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!
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 $m$.