# Easy character theory proof

• Apr 4th 2010, 11:30 AM
brisbane
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!
• Apr 4th 2010, 01:56 PM
NonCommAlg
Quote:

Originally Posted by brisbane
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 $\displaystyle c_3(g)=c_1(g)c_2(g).$ we have $\displaystyle c_1(g)=\text{tr}(p_1(g))=| \{s \in S: \ u(g,s)=s \}|$ and similarly $\displaystyle c_2(g)=\text{tr}(p_2(g))=| \{t \in T: \ v(g,t)=t \} |.$

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