This is a question on one step in a proof from Dummit and Foote pg. 597.
Let $p$ be an odd prime (there isn't much to say for the $p=2$ case) and let $G=\text{Gal}(\mathbb{Q(\zeta_p)}/\mathbb{Q})$ where $\zeta_p$ is a primitive $p$-th root of unity. Let $H\leq G$. Define $\alpha:=\sum_{\sigma\in H} \sigma(\zeta_p)$. The claim is that the fixed field of $H$ is $\mathbb{Q}(\alpha)$ e.g. $\text{Fix } H=\mathbb{Q}(\alpha)$.
We first show that $\mathbb{Q}(\alpha)\subseteq \text{Fix }H$.
The other direction: $\mathbb{Q}(\alpha)\supseteq \text{Fix }H\Longleftrightarrow G(\mathbb{Q}(\zeta_p)/\mathbb{Q}(\alpha))\subseteq H$ which by contrapositive is equivalent to $\sigma\not\in H\Longrightarrow \sigma\not\in G(\mathbb{Q}(\zeta_p)/\mathbb{Q}(\alpha))$ Now, every $\sigma\in G$ fixes $\mathbb{Q}$ so to show this last implication it is sufficient to check that $\sigma\not\in H$ implies that $\sigma$ does NOT fix $\alpha$.
Dummit and Foote says to next let $\tau\not\in H$. Then if we had $\tau(\alpha)=\alpha$ we get $\tau(\zeta_p)=\sigma(\zeta_p)$ for some $\sigma\in H$. From here I can complete the problem.
I know that $\{\zeta_p^k\}_0^{p-2}$ and $\{\zeta_p^k\}_1^{p-1}$ are both bases for $\mathbb{Q}(\zeta_p)$. I also know that automorphisms permute the set $\{\zeta_p^k\}_1^{p-1}$.
Might be useful to write (say) $(1,0,1,1,1,0)$ as $\alpha$ in $\{\zeta_p^k\}_1^{p-1}$-coordinates and say that any automorphism that fixes that element must "fix" the coordinates in that only powers with like coefficients are exchanged. (Else, we would have two different solutions which is impossible because we have a basis)
The question is, how exactly does the statement in the blockquote work.
Since the primitive roots of unity form a basis, if an automorphism fixes the sum of a certain subset of them, then by the uniqueness of an element as a linear combination of the basis, that automorphism must just permute those roots of unity, meaning it sends $\zeta_p$ to some $\sigma(\zeta_p)$.