# Quotient of the Galois group of a splitting field generated by two roots

abstract-algebragalois-theorygroup-theorysylow-theory

Let $$f \in \mathbb{Q}[x]$$ be irreducible of degree $$p$$, where $$p$$ is a prime. Let $$K$$ be the splitting field of $$f$$ and suppose that there are roots $$\alpha$$ and $$\beta$$ of $$f$$ such that $$K = \mathbb{Q}(\alpha,\beta)$$. We regard the Galois group $$G = G(K/\mathbb{Q})$$ as a subgroup of the symmetric group $$S_p$$. It is not hard to show that $$p$$ divides the order of $$G$$ and $$G$$ contains a $$p$$-cycle $$\sigma$$. Let $$H$$ the group generated by $$\sigma$$, one can use Sylow's third theorem to argue that $$H$$ is normal in $$G$$. Now, I'd like to show that $$G/H$$ is a cyclic group of order diving $$p-1$$. I'm trying to use results from cyclic extension but it is required that the base field contains the $$n$$-th roots of unity, which is not true in this case. Any help would be appreciated.

Let $$N$$ and $$C$$ be the normalizer and centralizer of $$H$$ in $$S_p$$, respectively. Then $$G$$ is a subgroup of $$N$$ since $$H$$ is normal in $$G$$. Therefore we can establish your result by showing $$N/H$$ is a cyclic group whose order divides $$p-1$$.
Since $$H$$ is cyclic of order $$p$$, it follows that $$C = H$$ and that $$\text{Aut}(H)$$ is cyclic of order $$p-1$$. The result now follows since $$N/C$$ is isomorphic to a subgroup of $$\text{Aut}(H)$$, which is a general property of normalizers and centralizers.