Representation of finite group question
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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|!
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|!
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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?
But why do you think this suffices? Is every |G|-dimensional FG-module isomorphic to FG?
wofsy
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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.
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.
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