I'm trying to calculate the Galois group of $\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}$. I'm not sure how to proceed since this is not a normal extension.
The normal closure of this is $\mathbb{Q}(\sqrt[5]{3},\zeta)$ where $\zeta=e^{2\pi i/5}$, and $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$ has degree 20. There is a unique subgroup of $S_5$ of order 20, so Gal$(\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q})$ is isomorphic to that group. I don't know if this is useful since $\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}$ is not normal. How do I calculate the Galois group?
Thanks for any help in advance.
The Galois group $G$ of $X^5-3$ is the Galois group of the extension $\Bbb Q(\sqrt[5]{3},\zeta_5)$ over $\Bbb Q$, which has degree $20$ and is a transitive subgroup of $S_5$. The only such group is the Frobenius group $Fr_{20}$, see here. This is the Galois group you are looking for, since the extension $\mathbb{Q}(\sqrt[5]{3})$ over $\Bbb Q$ is not normal.
References:
Galois group of solvable quintic is subgroup of Fr20
Galois group of $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$