Representation of finite group question

quasar987

Does anyone know how to prove that any irreducible representation of a finite group G has degree at most |G|?

Equivalently, that every representation of degree >|G| is reductible.

Thx!

Hurkyl

I suspect your conjecture is true -- for any vector v, the set Gv should span a direct summand of the entire G-vector space. But I haven't proven it yet, so grain of salt.

quasar987

If F is the field, write FG for the group algebra and call V the FG-module associated with a given representation of G. For any non zero v in V, Gv has at most |G| elements, and so the vector subspace W = span(Gv) has dimension at most |G| and it is clearly stable under the action of G (i.e., it is an FG-submodule of V). But W is non trivial and so if V is irreducible, it must be that W=V. Thus |G|>=dim(W)=dim(V).

I think this work, but according to Dummit & Foote Exercice 5 in the section on representation theory, we can do better and show that an irreducible representation has dimension strictly less than |G|!

Hurkyl

Oh! Silly me, I was looking for a direct sum decomposition -- I should have paid more attention to the definitions.


So... I think all we need to do now is to prove that FG is itself a reducible representation, right?

quasar987

It seems to me that the |G|-dimensional FG-module FG is reducible because if G={e,g_1,...g_r}, for v:=e+g_1+...g_r, we have that span(v) is a one dimensional FG-submodule of FG.

But why do you think this suffices? Is every |G|-dimensional FG-module isomorphic to FG?

Hurkyl

I thought it works as an addendum to the previous proof.

quasar987

In what sense?

wofsy

If the field is the complex numbers:

The restriction of the representation to a cyclic subgroup is a direct sum of 1 dimensional representations. Since the representation of the entire group is ireducible the number of these 1 dimensional representations in the decomposition must be less that the order of G.

Hurkyl

That it gives more information about the span of Gv.