Show that $Z(θ^G)≤H$ Suppose $H ≤ G$ and $θ \in Char(H)$. Show that $Z(θ^G)≤H$.  ($Z$ is the centre and $θ^G$ is the character induced by $G$)
 A: Well if $g \not\in H$, then $\theta^G(g)$ is the sum of fewer than $|G|/|H|$ character values of $\theta$, so its absolute value is less than $\theta^G(1)$.
A: If $K \leq G$, we write $core_G(K)$ for the largest normal subgroup of $G$ contained in $K$. It is easily seen that $core_G(K)=\bigcap_{x \in G}K^x$, the intersection of all the $G$-conjugates of $K$.
Proposition Let $\theta$ be a character of $H \leq G$. Then
$$Z(\theta^G)=core_G(Z(\theta)).$$
Proof Let $\chi=\theta^G$, then $g \in Z(\chi)$ if and only if 
$$\frac{1}{\#H}|\sum_{x \in G}\theta^{o}(x^{-1}gx)|=[G:H]\theta(1)$$ $\iff$ 
$$|\sum_{x \in G}\theta^{o}(x^{-1}gx)|=\sum_{x \in G}\theta(1).$$
Since $|\theta^{o}(x^{-1}gx)| \leq \theta(1)$ and using the triangle inequality, we see that  $g \in Z(\chi)$ if and only if $|\theta^{o}(x^{-1}gx)| = \theta(1)$ for all $x \in G$. But this is the case exactly when $g \in Z(\theta)^x$ for all $x \in G$.$\square$
Note that it now follows that $Z(\theta^G) \leq H.$ In addition: in a similar manner, $ker(\theta^G)=core_G(ker(\theta))$.
