I'm trying to find the Galois Group of a polynomial, while I wouldn't like to know the answer I'd like a push in the right direction if that's okay!
Let $P(X) = X^5 - 7 \in \Bbb Q[X]$. I'm aware that this has roots
$$\alpha_1 = 7^{1/5},\ \alpha_2 = 7^{1/5}\zeta_5,\ \alpha_3 = 7^{1/5}\zeta_5^2, \ \alpha_4 = 7^{1/5}\zeta_5^3,\ \alpha_5 = 7^{1/5}\zeta_5^4,$$
where $\zeta_5 = e^{2\pi i/5}$. I can see immediately that $P$ has splitting field $K \Bbb Q(7^{1/5}, \zeta_5)$, and that $P$ is a separable polynomial so this is a separable extension. Also, $[K:\Bbb Q] = 20$ so $\operatorname{Gal}(K/\Bbb Q)$ has $20$ elements.
I'm having trouble trying to find these elements though. A hint given to me was to try and find the minimal polynomial of $\alpha_1^2\alpha_2^3$ over $\Bbb Q$. I know this is a root of $X^5 - 7^5$, and this is reducible over $\Bbb Q$ as $X^5 - 7^5 = (X-7)(X^4 + 7X^3 + 49X^2 + 343X + 2401)$. Now, $\alpha_1^2\alpha_2^3$ does not satisfy $X - 7 = 0$ so it must be a root of the other factor. But now $m(X) = X^4 + 7X^3 + 49X^2 + 343X + 2401$ is Eisenstein over $\Bbb Q$ with prime $5$ after $X \mapsto X+2$, and is monic, so I think $m(X)$ is the minimal polynomial over $\Bbb Q$ of $\alpha_1^2\alpha_2^3$.
Now I can find all of the Galois Conjugates of $\alpha_1^2\alpha_2^3$ since these are just the roots of $m(X)$. These are $\alpha_1^2\alpha_2^3$, $\alpha_1^3\alpha_2^2$, $\alpha_1^4\alpha_2$, and $\alpha_1\alpha_2^4$. Now I'm having trouble seeing why this is useful for finding $\operatorname{Gal}(K/\Bbb Q)$. Clearly each of these elements is in $\Bbb Q(7^{1/5}, \zeta_5)$, so sending them to each other via automorphisms should comprise a set of elements of $\operatorname{Gal}(K/\Bbb Q)$, but I'm having trouble describing these automorphisms properly.
Thanks for the help!
The Galois automorphisms are determined by where they send $\newcommand{\al}{\alpha}\al=\sqrt[5]7$ and $\newcommand{\ze}{\zeta}\ze=\exp(2\pi i/5)$. Each can map $\al$ to any element of the set $\{\al,\ze\al,\ze^2\al,\ze^3\al,\ze^4\al\}$ and $\ze$ to any element of the set $\{\ze,\ze^2,\ze^3,\ze^4\}$.
To see the splitting field $K=\Bbb Q(\al,\ze)$ has degree $20$, note that $X^5-7$ is Eisenstein, so that $K$ contains a degree $5$ subfield, and the minimal polynomial of $\ze-1$ is also Eisenstein, so $K$ contains a degree $4$ subfield.
I don't see the relevance of the hint, though.