
#1
Feb110, 06:55 PM

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PF Gold
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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! 



#2
Feb110, 07:12 PM

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




#3
Feb210, 08:30 PM

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PF Gold
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If F is the field, write FG for the group algebra and call V the FGmodule 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 FGsubmodule 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! 



#4
Feb210, 09:02 PM

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PF Gold
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Representation of finite group question
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? 



#5
Feb310, 05:48 AM

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PF Gold
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It seems to me that the Gdimensional FGmodule 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 FGsubmodule of FG.
But why do you think this suffices? Is every Gdimensional FGmodule isomorphic to FG? 



#6
Feb310, 01:40 PM

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PF Gold
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#8
Feb310, 02:53 PM

P: 707

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. 


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