Solubility of a Galois Group going over some past papers with no answers and would like a bit of help if possible..
I've shown that for p a prime number then $x^p-1 \in K[x]$ is abelian where K is a subfield of $\mathbb{C}$.  I've now been asked to show that the Galois group of $x^p-a$ over K is soluble with $a \neq 0 \in K$.  I know any abelian group A is soluble, since ${1} \triangleleft  A$ is a subnormal series with its only factor A being abelian, so $x^p-1 \in K[x]$ is certainly soluble.  Not sure where to go from here and the question is worth quite a lot so guessing there is quite a bit more to do, any help appreciated.
 A: Hint: Let $c$ be a root of $X^p - a$ and let $\zeta$ be a primitive $p$-th root of unity. What can you say about $c, \zeta c, \dots, \zeta^{p-1} c$?
Now you can build up the splitting field of $X^p - a$ in two steps, by first adjoining $\zeta$ and then adjoining $c$. So you get fields $K\subset K(\zeta) \subset K(\zeta, c)$ and $K(\zeta, c)/K$ is Galois.
You already know what $\mathrm{Gal}(K(\zeta)/K)$ looks like (or at least that it is abelian). You should also be able to figure out the structure of $\mathrm{Gal}(K(\zeta, c)/K(\zeta))$. (One can write it down explicitely)
What is left to do: We need to find some sort of relationship between $\mathrm{Gal}(K(\zeta, c)/K)$, $\mathrm{Gal}(K(\zeta, c)/K(\zeta))$ and $\mathrm{Gal}(K(\zeta)/K)$.

Useful facts:

*

*Any subgroup of a solvable group is solvable.

*A finite product of solvable groups is solvable.

*If $N \lhd G$ is normal, then $G$ is solvable $\iff$ $G/N$ and $N$ are solvable.


Further Hint (although I feel this might be giving away too much):

  For instance you could look at $$\quad \mathrm{Gal}(K(\zeta, c)/K) \mapsto \mathrm{Gal}(K(\zeta)/K), \quad \sigma \mapsto \sigma|_{K(\zeta)} \quad $$ and use some isomorphism theorem.

