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This is regarding the proof of proposition 24(page 61) of Serre's Linear Representations of finite groups. Line 3 of the proof says that for $s\in G$, $\rho(s)$ permutes $V_i$? Can someone be kind enough to tell me why this is the case?

If you do not have the book, here is what I need. Suppose $\rho: G\to GL(V)$ is an irreducible representation of $G$. Let $A$ be a normal subgroup of $G$. Let $V=\bigoplus V_i$ be the canonical decomposition of the representation $\rho$ restricted to $A$ into a direct sum of isotypic representations. Show that for $g\in G$, $\rho(g)$ permutes $V_i$.

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2 Answers 2

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Hint: Observe $\rho(g):V_i\to\rho(g)V_i$ is an isomorphism of $A$-reps.

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  • $\begingroup$ yes I got it after an hour of posting it, but it took me a while. Thanks. $\endgroup$
    – user114539
    Commented Feb 8, 2015 at 5:40
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The morphisms $\rho(g): V_i\to \rho(g)V_i$ are not isomorphisms of $A$-representations. It does not need to be the case that $\rho(g)\rho(a) = \rho(a)\rho(g)$ for all $a\in A$. The $A$-representation on $\rho(g)V_i$ is in general isomorphic to a conjugate representation of $A\to GL(V_i)$.

You can work out the example with $G = S_3$ (symmetric group on three elements) and $A = C_3$ (cyclic subgroup of index 2). If you consider the irreducible representation $\rho: S_3\to GL(V)$ of degree 2 (there is a unique one up to iso), then the decomposition as in the proof of Serre is a decomposition into two 1-dim $A$-representations $V_1$ and $V_2$. An element $g\in G-A$ permutes those two lines but clearly $V_1$ and $V_2$ are not $A$-isomorphic (a non-trivial element of $A$ acts as an homothety on the lines but not with the same factor).

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