Question is to determine the Galois group of $x^p-2$ for an odd prime $p$.
For finding the 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 $a\in \mathbb{F}_p^\times$, 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.
Best Answer
Here it might be best to realize the Galois group as a group of permutations of the roots of the polynomial of interest (as in user64494's comment under the OP). You have already observed that the roots of $x^p-2$ are $x_i=\zeta_p^{i-1}\root p\of 2, i=1,2,\ldots,p.$ You also know that the splitting field is of degree $p(p-1)$, so that is also the order of the Galois group.
Let us consider the action of your automorphism $\tau$ defined by $\tau(\zeta_p)=\zeta_p$ and $\tau(\root p\of 2)=\zeta_p\root p\of 2$. So we see that $\tau(x_i)=x_{i+1}$, if $i<p$, and $\tau(x_p)=x_1$. The action of $\tau$ on the chosen indexing of the roots thus corresponds to the $p$-cycle $\tau=(123\cdots p).$
On the other hand the automorphism $\sigma_a:x_1\mapsto x_1, \zeta_p\mapsto \zeta_p^a,1\le a<p,$ keeps the real root $x_1$ fixed, and permutes the others according to the rule $x_i=x_1\zeta_p^{i-1}\mapsto x_1\zeta_p^{a(i-1)}=x_{1+a(i-1)}$, where the subscript is calculated modulo $p$. You see that all these share $x_1$ as a fixed point (this was also clear from your construction of $\sigma$:s as elements of the Galois group $G_2=Gal(\mathbb{Q}(x_1,\zeta_p)/\mathbb{Q}(\zeta_p))$.
You can either look at all these automorphisms as elements of $S_p$. This works beautifully, once you have found a generator of $G_2$. This is equivalent to finding a generator of the multiplicative group $\mathbb{Z}_p^*$, i.e. a primitive root. There is no general formula for such a generator, so I won't say much about that (this may be a cause of your difficulties). We simply know that one exists! But if $p$ is fixed, say $p=5$ or another smallish prime, then I recommend this way, as you can easily calculate with permutations.
PVAL is strongly hinting at the possibility that you may get a semi-direct product of $G_1$ and $G_2$. Indeed, you will see that one of the two subgroups is stable under conjugation by elements of the other. Which way does it work? I'm a mean dude and won't tell you! But you do remember that the fixed field of a normal subgroup is itself Galois over the base field. So which of the fields $\mathbb{Q}(\zeta_p)$ or $\mathbb{Q}(\root p\of2)$ is Galois over the rationals? The group of automorphisms associated to that field should be a normal subgroup of the big Galois group. After figuring that out, you can start studying the effect of either $\sigma_a\tau\sigma_a^{-1}$ or $\tau\sigma_a\tau^{-1}$ all according to which feels more interesting...