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Math Help - Easy character theory proof

  1. #1
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    Easy character theory proof

    Let u: G x S -> S and v: G x T -> T be two actions of a finite group G on two finite sets S and T. Let w: G x (S x T) -> (S x T) be defined by:

    w(g,(x,y))=(u(g,x),v(g,y)) for all (x,y) in (S x T) and all g in G.

    Let c1, c2, c3 be the characters of the permutation representations p1,p2,p attached to u,v,w respectively.

    Prove that c1(g)=c2(g)c3(g) for any g.

    So basically, prove the character of the action on the set (S x T) is equal to the product of the actions on the sets S and T.

    For the trivial representation, this is obvious. 1=(1)(1)

    But even for the sign representation, I don't get it. -1=/=(-1)(-1), for example. And then how to generalize to all representations of an abstract group G?

    Thanks!
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  2. #2
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    Quote Originally Posted by brisbane View Post
    Let u: G x S -> S and v: G x T -> T be two actions of a finite group G on two finite sets S and T. Let w: G x (S x T) -> (S x T) be defined by:

    w(g,(x,y))=(u(g,x),v(g,y)) for all (x,y) in (S x T) and all g in G.

    Let c1, c2, c3 be the characters of the permutation representations p1,p2,p attached to u,v,w respectively.

    Prove that c1(g)=c2(g)c3(g) for any g.

    So basically, prove the character of the action on the set (S x T) is equal to the product of the actions on the sets S and T.

    For the trivial representation, this is obvious. 1=(1)(1)

    But even for the sign representation, I don't get it. -1=/=(-1)(-1), for example. And then how to generalize to all representations of an abstract group G?

    Thanks!
    with your notation it should be c_3(g)=c_1(g)c_2(g). we have c_1(g)=\text{tr}(p_1(g))=| \{s \in S: \ u(g,s)=s \}| and similarly c_2(g)=\text{tr}(p_2(g))=| \{t \in T: \ v(g,t)=t \} |.

    therefore c_3(g)=\text{tr}(p(g))=| \{(s,t) \in S \times T: \ u(g,s)=s, \ v(g,t)=t \} |=c_2(g)c_3(g).
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