Determine whether the Galois group of the polynomial $x^5-2$ over $\mathbb Q$ is cyclic or NOT 
Let , $G$ be a Galois group of the splitting field of $x^5-2$ over $\mathbb Q$. Then $G$ is cyclic or NOT.

Attempt:
Roots of the equation $x^5-2=0$ are $2^{1/5}$ , $2^{1/5}\xi$ , $2^{1/5}\xi^2$ , $2^{1/5}\xi^3$ , $2^{1/5}\xi^4$ , where , $\xi=e^{2\pi i/5}$.
So , the splitting field of the polynomial $x^5-2$ over $\mathbb Q$ is $\mathbb Q(2^{1/5},\xi)$. Now , $$\left|Gal\left(\mathbb Q(2^{1/5},\xi)/\mathbb Q\right)\right|=\left(\mathbb Q(2^{1/5},\xi):\mathbb Q\right)=5.4=20.$$


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*Now how we can show that this group is cyclic or NOT ?


*Is it directly depend on $n$ and $a$ ? for the polynomial $x^n-a$.


*I know the Galois group of the polynomial $x^n-1$ over $\mathbb Q$ is abelian , but I don't know about cyclic.

 A: Hints: 


*

*An element of the Galois group can be faithfully identified by its action on the set of roots. Therefore we can view this Galois group as a subgroup of $S_5$. Are there elements of order $20$ in $S_5$?

*Assuming that $a$ has no multiple prime factors you always get a cyclic group of order $n$ as the Galois group $Gal(\Bbb{Q}(\root n\of a,\zeta_n)/\Bbb{Q}(\zeta_n))$. This will be a normal subgroup of the Galois group of $x^n-a$  over the rationals. The argument is a bit longer, but it turns out that the Galois group is a semi-direct product, of order $n\phi(n)$, and never cyclic when $n>2$.

*You probably also know that this Galois group is isomorphic to the group of units of modular integers, $\Bbb{Z}_n^*$. In an earlier course you have probably seen results telling for which choices of $n$ is this group cyclic?

A: You can take a couple of group elements, and see whether they commute. E.g., there's an automorphism $\sigma$ fixing $\root5\of2$, and taking $\xi$ to $\xi^2$, and there's another automorhism $\tau$ taking $\root5\of2$ to $\xi\root5\of2$, and fixing $\xi$. Do you get $\sigma\tau=\tau\sigma$?
