In the wikipedia page about Clifford theory, they are constructing a new character by conjugating another character. ie, Let N be a normal subgroup of G, If μ is a complex character of N, then for a fixed element g of G, another character, μ(g), of N may be constructed by setting
${\displaystyle \mu ^{(g)}(n)=\mu (gng^{-1})} \forall n\in N$
But how does this construct a new character of $N$? , my proof is: Let $\pi$ be the representation of G corresponding to $\mu$. Then, $\mu^g(n)= \mu(gng^{-1})= trace(\pi(gng^{-1}))=trace(\pi(g)\pi(n)\pi(g)^{-1})=trace(\pi(n))=\mu(n)\\$ Please help me figure out where am I going wrong. I think its beacause $\mu$ is a character of N , so $\pi$ is not a homomorphism on G. Is that right?
Thanks in advance.
Let $\rho:N\to \text{GL}_m(\Bbb C)$ be a representation (a homomorphism) with character $\chi$. Let $g\in G$. Then $\alpha:N\to N$ defined by $\alpha(h)=ghg^{-1}$ is a homomorphism. The composite $\rho\circ \alpha:N\to\text{GL}_m(\Bbb C)$ is a representation with character $h\mapsto\chi(ghg^{-1})$.