how to prove this lemma about sums on subgroups? 
Quoted from https://eprint.iacr.org/2019/601.pdf, but sadly with no proof.
 A: It’s sufficient to prove it for  $f=X^\alpha$, with $1\leq \alpha<n$.
Take $b\in H$. Observe that you can choose  $f(b)=b^\alpha\neq 1$ because the polynomial $X^\alpha-1$ has a finite number of roots, that is at most $ \alpha<n=|H|$. You can say the same if $\alpha\not \equiv 0 \mod n$ because if $\alpha=kn+\beta$, with $1\leq \beta<n$, then there is $b\in H$ such that $b^\beta\neq 1$. Then $b^\alpha=b^\beta\neq 1$.
Set $S:= \sum_{a\in H}f(a)$. Then
$f(b)S=f(b)\sum_{a\in H} f(a)=\sum_{a\in H}f(ab)= S $
So that
$(f(b)-1)S=0$ and so $S=0$
This proof is related to the representation theory. In fact $f=X^\alpha\in H^*$, where $H^*$ is the group of characters with values in $\mathbb{F}$. Now there is the usual formula
$\sum_{a\in H} \phi(a)=0$ for each $\phi\in H^*\setminus{1}$
In your case $f=X^\alpha$ is different from $1$ if and only if $\alpha$ is not a multiple of $|H|=n$. Thus you can generalise your result taking $f\in \langle X^\alpha: \alpha\not \equiv 0 \mod n \rangle_\mathbb{F}$. Observe that $\langle X^\alpha: \alpha\not \equiv 0 \mod n \rangle_\mathbb{F}$ contains $\mathbb{F}_{ 1\leq \deg<n}[X]$.
