a representation which is not completely reducible

If G is a finite groupMathworldPlanetmath, and k is a field whose characteristic does divide the order of the group, then Maschke’s theorem fails. For example let V be the regular representationPlanetmathPlanetmath of G, which can be thought of as functions from G to k, with the G action gφ(g)=φ(g-1g). Then this representationPlanetmathPlanetmath is not completely reducible.

There is an obvious trivial subrepresentation W of V, consisting of the constant functions. I claim that there is no complementary invariant subspace to this one. If W is such a subspacePlanetmathPlanetmath, then there is a homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath φ:VV/Wk. Now consider the characteristic functionMathworldPlanetmathPlanetmathPlanetmath of the identityPlanetmathPlanetmathPlanetmath eG


and =φ(δe) in V/W. This is not zero since δ generates the representation V. By G-equivarience, φ(δg)= for all gG. Since


for all ηV,




But since the characteristic of the field k divides the order of G, WW, and thus could not possibly be complementary to it.

For example, if G=C2={e,f} then the invariant subspace of V is spanned by e+f. For characteristics other than 2, e-f spans a complementary subspace, but over characteristic 2, these elements are the same.

Title a representation which is not completely reducible
Canonical name ARepresentationWhichIsNotCompletelyReducible
Date of creation 2013-03-22 13:31:47
Last modified on 2013-03-22 13:31:47
Owner bwebste (988)
Last modified by bwebste (988)
Numerical id 5
Author bwebste (988)
Entry type Example
Classification msc 20C15