$H \leq G$, $[G : H] = 2$, $\theta$ is an irreducible representation of $G$, $\alpha$ is an irreducible representation of $H$. ( field is $\mathbb{C}$ ).
$\beta$ is a map such that $\beta(h) = \alpha(ghg^{-1}) \forall h \in H$ for some fixed $g \in G \backslash H$.
I managed to prove that $\beta$ is also an irreducible representation of $H$, but I’m struggling with finding its character. I know that $\chi_{\beta}(h) = \chi_{\alpha}(ghg^{-1}) = Tr(\alpha(ghg^{-1}))$, but since $\alpha$ is a homomorphism on $H$ it is not true that $Tr(\alpha(ghg^{-1})) = Tr(\alpha(h))$. Maybe now I need to utilise the fact that $H$ is normal in $G$ but I don’t know how to.
How do I continue?
Thank you in advance!
This works quite generally, and you only need that $H$ is a normal subgroup of $G$.
For any $g \in G$ we have a conjugation automorphism $$ c_g \colon H \rightarrow H \qquad c_g(h) \colon= g h g^{-1}. $$ Therefore, given a representation $$ \alpha \colon H \rightarrow \text{GL}_n(\mathbb{C}), $$ we can obtain a new representation $$ \beta := \alpha \circ c_g \colon H \rightarrow H \rightarrow \text{GL}_n(\mathbb{C}) $$ This is exactly the $\beta$ you are given in the question.
If $\chi$ is the character of $\alpha$, then the character $\psi$ of $\beta$ is simply $$ \psi(h) = \chi(g h g^{-1}). $$ You can show using the formula for the inner product of characters that $$ (\chi, \chi) = (\psi, \psi), $$ and therefore $\psi$ is irreducible whenever $\chi$ is.
Note that if we have $g \in H$ then because the character is a class function, we obtain that the characters $\chi$ and $\psi$ are the same, and thus the representations $\alpha$ and $\beta$ are isomorphic. In fact, the isomorphisms between the representations $\alpha$ and $\beta$ is given by the matrix $\alpha(g)$.
However for $g \in G \setminus H$ these characters can be different.