Subgroup of Galois group of polynomial over $\mathbb{Q}$ Let $K$ be the splitting field of $x^5-3 \in \mathbb{Q}[x]$.
We can see $K = \mathbb{Q}(3^{1/5}, \zeta_5)$ where $\zeta_5 = e^{2 \pi i/5}$, and $[K: \mathbb{Q}] = 20$. It's easy to see $\sqrt{5} \in \mathbb{Q}(\zeta_5)\subseteq K$.
Let $H= \operatorname{Gal}(K/\mathbb{Q}(\sqrt{5})$). Does $H$ have to be abelian?
Edit:
The Galois group of the big extension $K/\mathbb{Q}$ is the group generated by $\sigma$ (order 5 element) and $\tau$ (order 4 element) defined by
$\sigma: 3^{1/5} \mapsto 3^{1/5} \zeta_5$ and $\tau: \zeta_5 \mapsto \zeta_5^2$.
I don't know what this group is, but it's definitely not abelian since if it's abelian the extension $\mathbb{Q}(3^{1/5})/\mathbb{Q}$ is Galois. 

The Galois group is of order 20, generated by $x$ of order 5 any $y$ of order 4, where $yxy^{-1}=x^{3}$. Its the Frobenius group of order 20. The subgroup of order 10 is generated by $x$ and $y^2$. It is a dihedral group.
