So I've been reading a proof of
Theorem : If E/K is a normal and radical field extension then Gal(E/K) is a solvable group.
The author then goes on a proof by induction but he shrugs off the base case saying it's trivial, while I've been trying to understand it.
Statement : If $ E = K(a_1, ..., a_n) / K $ is radical then Gal(E/K) is solvable for all $ n \in \mathbb{N} $
Base case : Let's show that $ E = K(a)/K $ is radical implies Gal(E/K) is solvable.
Without loss of generality we can suppose that $ p $ such that $, \; \: a^p \in K $ is prime.
If K contains the p-th roots of unity, then it follows that K(a)/K is a Kummer extension ( generated by the p-th root a ) and so Gal(E/K) is a cyclic group, so it's abelian which means it is solvable.
But my issue is that we don't have any hypothesis on whether K contains the p-th roots of unity or not. So how do we go about proving that Gal(E/K) is still an abelian group? We know that E/K is normal so the p-th roots of unity will be found in E.