I've been studying for an algebra qualifying exam. Any help with the following result would be appreciated.
Suppose $E$ is a splitting over $\mathbb{Q}$ of an irreducible polynomial $f(x)\in\mathbb{Q}[x]$. Assume $\deg(f(x))=p+1$ where $p$ is prime. Show that if $[E:\mathbb{Q}]=p(p+1)$ then there exists $\beta\in E$ such that $\mathbb{Q}(\beta)/\mathbb{Q}$ is Galois and $$ \mathrm{Gal}(\mathbb{Q}(\beta)/\mathbb{Q})\cong \mathbb{Z}/p. $$ Attempt: If $M/\mathbb{Q}$ is an extension with $M\subseteq E$ then $M/\mathbb{Q}$ is finite and separable. So by Artin's primitive element theorem, it suffices to construct a Galois extension $M/\mathbb{Q}$ with $M\subseteq E$ and $$ \mathrm{Gal}(M/\mathbb{Q})\cong \mathbb{Z}/p. $$ By The Fundamental Theorem of Galois Theory, such an extension exists if and only if $\mathrm{Gal}(E/\mathbb{Q})$ contains a normal subgroup of order $p+1$.
Fact: If $G$ is a group with $|G|=p(p+1)$, then $G$ contains a normal subgroup of order $p$ or $p+1$.
So it's enough to show that $\mathrm{Gal}(E/\mathbb{Q})$ does not contain a normal subgroup of order $p$. This is where I am stuck. My idea was that any subgroup of order $p$ is $p$-Sylow and so by Sylow's 2nd theorem I could just show that there are at least two subgroups of order $p$ in $\mathrm{Gal}(E/\mathbb{Q})$. Subgroup of order $p$ in $\mathrm{Gal}(E/\mathbb{Q})$ correspond to intermediary field extensions $M/\mathbb{Q}$ of degree $p+1$. I've tried showing there are at least two such intermediary field extensions but I haven't been able to do it.
If $\gamma$ is a root of $f$, then $|\mathbb{Q}(\gamma):\mathbb{Q}| = p+1$.
If ${\rm Gal}(E/\mathbb{Q})$ has a normal subgroup of order $p$, then it would be the unique subgroups of order $p$, so by the Galois correspondence, there would be a unique subfield $M$ of $E$ of degree $p+1$ over $\mathbb{Q}$, and it would be normal over $\mathbb{Q}$.
But then we would have $M = \mathbb{Q}(\gamma)$ and so $M$ would be a splitting field of $f$, contradiction.