find the Galois group of $x^4-3$ over $\mathbb{Q}$ and show that is isomorphic to $D_4$ I have to determine the Galois group of $x^{4}-3$. If I have not made any mistakes, then the Galois group is:
$$Gal(\mathbb{Q}(\sqrt[4]{3},i)/\mathbb{Q})=\{ \sigma_{mn}\in Aut(\mathbb{Q}(\sqrt[4]{3},i)/\mathbb{Q})\mid \sigma_{mn}(i) = i^m, m=\{1,3\}, \sigma_{mn}(\sqrt[4]3)=i^n \sqrt[4]{3}, n=\{0,1,2,3\} \}$$
I would like to calculate all elements of the Galois group in the following table:

I think I have made a mistake. Because the group $D_4$ has the following elements $D_4 = \{id,(1234),(13)(24),(1432),(24),(13),(12)(34),(14)(23)\}$. When I compare this with the last line, the permutations are not the same. So I have a mistake there.
Does anyone see my mistake? Thanks a lot.

I think it will be easier to construct $D_4$ by thinking about generators rather than all elements together. It is known that $D_4$, as the symmetry group of the square, is generated by a flip and a $90^\circ$ rotation. Do we have something like that here? Yes, we do.
For the flip, the most likely candidate is the complex conjugate. For a rotation, we would like $i^n\sqrt[4]3\mapsto i^{n+1}\sqrt[4]3$, and there is an element in the Galois group which does this (you have called it $\sigma_{11}$).
We see that these two elements together act on the square defined by the four roots of the polynomial exactly the way we would like $D_4$ to act. Together with an argument for why the Galois group can't have more than 8 elements, and we are done.
If we want to follow your approach to the end, though, we can do that too. But we need to be careful to not make any mistakes the way you seem to have done (for instance, your $\sigma_{11}$ is tagged as $(1432)$ but really it's $(1324)$ and your $\sigma_{13}$ really corresponds to $(1423)$). Also, ordering the roots along the square rather than in diagonal pairs will help the correspondence with your standard $D_4$ presentation:
$$
\begin{array}{c|c|c|c|c|c|c|c|c|c|}
&\text{root}&\sigma_{10}&\sigma_{11}&\sigma_{12}&\sigma_{13}&\sigma_{30}&\sigma_{31}&\sigma_{32}&\sigma_{33}\\\hline
1 & \sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3\\\hline
2 & i\sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3\\\hline
3 & -\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3\\\hline
4 & -i\sqrt[4]3& \sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3\\\hline
\text{permutation} && \operatorname{id}&(1234)&(13)(24)&(1432)&(24)&(12)(34)&(13)&(14)(23)\end{array}
$$
