Fourier transform of subgroups of $(\mathbf{Z}/n\mathbf{Z})^*$ Let $\zeta$ be a primitive $n$-th root of unity, and for each function $f : \mathbf{Z}/n\mathbf{Z} \to \mathbf{C}$ define its Fourier transform $\widehat{f} : \mathbf{Z}/n\mathbf{Z} \to \mathbf{C}$ by
$$\widehat{f}(a) = \sum_{x \in \mathbf{Z}/n\mathbf{Z}} f(x) \zeta^{ax} .$$
Further, for any $A \subseteq \mathbf{Z}/n\mathbf{Z}$ put $\widehat{A}(a) = \widehat{\mathbf{1}_A}(a)$, where $\mathbf{1}_A$ is the indicator function of $A$.
If $G$ is a subgroup of $(\mathbf{Z}/n\mathbf{Z})^*$, what can we say about the dimension of the $\mathbf{Q}$-vector space spanned by $\{\widehat{G}(a) : a \in \mathbf{Z}/n\mathbf{Z}\}$ ?
Thanks.
 A: The dimension is $\phi(n)/|G|$.
Denote the $\mathbb{Q}$-span of $\{\hat{G}(a)\mid a\in\mathbb{Z}/n\mathbb{Z}\}$ in $\mathbb{C}$ by $V$.
Observe that $V<\mathbb{Q}(\zeta)$, which is a Galois extension of $\mathbb{Q}$ with Galois group $\Gamma\simeq (\mathbb{Z}/n\mathbb{Z})^*$.
Identifying $G$ with a subgroup of $\Gamma$, I claim that $V$ is the space of $G$-invariants, which leads to the above expression by Galois' theorem.
Observe that $v\mapsto \frac{1}{|G|}\sum_{x\in G} x(v)$ is a projection of $\mathbb{Q}(\zeta)$ on $\mathbb{Q}(\zeta)^G$, hence also the image of the operator $P(v)=\sum_{x\in G} x(v)$ is $\mathbb{Q}(\zeta)^G$. We therefore need to show that $P(\mathbb{Q}(\zeta))=V$. Since $\mathbb{Q}(\zeta)$ is spanned by $\{\zeta^a\}$, it is enough to show that the span of $\{P(\zeta^a)\}$ is $V$.
This indeed holds since, under our identification $x(\zeta)=\zeta^x$, we have 
$$ P(\zeta^a)=\sum_{x\in G}x(\zeta^{a}) =\sum_{x\in G}\zeta^{ax} =\sum_{x\in\mathbb{Z}/n\mathbb{Z}}\mathbf{1}_G(x)\zeta^{ax}  =\hat{G}(a). $$
