Let $E$ be a Galois extension of a field $F$. Suppose that the Galois group $Gal(E/F)$ is an abelian group of order $p^2$ which is not cyclic.
Then I want to show
$\exists K$ a proper subfield of $E$, such that $K=F(\alpha)$ for some $\alpha\in E$ when $K$ is a proper subfield of $E$,
and want to know
how many numbers of subfields $K$ such that $ F\subsetneq K\subsetneq E$
From group theory what I know is $\operatorname{Gal}(E/F) \simeq \mathbb{Z}_p \times \mathbb{Z}_p$ and it has following subgroup : $1, \mathbb{Z}_p, \mathbb{Z}_p\times \mathbb{Z}_p$. I am trying to see via Galois correspondence, but it makes me more complicated.
Best Answer
The existence of an intermediate field $K$ is guaranteed by the Galois correspondence. The fact that $K$ is generated by exactly one element over $F$, i.e $K = F(\alpha)$, is guaranteed from the fact that $[K:F] = p$.
The Galois correspondence gives you a different intermediate field for each subgroup of $\mathbb{Z}_p \times \mathbb{Z}_p$. So, to find the number of these intermediate fields, you just need to find the number of subgroups (non-trivial subgroups obviously). Moreover, any subgroup of $\mathbb{Z}_p \times \mathbb{Z}_p$ has order $p$ so it suffices to count the subgroups of order $p$.
Any two subgroups of order $p$ intersect trivially. There are $p-1$ non-identity elements in any subgroup, so the number of subgroups is given by $x$ where $(p-1)x = p^2 -1$. We conclude $x = p+1$. Therefore, there are $p+1$ intermediate fields.