I'm working on the following proof, and think that what I've got so far is okay, but was hoping someone could help give a hint with the remainder.
Question: let $E/F$ be a Galois extension with Galois group $G = G(E/F)$ and $\alpha \in E$. Prove that $E = F(\alpha)$ if and only if for all $g\in G$, $g \neq$ id we have $g(\alpha)\neq \alpha$, and hence prove the minimal polynomial for $\alpha$ is given by: $$ (x-g_1(\alpha))\ldots(x-g_n(\alpha)) $$ where $G = \{g_1, \ldots, g_n\}$.
Proof (so far): We first prove the $\Rightarrow$ direction. Suppose $E = F(\alpha)$, $g\in G$ with $g\neq$ id. Then note that for any $g \in G$ we have $g(x)=x$ for all $x\in F$, so $g$ must always be entirely determined by $g(\alpha)$, and hence if $g(\alpha)=\alpha$ we must have $g = $id, a contradiction. Thus we conclude $g(\alpha)\neq \alpha$ for all $g\in G$, $g\neq$ id. We now prove the $\Leftarrow$ direction. Suppose that for all $g\in G$ with $g\neq$ id we have that $g(\alpha) \neq \alpha$. Note that since $F \subseteq E$ and $\alpha \in E$, clearly $F(\alpha) \subseteq E$, and so it remains to show that $[F(\alpha):F] = [E:F]$ (or equivalently $[E:F(\alpha)]=1$), since then we would have a subspace of equal dimension, thus implying that $E = F(\alpha)$. (Here is where I am stuck for this part of the proof)
Now let $f$ be a polynomial, say of degree $n$, with $\alpha \in E$ as a root, and let $\alpha = \alpha_1, \ldots, \alpha_n \in E$ denote the roots of $f$, possibly not distinct. Now let $\alpha_1, \ldots, \alpha_k$ denote the $k$ distinct roots, and suppose $g \in G = \{g_1, \ldots, g_n\}$ with $g \neq$ id. Then since $G$ forms a group we know that there is a bijective correspondence between the sets $\{\alpha_1, \ldots, \alpha_n\} \leftrightarrow \{g(\alpha_1), \ldots, g(\alpha_n)\}$, that is $g$ permutes the roots. Thus restricting any $g \in G$ to the set of distinct roots $\{\alpha_1, \ldots, \alpha_k\}$ will also be a permutation, since $g$ is injective (it is an automorphism) so its restriction to a finite set is a bijection. Let $p$ be the following polynomial: \begin{align} p(x) = (x-g(\alpha_1))\ldots(x-g(\alpha_k)) \end{align} Then $p$ has coefficients fixed by any element $g \in G$ since these permute the factors by the homomorphism property and the fact that $g$ would just permute the $\alpha_i$, and so the coefficients must lie in the fixed field of $G$, that is, $p \in F[x]$. Note that when applying $g$ to $p$ we do not "gain" any repeated factors, since $g(\alpha)\neq \alpha$ for all $g \in G$ with $g \neq$ id, by our earlier proof. Clearly $p$ is monic, and we claim that it must also be irreducible. To see this, note that if it factored non-trivially, then we must have some other polynomial $h \in F[x]$ with $\alpha$ as a root, and $g(\alpha)$ not a root for some $g \in G$. But then this $h$ will not have all coefficients in $F$, since the automorphisms in $G$ will not simply permute the factors as they did for $p$, and so we conclude no such $h$ can exist. Hence we deduce that in fact $p$ must be the minimal polynomial for $\alpha$.
I'm having trouble with the $\Leftarrow$ direction of the if and only if statement, but think the rest is okay. Any help or feedback would be great!