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If $G$ is a group and $H$ a subgroup of $G$, would it be possible to have a simple $\mathbb{C}G$-module $V$ for which the restriction $\operatorname{Res}^G_HV$ is not a simple $\mathbb{C}H$-module? What would be examples of such a behaviour?

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    $\begingroup$ More or less every triple $(G,H,V)$ satisfies this. For example, any simple module of dimension greater than $1$ and any abelian subgroup. $\endgroup$ – David A. Craven Jul 31 at 21:03
  • $\begingroup$ Okay, lets say $V$ is an irreducible $\mathbb{C}G$-module with $dim(V)=2$ for example. Now lets say $H$ is an abelian subgroup of $G$. We can consider $H$ as a $\mathbb{C}H$-module. Since $dim(V)>1$ we can decompose it in two vector spaces $V=V_1\oplus V_2$. But why are they submodules? If $v_1\in V_1$, do we necessarily have $h\cdot v_1\in V_1$? $\endgroup$ – roi_saumon Jul 31 at 21:34
  • $\begingroup$ All simple modules for abelian groups are $1$-dimensional. Not all subspaces are submodules. But they are there somewhere. $\endgroup$ – David A. Craven Jul 31 at 21:39
  • $\begingroup$ Oh, I get it now. Thanks $\endgroup$ – roi_saumon Jul 31 at 21:55
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If we take for $H$ the trivial subgroup of $G$ then the restriction $\operatorname{Res}^G_H V$ is simple if and only if $V$ is one-dimensional. In this way every simple $\mathbb{C}G$-module of dimension at least two gives a counterexample.

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