If $Gal(f/\mathbb{Q}) \cong S_n$ then $Gal(f/\mathbb{Q}(\alpha_1)) \cong S_{n-1}$?

Question

Suppose I denote by $Gal(f/\mathbb{Q})$ the Galois group of the extension given by the splitting field of a separable polynomial $f$ of degree 6 over field $\mathbb{Q}$.

Suppose that, $Gal(f/\mathbb{Q}) \cong S_6$ and that the roots of $f$ are given by $\alpha_1, \ldots, \alpha_6$. How would one prove that $Gal(f/\mathbb{Q}(\alpha_1)) \cong S_5$?

Would this still hold for $Gal(f/\mathbb{Q}) \cong S_7$?

My thoughts

For the first case, I thought of applying the natural irrationalities theorem that tells that $Gal(KL/K) \cong Gal(L/K \cap L)$ with $K = \mathbb{Q}(\alpha_1), L = \mathbb{Q}(\alpha_2,\ldots,\alpha_6)$ and then the isomorphism gives $$Gal(E/\mathbb{Q}(\alpha_1)) \cong Gal(\mathbb{Q}(\alpha_2,\ldots,\alpha_6)/\mathbb{Q}(\alpha_1))$$ Since I realize that if I have 5 roots of $f$ then I have the sixth one .

Answer 1

$\DeclareMathOperator{Gal}{Gal}$ If $f\in F[x]$ is of degree $n$, denote $K$ as its splitting field, one of its root $\alpha$.

Note that $\Gal(K/F) \cong S_n$ implies $K/F$ is separable, $f$ is irreducible over $F$ and $[K:F] = n!$, hence $[F(\alpha):F] = n \implies [K:F(\alpha)] = (n-1)!$, so $|\Gal(K/F(\alpha))|= | S_{n-1}|$. Since $\Gal(K/F(\alpha))$ can be embedded into $S_{n-1}$, this shows that $\Gal(K/F(\alpha))\cong S_{n-1}$.