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If $G$ is a finite group, $V$ is an irreducible $G$-representation and $W$ is any 1-dimensional $G$-representation (both over an algebraically closed field of characteristic zero), show that $V \otimes W$ is an irreducible $G$-representation.

This is the second time I've come across this problem this month (2 different classes), and I'm unsure of the solution I tried in the first class. Essentially my solution was to fix bases $v_1, \ldots, v_k$ of $V$ and $w$ of $W$, show that $v_i \otimes w$ is a basis of $V \otimes W$ and then through brute force show that a submodule of $V \otimes W$ would imply a submodule of $V$, which gives the contradiction. I feel like there is an easier way - any hints would be great!

itsricobitches
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  • Presumably, you've look at these answers to this question? http://math.stackexchange.com/questions/355027/show-u-otimes-v-is-an-irreducible-g-module/355040#355040 – Sammy Black Apr 15 '13 at 16:11
  • Small comment about proof-writing style: a direct proof seems cleaner (to me) than a contradiction proof. In your example, rather than assuming that no nontrivial, proper submodule of $V \otimes W$ exists and then arriving at a contradiction, you can just begin with an arbitrary submodule of $V \otimes W$ and argue that it must be trivial or the whole module. – Sammy Black Apr 15 '13 at 16:14
  • You might consider the fact that showing irreducibility requires showing that there is no proper non-zero linear subspace invariant under the group, and tensoring by a $1$-dimensional representation just multiplies the matrix representing each group element by a scalar. – Geoff Robinson Apr 15 '13 at 16:40

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