Galois group and solvable by radicals

Question

I came across the following problem in an old qualifying exam which states:

Show that the irreducible $h(x)\in \mathbb{Q}[x]$ is solvable by radicals if $[K:\mathbb{Q}]=25$ where $K$ is the splitting field of $h(x)$ over $\mathbb{Q}$.

My approach was to consider the field $\mathbb{Q}(\alpha)$ where $\alpha$ is a real root of $h(x)$. Now we get that $25=[K:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$. How do we proceed from here without knowing what the degree of $\alpha$ is? I want to eventually show that the Galois group is solvable.

Answer 1

A group with $25$ elements is abelian, hence solvable.