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?
semisimplicity for subgroup
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$\begingroup$
group-theory
representation-theory
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0You 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
1 Answers
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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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0The first sentence names the theorem. The first two sentences after "Roughly" state the theorem. The rest indicates why it is true. – 2011-02-10