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.
 A: I have to write an answer, since as a comment it is harder to get the right display. To have an easy notation, let $a$ be $\sqrt[4]3$. We are working in the field
$$
K = \Bbb Q(a,i)\ .
$$
It is also good to have two generators of the Galois group written explicitly,
they act on the generators $i,a$ of $K$ as follows:

*

*$s$ (simpler to type than $\sigma$) of order four, that invariates $i$ and maps $a$ to $ai$,

*and $t$ (simpler to type than $\tau$) of order two, that maps $i$ to $-i$ and invariates $a$.
Then the conjugation $tst^{-1}=tst$ is mapping
$$
\begin{aligned}
tst(i) &=ts(-i)=t(-i)=i=s^{-1}(i)\ ,\\
tst(a) &=ts(a)=t(ai)= t(a)\cdot t(i)=a(-i)=s^{-1}(a)\ .
\end{aligned}
$$
The Galois group $G$ of $K$ over $\Bbb Q$ has thus the subgroup of order four, index two (thus normal), generated by the element of order four $s$.
And the action of the other generator realizes $G$ as a semiproduct with the (sub)group of order two generated by $t$. (To be precise, take an abstract copy of it instead.) So $G$ is isomorphic to the dihedral group $D_4$ with eight elements.


The question is but: Where is the mistake. The mistake is hidden in the notation. You describe the map $\sigma_{mn}$ with $m$ "working on $i$", it is either $1$ or $3$, and $n$ "working on a", it is among $0,1,2,3$.
The morphism $\sigma_{1n}$ is determined, it is in my notation above
$$\sigma_{1n}=s^n\ .$$
Now you have still to decide the following regarding the case $m=3$:
$$
\color{blue}{
\text{Is $\sigma_{3n}=s^nt$ or is it $\sigma_{3n}=ts^n$ ?}
}
$$
Fix one choice and redo the computation. (Although not needed for the initial problem, but needed to fix the displayed table.)

Also i do no understand how $\sigma_{11}=s$ works.
It maps $a\to ai\to ai^2=-a\to ai^3\to a$.
Given the order of the rows this is the permutation $1\to 3\to 2\to 4\to 1$, so in cycle notation $(1324)$.
A: 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:
$$
\begin{array}{c|c|c|c|c|c|c|c|c|c|}
&\text{root}&\sigma_{10}&\sigma_{11}&\sigma_{12}&\sigma_{13}&\sigma_{30}&\sigma_{31}&\sigma_{32}&\sigma_{33}\\\hline
1 & \sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3\\\hline
2 & i\sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3\\\hline
3 & -\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3& i\sqrt[4]3\\\hline
4 & -i\sqrt[4]3& \sqrt[4]3& \sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& i\sqrt[4]3& -\sqrt[4]3& -i\sqrt[4]3& \sqrt[4]3\\\hline
\text{permutation} && \operatorname{id}&(1234)&(13)(24)&(1432)&(24)&(12)(34)&(13)&(14)(23)\end{array}
$$
