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Maschke's theorem (Theorem)

Let $ G$ be a finite group, and $ k$ a field of characteristic not dividing $ \vert G\vert$. 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

$\displaystyle \pi'(v)=\frac 1{\vert G\vert}\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 3680 times total.

Classification:
AMS MSC: 20C15 (Group theory and generalizations :: Representation theory of groups :: Ordinary representations and characters)
Pending Errata and Addenda
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