Galois group of $X^5-n$ In the following situation I want to find the Galois group of a specific polynomial:
Let $n > 1$ be a square-free integer, $f =X^5 - n \in \mathbb{Q}[X]$, $x:=\sqrt[5]{n}$ and $\zeta=\text{e}^{2 \pi i /5}$.
I could already show that $L=\mathbb{Q}[x,\zeta]$ is the splitting field of $f$ over $\mathbb{Q}$ with $[L:\mathbb{Q}]=20$ because $[\mathbb{Q}[x]:\mathbb{Q}]=5$ and $[\mathbb{Q}[\zeta]:\mathbb{Q}]=4$.
Now we define $x_j = x \zeta^j$ for $j=1,...5$ and want to determine the Galois group $G=\text{Aut}(L|\mathbb{Q})$ with the symmetric group $S_4$. Therefor we identify $\widetilde{G} \leq S_4$ with $G$. 
For $\sigma \in G$ we get an action on $S_4$ by defining $\sigma(x_j)=x_{\widetilde{\sigma}(j)}$ for $j=1,...,4$ where ${\widetilde{\sigma}} \in S_n$.
How can I show that $$ \widetilde{G}=\langle s,t \rangle\ \text{with} \ s=(12345),t=(1243). $$ 
 A: $\def\Q{\mathbb{Q}}$
Since you've stated that $\mathbb{Q}[x]$ is degree 5 over $\mathbb{Q}$, I'm assuming that $n$ is such that $p(x) = x^{5} - n$ is irreducible. 
In this case, $x_{1}, \ldots, x_{5}$ are the five roots of $p$. So, what you're asking is to essentially show that there exists automorphisms that just permute the first 4 roots cyclically and that permute all 5 roots cyclically. Once you have such automorphisms, by order considerations $\tilde{G}$ is what you want it to be.
So, first note that $\Q[x, \zeta]$ is also a Galois extension over $\Q[x]$ and is the splitting field of the irreducible polynomial $1 + X + X^{2} + X^{3} + X^{4}$ with roots $\zeta, \zeta^{2}, \zeta^{3}, \zeta^{4}.$ This we know because $\Q[x, \zeta]$ is degree 4 over $\Q[x]$ and $\zeta$ satisfies such a polynomial. 
Now, the Galois groups of such an extension acts transitively on the set of roots. Hence, there is an automorphism $\sigma$ of $\Q[x, \zeta]/\Q[x]$, which fixes $x$ and sends $\zeta$ to $\zeta^{2}.$
It is now easy to check that $\sigma$ acts as $(1, 2, 3, 4)$ because $x\zeta^{i}$ goes to $x \zeta^{i+1}$. So, we have one of the desired automorphisms.
To get the other automorphism, we need to repeat the above argument with the roles of $\zeta$ and $x$ reversed. Note that $\Q[x, \zeta]$ is also a Galois extension over $\Q[\zeta]$ of degree $5$ and is hence the splitting field of the polynomial $X^{n} - 5$ (as $x$ satisfies such a polynomial.) The roots of this polynomial are $x_{1}, \ldots, x_{5}$ (with $x = x_{5}$) and hence, as the Galois group acts transitively, we have an automorphism $\tau$ which fixes $\zeta$ and sends $x = x_{5}$ to $x_{1}$. Now, you can check using the fact that $\zeta$ is fixed that $\tau = (1, 2, 3, 4, 5).$
So, you have both the automorphisms you need.
