# Representation Theory

• Apr 7th 2011, 12:34 AM
Turloughmack
Representation Theory
Im stuck on these problems and could use some help.

1. Show that if p is a representation of a group G, then p* defined as

p*(x) = p(x^-1)^T

is again a representation of G. (Here T means transpose.)

2. Let G be a finite group with normal subgroup N and v be the canonical homomorphism from G to G* := G/N via v : g -> Ng.

Find the kernel of the algebra homomorphism induced by v from F(G) to F(G*) = f(G/N).

Also show that

e:= (1/|N|)sum(n)

is an idempotent in F(G).
Show that F(G) - e.F(G) + (1-e)F(G)
with e.F(G) isomorphic to F(G*).

Thanks
• Apr 7th 2011, 02:54 AM
Deveno
1. we need to show that p* is a homomorphism (it should be clear that the matrix p(x^-1)ᵀ is in GLn(F) if p(x) is).

so: p*(xy) = [p((xy)^-1)]ᵀ = [p(y^-1x^-1)]ᵀ = [p(y^-1)p(x^-1)]ᵀ (since p is a homomorphism)

= [p(y)^-1p(x)^-1]ᵀ = (p(x^-1))ᵀ(p(y^-1))ᵀ = p*(x)p*(y)

2. is F(G) the group algebra of G over F? if so then the induced homomorphism (which i shall call v') takes Σajgj to Σajv(gj).

so ker(v') = {Σajgj in F(G) : Σajv(gj) = v(e) = N}. this implies that v(gj) = N for all j (so gj is in N), and that Σaj = 1.
• Apr 7th 2011, 03:46 AM
Turloughmack
Quote:

Originally Posted by Deveno
1. we need to show that p* is a homomorphism (it should be clear that the matrix p(x^-1)ᵀ is in GLn(F) if p(x) is).

so: p*(xy) = [p((xy)^-1)]ᵀ = [p(y^-1x^-1)]ᵀ = [p(y^-1)p(x^-1)]ᵀ (since p is a homomorphism)

= [p(y)^-1p(x)^-1]ᵀ = (p(x^-1))ᵀ(p(y^-1))ᵀ = p*(x)p*(y)

2. is F(G) the group algebra of G over F? if so then the induced homomorphism (which i shall call v') takes Σajgj to Σajv(gj).

so ker(v') = {Σajgj in F(G) : Σajv(gj) = v(e) = N}. this implies that v(gj) = N for all j (so gj is in N), and that Σaj = 1.

Thanks!
• Apr 7th 2011, 12:55 PM
Drexel28
Quote:

Originally Posted by Turloughmack
Also show that

e:= (1/|N|)sum(n)

is an idempotent in F(G).
Show that F(G) - e.F(G) + (1-e)F(G)
with e.F(G) isomorphic to F(G*).

Thanks

Too many questions in one thread. I assume this is supposed to be $\displaystyle \displaystyle e=\frac{1}{|N|}\sum_{n\in N}n$ note then that

$\displaystyle \displaystyle e^2=\frac{1}{|N|^2}\sum_{n,n'\in N}nn'=\frac{1}{|N|^2}\sum_{n\in N}\sum_{n'\in N}nn'=\frac{1}{|N|^2}\sum_{n\in N}\sum_{n'\in N}n'=\frac{|N|}{|N|^2}\sum_{n'\in N}n'=e$