Let $f(x) \in k[x]$ be a separable irreducible polynomial of degree n over a field k, and let F be its splitting field. Assume $Aut_k(F) \cong S_n$ , and let $\alpha$ be a root of $f(x)$ in F. Prove that $Aut_{k(\alpha)}(F) \cong S_{n-1}$.
I am saying that it suffices to prove that $k(\alpha)$ contains no other roots of $f(x)$ since our assumption is that our Automorphism group is the set of all permutations of roots.
$\textrm{Aut}_k(F) \cong S_n$ consists of every permutation of the roots of $f$. So $\textrm{Aut}_{k(\alpha)}(F)$ is the group of permutations which fix $\alpha$. This is clearly isomorphic to $S_{n-1}$.