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If $G$ is a group and $\mathbb K$ is a field let $\mathbb KG$ be the usual group ring. We know a representation $\rho:G\longrightarrow GL(V)$, where $V$ is a $\mathbb K$-vetor space, is the same as a structure of $\mathbb KG$-module on $V$.

Suppose $G$ is a finite abelian group and let $\mathbb K=\mathbb C$. How can I show that if $V$ is a $\mathbb CG$-module then we can choose a basis for $V$ (as $\mathbb K$-vector space) such that $\rho(g)$ is a diagonal matrix, where $\rho$ is the afforded representation?

Obs: I'm always supposing the $\mathbb KG$-modules are finite dimensional $\mathbb K$ vector space.

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  • $\begingroup$ Related: math.stackexchange.com/questions/354926/… $\endgroup$
    – Rasmus
    Commented Nov 25, 2014 at 13:50
  • $\begingroup$ Are you familiar with the fact that a matrix of finite order is diagonalizable? And that a set of pairwise commuting diagonalizable matrices is simultaneously diagonalizable? $\endgroup$ Commented Nov 25, 2014 at 13:50
  • $\begingroup$ Well @TobiasKildetoft, I'm not, but I might be =) Where can I read about it? $\endgroup$
    – PtF
    Commented Nov 25, 2014 at 14:03
  • $\begingroup$ Those are both "standard" results from linear algebra. Not sure what a good reference is, but they are usually assumed known when you start studying group representations. $\endgroup$ Commented Nov 25, 2014 at 14:06
  • $\begingroup$ Ok, I'll look for them in a linear algebra book. Thanks =) $\endgroup$
    – PtF
    Commented Nov 25, 2014 at 14:08

2 Answers 2

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It is a standard fact that irreducible representations of finite abelian groups are $1$-dimensional. Every finite-dimensional (you forgot this assumption!) representation is a direct sum of irreducible representations. The claim follows.

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This is actually a direct consequence of Schurs lemma. I recommend you to study Schurs lemma and see if you can deduce the result.

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