I'm trying to determine the Galois group of $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$, where $\zeta=e^{2\pi i/5}$. I have the following information:
- $\mathbb{Q}(\sqrt[5]{3},\zeta)$ is the splitting field for $x^5-3$, whose roots are $\sqrt[5]{3}, \zeta \sqrt[5]{3}, \zeta^2 \sqrt[5]{3}, \zeta^3 \sqrt[5]{3},$ and $\zeta^4 \sqrt[5]{3}$.
- $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$ has degree 20.
- If $\sigma$ is an element of the Galois group, then $\sigma$ must send a root of $x^5-3$ to a root of $x^5-3$. So the Galois group is isomorphic to a subgroup of $S_5$ with order 20.
- $\sigma$ is completely determined by where it sends $\sqrt[5]{3}$ and $\zeta$ (the basis elements for $\mathbb{Q}(\sqrt[5]{3},\zeta):\mathbb{Q}$).
There is a unique subgroup of $S_5$ of order 20, which is $\langle (1\,2\,3\,4\,5),(1\,2\,4\,3)\rangle$, but without using this, how can I determine the Galois group structure?
Also, how can I show it's not abelian?
This is the Galois group of $x^5-3$. In general, the Galois group of $x^n-a$ is explicitly known, see here:
Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$
Proposition [Jacobson, Velez]: One has $G\cong \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^{\times}$ if and only if $n$ is odd, or $n$ is even and $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$.
Here $n=5$ is odd, so we know that $G\cong \Bbb Z/5\rtimes \Bbb Z/4$ is a solvable, non-abelian group of order $20$.