I'm stuck somewhere in the following claim, I would appreciate if you could help.
( Recall : Let $G$ be a group and $H \leq G$. If $\alpha$ is a character of this subgroup $H$, we define
$\alpha^{G}(g)=\frac{1}{\vert H \vert}\sum_{x\in G}\alpha^{\circ}(xgx^{-1})$, where $\alpha^{\circ}$ is the function on $G$ equal to $\alpha $ for elements of $H$ and equal to zero otherwise. This $\alpha^{G}$ is also character of $G$ called induced character. )
Our claim is that Ker($\alpha^{G}$)$\subseteq$Ker($\alpha$):
If we take an element $k \in$ Ker($\alpha^{G}$), then $\alpha^{G}(k)= \frac{1}{\vert H \vert}\sum_{x\in G}\alpha^{\circ}(xkx^{-1}) = \alpha^{G}(1)= \frac{1}{\vert H \vert}\sum_{x\in G}\alpha^{\circ}(x1x^{-1})= \frac{1}{\vert H \vert}\sum_{x\in G}\alpha(1).$
What should I do after this equation? I feel like we're going to have to see that $\alpha^{\circ}(xkx^{-1})= \alpha(1)$ for any $x$. Then the result immediately will follow.
Best Answer
Let $\chi$ be a character of $H \leq G$. Then $${\rm core}_G(\ker(\chi))=\ker(\chi^G)=\bigcap_{g \in G}g^{-1}(\ker(\chi))g.$$ Proof If $x \in G$ then $x \in \ker(\chi)$ if and only if $\sum_{g \in G}\chi^{\circ}(gxg^{-1})=\sum_{g \in G}\chi(1)$. (Here (induction definition) $\chi^{\circ}(t)=0$ if $t \notin H$ and $\chi^{\circ}(t)=\chi(t)$ if $t \in H$)
Since $|\chi^{\circ}(gxg^{-1})| \leq \chi(1)$, we see that $x \in \ker(\chi)$ iff $\chi^{\circ}(gxg^{-1})=\chi(1)$ for all $g \in G$. And this is the case iff $x \in g^{-1}(\ker(\chi))g$ for all $g$. $\square$
Example If $\chi$ is faithful, that is, if $\ker(\chi)=1$, then also $\ker(\chi^G)=1$.
Note In general, if $X \leq G$, the ${\rm core}_G(X)$ is the largest normal subgroup of $G$ contained in $X$.