Galois Group of $X^p - 2$, p-an odd prime

Galois Group of $X^p - 2$, p-an odd prime

Question is to Determine the Galois group of $x^p-2$ for an odd prime p.
For finding Galois group, we look for the splitting field of $x^p-2$ which
can be seen as $\mathbb{Q}(\sqrt[p]{2},\zeta)$ where $\zeta$ is a
primitive $p^{th}$ root of unity.
Consider $\mathbb{Q}\subset \mathbb{Q}(\zeta) \subset
\mathbb{Q}(\sqrt[p]{2},\zeta)$.we know that
$\mathbb{Q}(\sqrt[p]{2},\zeta)$ is Galois over $\mathbb{Q}(\zeta)$, we
find Corresponding Galois Group say $G_1$.
Consider $\mathbb{Q}\subset \mathbb{Q}(\sqrt[p]{2}) \subset
\mathbb{Q}(\sqrt[p]{2},\zeta)$. we know that
$\mathbb{Q}(\sqrt[p]{2},\zeta)$ is galois over $ \mathbb{Q}(\sqrt[p]{2})$,
we find Corresponding Galois Group say $G_2$.
Then Galois Group of $\mathbb{Q}(\sqrt[p]{2},\zeta)$ would possibly be
Product of these two subgroups $G_1$ and $G_2$ with some relation between
the generators.
For $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\zeta))$, consider
$\tau: \mathbb{Q}(\sqrt[p]{2},\zeta) \rightarrow
\mathbb{Q}(\sqrt[p]{2},\zeta)$ fixing $\zeta$ and sending $\sqrt[p]{2}
\rightarrow \sqrt[p]{2}\zeta$.
$\tau(\sqrt[p]{2})=\sqrt[p]{2}\zeta$,
$\tau^2(\sqrt[p]{2})=\tau(\tau(\sqrt[p]{2}))=\tau(\sqrt[p]{2}\zeta)=\tau(\sqrt[p]{2})\tau(\zeta)=\sqrt[p]{2}\zeta^2$,
For similar Reasons, $\tau^{p}(\sqrt[p]{2})=\sqrt[p]{2}\zeta^p=\sqrt[p]{2}$.
No power of $\tau$ less than $p$ gives identity as no power of $\zeta$
less than $p$ gives identity.
So, $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\zeta)) \cong
\mathbb{Z}_p \cong \big< \tau \big>$.
For $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\sqrt[p]{2}))$, consider
$\sigma : \mathbb{Q}(\sqrt[p]{2},\zeta) \rightarrow
\mathbb{Q}(\sqrt[p]{2},\zeta)$ fixing $\sqrt[p]{2}$ and sending $\zeta
\rightarrow \zeta^2$
$\sigma(\zeta)=\zeta^2$
$\sigma^2(\zeta)=\sigma(\sigma(\zeta))=\sigma(\zeta^2)=\zeta^{(2^2)}$
For similar reasons, $\sigma^{p-1}(\zeta)=\zeta^{(2^{p-1})}$, as for every
$a\in \mathbb{F}_p$, we have $a^{p-1}=1$ we have in particular $2^{p-1}
\equiv~1~mod~p$.
So, $\sigma^{p-1}(\zeta)=\zeta^{(2^{p-1})}=\zeta$ (as $\zeta$ is a
$p^{th}$ root of unity).
No power of $\sigma$ less than $p-1$ gives identity as $2\in \mathbb{F}_p$
generates Multiplicative group, no power of $2$ less than $p-1$ can be
equal to $1~mod~p$.
So, $Gal(\mathbb{Q}(\sqrt[p]{2},\zeta)/\mathbb{Q}(\sqrt[p]{2})) \cong
\mathbb{Z}_{p-1} \cong \big< \sigma\big>$.
As $[\mathbb{Q}(\sqrt[p]{2},\zeta):\mathbb{Q}]=p(p-1)$ and $|\sigma|=p-1$
and $|\tau|=p$ i strongly feel Galois group should be possibly generated
by $\sigma$ and $\tau$ with "Some extra related conditions"But not very
sure to confirm this.
I am not able to go any further, I can see that $\sigma$ and $\tau$ do not
commute with each other. I am unable to produce a know group which contain
isomorphic copies of $\mathbb{Z}_{p-1}$ and $\mathbb{Z}_p$ as subgroups.
I would be thankful if some one can help me out in this case.
Thank You.