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Maschke's theorem (Theorem)

Let $G$ be a finite group, and $k$ a field of characteristic not dividing $|G|$ . Then any representation $V$ of $G$ over $k$ is completely reducible.

Proof. We need only show that any subrepresentation has a complement, and the result follows by induction.

Let $V$ be a representation of $G$ and $W$ a subrepresentation. Let $\pi:V\to W$ be an arbitrary projection, and let $$ \pi'(v)=\frac 1{|G|}\sum_{g\in G} g^{-1}\pi (gv) $$ This map is obviously $G$ -equivariant, and is the identity on $W$ , and its image is contained in $W$ , since $W$ is invariant under $G$ . Thus it is an equivariant projection to $W$ , and its kernel is a complement to $W$ . $ \qedsymbol$




"Maschke's theorem" is owned by bwebste.
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a representation which is not completely reducible (Example) by bwebste
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Cross-references: kernel, equivariant, invariant, contained, image, identity, map, projection, induction, complement, subrepresentation, completely reducible, representation, characteristic, field, finite group
There are 3 references to this entry.

This is version 6 of Maschke's theorem, born on 2003-01-04, modified 2005-03-10.
Object id is 3874, canonical name is MaschkesTheorem.
Accessed 4361 times total.

Classification:
AMS MSC20C15 (Group theory and generalizations :: Representation theory of groups :: Ordinary representations and characters)

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