How to construct Automorphisms in Galois Group I am studying Galois theory at the moment and having some serious troubles in calculating the Galois Groups in certain problems
For eg : I need to show to that $\mathbb{Q} \sqrt{2+ \sqrt2}$ is a cyclic quartic field of over $\mathbb{Q}$, meaning it is a cyclic Galois Group.
Now, the irreducible polynomial corresponding to this is  $x^4- 4x^2 +2$, this has $4$ roots
$\pm \sqrt{2 \pm \sqrt{2}}$
Now, I know that the Galois Group will contain $4$ elements and to prove that it is cyclic it is enough to produce an automorphism of order $4$.
Beyond this stage I am totally lost, I don't really get how exactly do we define the automorphisms in these kinds of problems, although the questions has been answered here Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$ still I don't get why did we define $f(\sqrt{(2+ \sqrt2)}) =  \sqrt{2 -\sqrt2}$ only. Why can't it map to any other element?
It would be very helpful if someone can clear these doubts.
Thank You.
 A: How many choices do you have to map $\sqrt{2+\sqrt{2}}$ ? Well you obviously have the identity map $1$.
You can map it to $-\sqrt{2+\sqrt{2}}$. In that case $\sigma : \sqrt{2+\sqrt{2}}\mapsto -\sqrt{2+\sqrt{2}}$ .
$\sigma^{2}(\sqrt{2+\sqrt{2}})= -\sigma(\sqrt{2+\sqrt{2}})=\sqrt{2+\sqrt{2}}\implies \sigma^{2}=1$ (identity).
So $\sigma^{2}=1$ is of order hence $|\sigma|=2$
So this cannot generate a cyclic group of order $4$.
Also you can map $\tau(\sqrt{2+\sqrt{2}})=-\sqrt{2-\sqrt{2}}$.
In that case again following what it did in the linked answer you can prove that $|\tau|>2$.
The key fact is that there are only two groups of order $4$ upto isomorphism. The Klein 4 group which is not cyclic ( hence every element is of order $2$) and the Cyclic group of order $4$. Since you have already found an element of order $>2$. You can conclude that $Gal(\Bbb{Q}\sqrt{2+\sqrt{2}}/\Bbb{Q})\cong \frac{\mathbb{Z}}{4\mathbb{Z}}$ .
Also it is first critical to prove that $\Bbb{Q}\sqrt{2+\sqrt{2}}$ is the splitting field of the polynomial $x^{4}-4x^{2}+2$ to at all conclude that the extension is indeed Galois . To do that you need to do as the answer in the link does .
$$\frac{\sqrt 2}{\sqrt{2+\sqrt{2}}}=\frac{\sqrt 2 \cdot\sqrt{2-\sqrt 2}}{\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt 2}}=\frac{\sqrt 2\cdot\sqrt{2-\sqrt 2}}{\sqrt{4-2}}=\sqrt{2-\sqrt 2}$$
And say that all roots of the polynomial are in this extension.
