Q: Is my proof right?(Galois extension) Q) $E$ is a Galois extension of the $F$
Let $G=Aut(E/F)$ and $S_\alpha = \{g \in G \vert g(\alpha)=\alpha \}$ for $ \alpha \in E$
Say the $i : E \to E$ is a identity mapping
Prove "If $S_\alpha = \langle i \rangle$, then  $ E=F(\alpha)$"

I solved this question by myself. But there is not any solution in my workbook. So I can't check my proof is right or not. Here is my attempt.
$pf)$ Since the $E$ is a Galois extension, $G(E/F)$ is a Galois group which means $G(=Aut(E/F)) = G(E/F)$
By the Galois correspondence, only have to show $S_\alpha = G(E/F(\alpha))$. Because it is equal to $[E;F(\alpha)] =1$
First it is clear $i \in G(E/F(\alpha))$, so $\langle i \rangle \subset G(E/F(\alpha))$
Therefore  $S_\alpha(=\langle i \rangle) \subset G(E/F(\alpha))$
On the other hand, Let the $\forall h \in G(E/F(\alpha))$
Then $h \in S_\alpha$ because it fixes $F(\alpha) $ and $h \in Aut(E)$.
So $G(E/F(\alpha)) \subset S_\alpha$
Any comments or advice would be appreciated. Thanks.

A: You can prove this a bit quicker using the subfield of $E$ fixed by a subgroup of $G$:

*

*$E^{\langle i \rangle} = E$


*$E^{S_\alpha} = F(\alpha)$
therefore if $\langle i \rangle = S_\alpha$ then $E = F(\alpha)$.
