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Math Help - Representation Theory

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by Deveno View Post
    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!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Turloughmack View Post
    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 e=\frac{1}{|N|}\sum_{n\in N}n note then that

    \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
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