Question:
Let $G \leq \text{Gal}(K/F)$ be a subgroup of the Galois group of the extension $K/F$ and suppose $\sigma_1, …, \sigma_k$ are generators for $G$. Show that the subfield $E/F$ is fixed by $G$ if and only if it is fixed by the generators $\sigma_1, …, \sigma_k$.
My attempt:
First suppose that the subfield $E/F$ is fixed by $G$. This means that for any automorphism $\tau \in G$ one has $\tau(x) = x$ for all $x \in E$. Since $G$ is generated by $\sigma_1, …, \sigma_k$ we certainly have that each of the automorphisms $\sigma_1, …, \sigma_k$ are elements in $G$ and therefore $\sigma_i(x) = x$ for all $x \in E$ and for each of $i = 1, 2, …, k$. In other words, the subfield $E$ is fixed by the generators $\sigma_1, …, \sigma_k$.
For the converse suppose hat the subfield $E$ is fixed by the generators $\sigma_1, …, \sigma_k$. Now let $\tau \in G$ be an arbitrary automorphism. Since $G$ is generated by the set $X = \sigma_1, …, \sigma_k$ we have that
$$
\tau = \tau_1^{\epsilon_1} \circ \tau_2^{\epsilon_2} \circ … \circ \tau_r^{\epsilon_r}
$$
where $r \in \mathbb{N}, \tau_1, …, \tau_r \in X$ and $\epsilon_1, …, \epsilon_r \in \{-1, 1\}$. Now, let $x \in E$ be any arbitrary element. We have
$$
\tau(x) = \big(\tau_1^{\epsilon_1} \circ \tau_2^{\epsilon_2} \circ … \circ \tau_r^{\epsilon_r}\big)(x).
$$
First, consider $\tau_r^{\epsilon_r}(x)$. If $\epsilon_r = 1$ then $\tau_r^{\epsilon_r}(x) = \tau_r(x) = x$ since we are now assuming that the subfield $E$ is fixed by all of the generators, and $\tau_r$ is an automorphism which is precisely one of these generators $\sigma_1, …, \sigma_k$. If instead $\epsilon_r = -1$, then we still have $\tau_r^{-1}(x) = x$ since $\tau_r$ is an automorphism (hence a bijection), and we know that $\tau_r(x) = x$. Thus the above equality reduces to
$$
\tau(x) = \big(\tau_1^{\epsilon_1} \circ \tau_2^{\epsilon_2} \circ … \circ \tau_{r-1}^{\epsilon_{r-1}}\big)(x),
$$
and after repetition of the same argument we eventually arrive at $\tau(x) = x$. Since $\tau \in G$ was an arbitrary automorphism it follows that every automorphism from the subgroup $G$ fixes the subfield $E$.
Best Answer
Your proof is correct. Here is also a shorter proof. Let:
$H=\{\sigma\in G: \sigma(x)=x \ \ \text{for all} \ \ x\in E\}$
It's very easy to show that $H$ is a subgroup of $G$. Since by assumption it contains a generating set of $G$, we must have $H=G$.