What is the Galois group of the splitting field of $X^8-3$ over $\mathbb{Q}$? I've computed the splitting field of $x^8-3$ over $\mathbb{Q}$ to be $\mathbb{Q}(\sqrt[8]{3},\zeta_8)=\mathbb{Q}(\sqrt[8]{3},\sqrt{2},i)$, which is of degree 32 over $\mathbb{Q}$. 
The possible automorphisms are the maps fixing $\mathbb{Q}$ of form
$$
\sqrt[8]{3}\mapsto \zeta_8^i\sqrt[8]{3}\quad (0\leq i\leq 7),\qquad \sqrt{2}\mapsto\pm\sqrt{2},\qquad i\mapsto\pm i.
$$
There are 32 automorphisms, and thus these are all automorphisms. So I have an explicit description of the automorphisms in the Galois group $G$, but if I wanted to actually say what  $G$ is isomorphic to, how do I find that? I looked on groupprops subwiki, and there seem to be 51 groups of order 32 up to isomorphism, at least.
I made some little observations, like that there are 7 elements of order 2, but not sure how to actually classify the Galois group.
I've also noticed that the maps fixing $\sqrt[8]{3}$ will form a subgroup isomorphic to the Klein-4 group, and the maps fixing $\sqrt{2}$ and $i$ will form a cyclic subgroup of order 8. Does this narrow it down?
 A: Your description of $G$ is perfectly fine as it is.
But maybe a representation of $G$ in $GL_2(\mathbb{Z}/8\mathbb{Z})$, would be more to your taste :
if $\sigma \in G$ satisfies $\sigma(\zeta_8) = \zeta_8^a$ and $\sigma(\sqrt[8]{3}) = \zeta_8^b \sqrt[8]{3}$, it is represented by the matrix
$ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}$
You can observe that the automorphisms fixing $\sqrt[8]{3}$ are those with $b=0$, and they form a (non normal) subgroup $H$ isomorphic to $(\mathbb{Z}/8\mathbb{Z})^*$.
The automorphisms fixing $\zeta_8$ are those with $a=1$, and they form a normal subgroup $N$ isomorphic to $\mathbb{Z}/8\mathbb{Z}$, and $G$ can also be seen as a semi-direct product of those two groups :
In the exact sequence 
$0 \rightarrow N = Gal_{\mathbb{Q}(\zeta_8)}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) \rightarrow G = Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) \rightarrow G/N = Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8)) \rightarrow 0$
there is a section $s : Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8)) \rightarrow H \subset Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) $, defined simply by extending an automorphism $\sigma$ with $s(\sigma)(\sqrt[8]{3}) = \sqrt[8]{3}$.
$Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8))$ is isomorphic to $(\mathbb{Z}/8\mathbb{Z})^*$, so that the canonical surjection $f$ is simply picking $f(\sigma) = a$, and the section $s : (\mathbb{Z}/8\mathbb{Z})^* \rightarrow H$, 
is $s(a) = \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$
This gives an isomorphism between $G/N$ and $H$ and shows that $G$ is the semidirect product of $H$ acting on $N$
