In my last algebra exam, I had this exercise that I wasn't able to solve:
Let $H$ be a subgroup of the group $G$. Let $\chi$ be a character of $H$.
- Assuming that $\chi$ is faithful , prove that the induced character $\chi\uparrow G$ is faithful
- What is the link between $\ker(\chi)$ and $\ker (\chi\uparrow G)$ ?
To clarify the notation, this exam is about group representations and in particular about representations of finite groups on a $\mathbb{C}$ vector space. A faithful character is a character of an injective representation.
I know that if $\chi$ is the character of a representation $\rho$ then:
$$ \ker \rho = \ker \chi := \{g\in G\ |\ \chi(g) = \chi(1)\} $$.
I also know that
$$\chi\uparrow G(g) = \frac{1}{|H|}\sum_{x\in G}\dot\chi(xgx^{-1})$$
where
$$\dot\chi(g) =
\begin{cases}
\chi(g) & \text{if} & g\in H \\
0 & \text{if} & g\not\in H
\end{cases}$$
.
Using these formulae is easy to show that:
$$\ker \chi\uparrow G = \left\{g\in G\ \middle|\ \frac{1}{|H|}\sum_{x\in G}\dot\chi(xgx^{-1}) = |G:H|\chi(1)\right\}$$
but i don't know how to continue from here.
Best Answer
I do not use the arrow for induction but a superscript of the group the character is induced to.
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 $g$. $\square$
Hence 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 $X$.