Intermediate fields of $X^p - 2 $ I've been working on an exercise I have to do for my algebra course.
Exercise:
Let p be prime and $L$ the splitting field of $ f = X^p - 2$ over $\mathbb{Q}$.
a) Show that $ Gal(L/\mathbb{Q})$ is isomorphic to the subgroup
$$ H = \left \{
\begin{pmatrix}
a && b  \\
0 && 1
\end{pmatrix} \; \rvert \; a \in \mathbb{F}_p^* , b \in \mathbb{F}_p \right \} \subseteq GL_2(\mathbb{F}_p) $$
b) Determine all intermediate fields $E$ of $L/\mathbb{Q}$ such that $ [ E : \mathbb{Q} ] = p - 1$.
I was able to do a) using that f is irreducible and we can get the degree of the extension by looking at:
$$ \mathbb{Q} \subseteq \mathbb{Q}(\sqrt[p]{2} \subseteq \mathbb{Q}(\sqrt[p] 2 , \xi) = L $$
Where $\xi$ is the p-th root of unity. We can now see that $[L : \mathbb{Q} ] = p(p-1) $ since $\xi \notin \mathbb{Q}(\sqrt[p] 2) $.
Therefore $ |Gal(L / \mathbb{Q})| = p(p-1) = |H| $.
I was now able to show that these groups are isomorphic. I did it by using that all automorphisms $\tau$ are given by 
$$ \xi \mapsto \xi^k \; , \; k \in [p-1] \\
\sqrt[p] 2 \mapsto \xi^j \cdot \sqrt[p] 2 \; , \; j \in [p] $$
and we can easily declare an isomorphism through $ j $ and $ k$.
With b) I got stuck. I am not able to match subgroups for these intermediate fields $E$.
How can I do that?
 A: $\newcommand{\Size}[1]{\lvert #1 \rvert}$$\newcommand{\Gal}{\mathrm{Gal}}$$\newcommand{\Q}{\mathbb{Q}}$Let $L'= H$ be the subgroup of $G = \Gal(L/\Q)$ corresponding to $L$. Then
$$
\Size{H} = \Size{H : \{ 1 \}} = \Size{ \{1\}' : H'} = \Size{L : E} = \frac{\Size{L : \Q}}{\Size{E : \Q}} = \frac{p(p-1)}{p-1} = p.
$$
Now show that $G$ has a unique subgroup of order $p$.
Here I am writing $X'$ for the subgroup of $G$ corresponding to the intermediate field $X$, and also for the intermediate field corresponding to the subgroup $X$.
To show that there is a unique subgroup of order $p$, just note that if $a \ne 1$, then
$$
\begin{bmatrix}
a && b  \\
0 && 1
\end{bmatrix}^{p}
=
\begin{bmatrix}
a^{p} && (a^{p-1} + a^{p-2} + \dots + a + 1) b  \\
0 && 1
\end{bmatrix}
=
\begin{bmatrix}
a && b  \\
0 && 1
\end{bmatrix},
$$
so a matrix of order $p$ must have $a = 1$. (While for $a \ne 1$ the matrix has order dividing $p-1$.)
Here I have used that if $a \ne 1$, then
$$
a^{p-1} + a^{p-2} + \dots + a + 1
=
\frac{a^{p} - 1}{a - 1} = \frac{a -1}{a-1} = 1.
$$
