Let $p$ be a prime number and $a$ a positive integer that is not a p-th power. Let $L$ be the splitting field of the polynomial $x^p-a$ over $\mathbb{Q}$. Show that $Gal(L/\mathbb{Q})$ is isomorphic to the subgroup of the invertible $2$ by $2$ matrices with entries in $\mathbb{F}_p$,
\begin{bmatrix} 1 & r\\ 0 & s \end{bmatrix}
, where $r \in \mathbb{F}_p$ and $s \in \mathbb{F}_p^*$. (Hint: Consider $Gal(L/K)$ and $Gal(K/\mathbb{Q})$, where $K = \mathbb{Q}(\mu_p)$, the field of p-th roots of unity.)
Here is my progress so far, which has been verified as correct :
One can show that $Gal(L/K)$ and $Gal(K/\mathbb{Q})$, occuring as subgroups of $Gal(L/\mathbb{Q})$, are isomorphic to $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/(p-1)\mathbb{Z}$, respectively. Furthermore, one can show that $Gal(L/\mathbb{Q})$ is a group of order $p(p-1)$. Lastly, the subgroup $Gal(L/K) \cong \mathbb{Z}/p\mathbb{Z}$ can be shown to be a normal subgroup of $Gal(L/\mathbb{Q})$ using the Sylow Theorems.
Using the data above, one may believe that $Gal(L/\mathbb{Q})$ is isomorphic to the direct product $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/(p-1)\mathbb{Z}$. However, this is not possible -- if it were, this would mean that $Gal(L/\mathbb{Q})$ is abelian, but when $p = 3$ and $a = 2$, $Gal(L/\mathbb{Q})$ is in fact isomorphic to $S_3$, a non-abelian group. So I believe that $Gal(L/\mathbb{Q})$ must be the semi-direct product of $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/(p-1)\mathbb{Z}$. Nevertheless, we try our hand at constructing the isomorphism between $Gal(L/\mathbb{Q})$ and the specified matrix group below.
First, we can write down the $p(p-1)$ automorphisms occuring as elements of $G = Gal(L/\mathbb{Q})$. We note that $L$ is $\mathbb{Q}(a^{\frac{1}{p}}, \zeta)$, where $\zeta$ is a primitive p-th root of unity. Thus, a map $\phi \in G$ is determined by $\phi(a^{\frac{1}{p}})$ and $\phi(\zeta)$ : $\phi$ maps
$a^{\frac{1}{p}} \longmapsto \zeta^b a^{\frac{1}{p}} \quad (b = 0,1, ... , p-1)$
$\zeta \longmapsto \zeta^c \quad (c = 1, 2, ... , p-1)$
Now, let $\chi : G \longrightarrow G' : = \{ \begin{bmatrix} 1 & b\\ 0 & c \end{bmatrix} \quad | \quad c \in \mathbb{F}_p* , b \in \mathbb{F}_p \}$, where $\chi : \phi \longmapsto \begin{bmatrix} 1 & b\\ 0 & c \end{bmatrix}$.
I'd like to show the map $\chi$ is the desired isomorphism between $Gal(L/\mathbb{Q})$ and the given matrix group $G'$. It is certainly a bijective map between the two groups. However, I run into trouble when I verify the homomorphism property :
Let $\phi_1 $ map $a^{\frac{1}{p}} \longmapsto \zeta^{b_1} a^{\frac{1}{p}}$ and $\zeta \longmapsto \zeta^{c_1}$, and let $\phi_2 $ map $a^{\frac{1}{p}} \longmapsto \zeta^{b_2} a^{\frac{1}{p}}$ and $\zeta \longmapsto \zeta^{c_2}$. Then, we have
$\chi(\phi_1) \chi(\phi_2) = \begin{bmatrix} 1 & b_1\\ 0 & c_1 \end{bmatrix}$ $\begin{bmatrix} 1 & b_2\\ 0 & c_2 \end{bmatrix} = \begin{bmatrix} 1 & b_2 + b_1c_2\\ 0 & c_1c_2 \end{bmatrix}$.
On the other hand, since $\phi_1 \phi_2$ maps $a^{\frac{1}{p}}$ to $\zeta^{b_2c_1 + b_1}a^{\frac{1}{p}}$ and maps $\zeta$ to $\zeta^{c_1c_2}$ -- so $\chi(\phi_1)\chi(\phi_2)$ above will not quite agree with $\chi(\phi_1 \phi_2)$. Instead, it looks like $\chi$ is actually an anti-isomorphism.
How can I adjust my map $\chi$ to be the desired isomorphism ? Could it be that there is a mistake in the writing of the problem, and these two groups are actually anti-isomorphic ?
Thanks!