite dimensional $G$-representations as direct sums of irreducible representations in which the number of times a summand occurs is independent of the decomposition chosen (provided char $k \nmid |G|$, where $k$ is the underlying field).

My question is, what happens when we try to do all this with the direct sum replaced by the tensor product? Are there finitely many $G$-representations which cannot be decomposed as the tensor product of other $G$-representations? Is there a $G$-representation for some $G$ which can be decomposed into a tensor product of representations atomic with respect to the tensor product in two different ways?

  • $\begingroup$ I very much appreciate Rickard's answer. I'd liked to add a follow-up question to Chaturvedi's question --- if it's OK I'll add it as a comment. Suppose $U$ and $V$ are two irreducible complex representations of a finite group $G$ such that the tensor product $U \otimes V$ is also irreducible. Must it be the case that either $U$ or $V$ is one dimensional ? Thanks, Ines. $\endgroup$ Aug 11, 2017 at 1:46

1 Answer 1


There are infinitely many "tensor irreducible" representations of any group. Just take the direct sum of $p$ copies of the trivial module, for any prime $p$.

One-dimensional representations give non-unique tensor decompositions in trivial ways (if $U$ and $V$ are representations with $U$ one-dimensional, then $U\otimes U^*\otimes V\cong V$), so you probably want uniqueness up to one-dimensional factors.

But even then, the alternating group $A_5$ has no one-dimensional representation apart from the trivial one, but it has two three-dimensional representations $U$ and $U'$ and a five-dimensional representation $V$ with $U\otimes V\cong U'\otimes V$.


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