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Let $G$ be a finite group. Suppose that $k$ is a splitting field for all subgroups of $G$ and that $|G|$ is invertible in $k$. Let $N$ be a normal subgroup of $G$. Let $χ ∈ \operatorname{Irr}(kG)$ and $ψ ∈ \operatorname{Irr}(kN)$ such that $\langle\, \operatorname{Res}^G_N(χ), ψ\,\rangle_N= 0$. Denote by $H$ the subgroup of all $x ∈ G$ satisfying $ψ(xyx^{−1}) = ψ(y)$ for all $y ∈ N$.

Let $V$ be a simple $kG$-module with character $χ$, and let W be the isotypic component of $\operatorname{Res}^G _N(V)$ consisting of the sum of simple $kN$-submodules with character $ψ$; Note that the action of G on V permutes the isotypic components of $\operatorname{Res}^G_N(V)$, and it permutes them transitively because $V$ is simple. Thus $\dim_k(V) = |G : H|\dim_k(W)$. Since the elements in $H$ stabilise $ψ$, W is in fact a kH-submodule of $\operatorname{Res}^G_N(V)$.

This $\dim_k(V) = |G : H|\dim_k(W)$ appears abruptly to me. Not sure where it come from. Also, from this equation, can we say that $W$ is simple as a $kH$-module? thank you

Bernard
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scsnm
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  • Doesn't $\langle{\rm Res}_N^G\chi,\psi\rangle_N=0$ imply ${\rm Res}_N^GV$ has no $\psi$-components, i.e. $W=0$? – anon Sep 18 '20 at 15:51

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If $G$ permutes the isotypic components of $V$ (as $N$-reps), then the components must all be the same dimension as $W$, and by orbit-stabilizer theorem if $G$ permutes them transitively then the number of components is the index $[G:H]$ (since $H$ is the stabilizer of $W$), thus

$$ \dim V=\underbrace{\dim W+\cdots+\dim W}_{[G:H]} $$

because $V$ is a direct sum of its isotypic components.

anon
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  • thank you! this makes sense. another question is that if W is a kH simple module... I feel it is because $dim_k(V) = |G|$ and by the equation on dim, we have $dim_k(W) = |H|$. not sure at all... – scsnm Sep 18 '20 at 16:02
  • How is $\dim V=|G|$ possible? Isn't $V$ an irrep of $G$, so its dimension is strictly smaller than $|G|$ (unless $G$ is trivial of course)? – anon Sep 18 '20 at 16:12
  • yeah that is right. Then no idea again, then... – scsnm Sep 18 '20 at 16:15
  • it is supposed to be simple. I just cannot deduce this. – scsnm Sep 18 '20 at 16:16