Problem 2.16 - Character theory by Isaacs Let $H\leq G$ and let $\chi$ be a (possibly reducible) character of $G$. Suppose $\chi$ vanishes on $G-H$. Assume that either $H=\{e\}$  or $G$ is abelian. Show that $[G:H]$ divides $\chi(1)$.
I have proved it partially, in the case when $H$ is the trivial subgroup. But I haven't been able to do the abelian case.
Thanks in advance!
 A: 
Lemma Let $H \le G$ and let $\phi$ be a (not necessarily irreducible) character of $G$ vanishing on $G-H$. Assume that either $H=1$ or $G$ is abelian. Then $|G:H| | \phi(1)$. 


Proof Assume first that $H=1$. Now $|G|[\phi,1_G]=\sum_{g \in G}\phi(g)=\phi(1)$. Since $\phi$ is a character, $\phi(1) \neq 0$ and $[\phi,1_G]$ is a positive integer, it follows that $|G|$ divides $\phi(1)$. In fact, in this case $\phi=b\rho$, where $b=\frac{\phi(1)}{|G|}$ and $\rho$ is the regular character of $G$, see Exercise (3.8) of the same book.
Now assume that $G$ is abelian and $\phi\equiv0$ outside $H$. We can write $\phi=\sum_{\mu \in Irr(G)} a_{\mu}\mu$, with $a_{\mu}$ non-negative integers and not all equal to $0$. Note that the $\mu$ are of course all linear since $G$ is abelian. Hence their restrictions $\mu_H$ are also linear and irreducible. This implies that $\phi_H=\sum_{\mu \in Irr(G)} a_{\mu}\mu_H$ is the decomposition of $\phi_H$ as a sum of irreducible characters of $H$. Note that different $\mu's$ can be equal when restricted to $H$. So in this last sum some of the irreducible constituents may coincide. Put $\phi_H=\sum_{\lambda \in Irr(H)} b_{\lambda}\lambda$ with $b_{\lambda}$ non-negative integers. Fix such $\lambda$, then $\lambda=\mu_H$ for some irreducible constituent $\mu$ of $\phi$. Since $\phi$ vanishes off $H$, $$b_{\lambda}|H|=\sum_{h \in H}\phi(h)\overline{\lambda(h)}=\sum_{g \in G}\phi(g)\overline{\mu(g)}=|G|a_{\mu}$$So we see that each of the $b_{\lambda}=|G:H|a_{\mu}$ is hence divisible by $|G:H|$. Since $\phi(1)=\sum_{\lambda \in Irr(H)} b_{\lambda}\lambda(1)=\sum_{\lambda \in Irr(H)} b_{\lambda}$, we now see that $|G:H|$ divides $\phi(1)$, as wanted.$\square$
Note 1 A natural common generalization of the situation $H=1$ and $G$ is abelian is $H \subseteq Z(G)$. However, under these hypotheses the above result is not true. By the way, in this situation, owing to Lemmata (2.27)(c) and (2.29) of Isaacs' book, $\phi(1)^2=|G:H|[\phi,\phi]$, hence $|G:H|$ divides $\phi(1)^2$.
Here is a counterexample that shows that $|G:H|$ not necessarily divides $\phi(1)$. The character table of the quaternion group $Q$ of order $8$ looks as follows. Here $Q=\{\pm 1,\pm i, \pm j,\pm k \}$, where $i^2=j^2=k^2=-1, ij=k,jk=i,ki=j$. $Q$ has five conjugacy classes.
$$ \begin{array}{|c|c|c|c|}
\hline
&1& -1 & \{\pm i \} & \{\pm j \} & \{\pm k \} \\ \hline
1_Q & 1 & 1  & 1 & 1 & 1 \\ \hline
\chi_2 & 1 & 1 & 1 & -1 & -1 \\ \hline
\chi_3 & 1 & 1 & -1 & -1 & 1 \\ \hline
\chi_4 & 1 & 1 & -1 & 1 & -1 \\ \hline
\chi_5 & 2 & -2 & 0 & 0 & 0 \\ \hline
\end{array}
$$
So, $\chi_5$ vanishes off the center $Z(Q)=\{\pm 1\}$. But $|Q:Z(Q)|=4$ does not divide $\chi_5(1)=2$.
Note 2 The above lemma is being used in the proof of Theorem (3.13) of Isaacs' book.
A: Assume $G$ is abelian. Let $\psi_1,\psi_2,\ldots,\psi_m, m=[G:H],$ be the (1-dimensional) irreducible characters of $G/H$ viewed as characters of $G$ by precomposing with the projection $G\to G/H$. 
Observe that as $\chi(x)=0$ whenever $x\notin H$, we have $\chi(x)\psi_j(x)=\chi(x)$ for all $x\in G$ and all $j=1,\ldots,m$. We take advantage of this as follows.
If $\chi_i$ is any of the irreducible characters (also 1-dimensional) of $G$, then 
$$
\begin{aligned}
\langle \chi,\chi_i\rangle&=\frac1{|G|}\sum_{x\in G} \chi(x)\overline{\chi_i(x)}\\
&=\frac1{|G|}\sum_{x\in G} \bigg(\chi(x)\overline{\psi_j(x)}\bigg)\overline{\chi_i(x)}\\
&=\frac1{|G|}\sum_{x\in G} \chi(x)\bigg(\overline{\psi_j(x)}\overline{\chi_i(x)}\bigg)\\
&=\langle\chi,\chi_i\overline{\psi_j}\rangle.
\end{aligned}
$$
All the products $\chi_i\psi_j$ are themselves irreducible. Call the irreducible characters $\chi_i$ and $\chi_{i'}$ $H$-similar, if 
$\chi_{i'}=\chi_{i}\psi_j$ for some $j$. This is an equivalence relation of the set $\hat G$ of irreducible characters of $G$ with each equivalence class containing $m$ characters. Let $D$ be a set of representatives of $H$-similarity classes. The above calculation shows that the inner product
$\langle \chi,\chi_i\rangle$ only depends on the $H$-similarity class of $\chi_i$.
This lets us conclude as follows
$$
\begin{aligned}
\chi(1)&=\sum_{\chi_i\in\hat{G}}\langle \chi,\chi_i\rangle\,\chi_i(1)\\
&=\sum_{\eta\in D} m\langle\chi,\eta\rangle\,\eta(1).
\end{aligned}
$$
This number is clearly a multiple of $m$.
