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let $G$ be a group (can assume finite) acting on a vector space $V$, and assume that the action is semisimple. Let $H$ a normal subgroup of $G$. Why also the action of $H$ on $V$ is semisimple too?

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    Is there any condition on the characteristic of the field? because Maschke's theorem shows that any $F[G]$ module is semisimple when $|G| \in F^{\times}$2011-02-10
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    actually not. The problem is that these group arise as galois group acting on Tate modules of an abelian variety.2011-02-10
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    You should write in this format (with @) in order for the person to whom you're replying to receive a notification about your comment. (Of course since you're unregistered, *my* formatting won't make a difference :) )2011-02-10

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This is called Clifford's theorem. Roughly:

Theorem. Take a simple G-submodule W of V. Then W is a direct sum of simple H-submodules U1⊕…⊕Un.

Let U be any simple H-submodule of W. Then the G-conjugates of U are also H-submodules, since H is normal. They intersect as proper H-submodules of U, and so intersect trivially. They span W since their span is a G-submodule of W. Since every H-submodule of V is a direct sum of simple G-submodules (G acts semisimply), and each simple G-submodule is a direct sum of simple H-submodules (as we just showed), V is a direct sum of simple H-submodules, and so is semi-simple.

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    In your first two sentences, you are assuming that simple $G$-modules are semisimple when restricted to $H$, no?2011-02-10
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    The first sentence names the theorem. The first two sentences after "Roughly" state the theorem. The rest indicates why it is true.2011-02-10