Given the polynomial $x^4-2$ we know that a splitting field is $\Bbb Q(i,\sqrt[4]{2})$ and that the Galois group is given by $D_8=\langle r,s|r^4=s^2=1,srs=r^{-1}\rangle$. Now take the subgroup $\langle r^2 \rangle$ for instance. It's corresponding fixed field is $\Bbb Q(i, \sqrt[4]{2})$. Now for a long time I had been under the impression that this meant $\langle r^2\rangle$ was the subgroup of $D_8$ isomorphic to $Gal(\Bbb Q(i,\sqrt[4]{2})/\Bbb Q)$. The reason being that I misunderstood the definition of a Galois extension. But after rereading my college notes and in particular looking over the fundamental theorem of Galois I'm sure the subgroup of automorphisms of the Galois group corresponding to a fixed fields are not the same as the Galois groups themselves. Here is some work I did realising that and I was hoping someone could hopefully look over it and verify it and tell me if there are many mistakes:

So We have our splitting field $\Bbb Q(\alpha,i)$ ($\alpha=\sqrt[4]{2}$), and we know it's Galois group is isomorphic to $D_8$. The subgroups and corresponding fixed fields are given by :

$\langle r \rangle$ which fixes $\Bbb Q(i)$

$\langle r^2,rs \rangle$ which fixes $\Bbb Q(i\sqrt{2})$

$\langle r^2,s \rangle$ which fixes $\Bbb Q(\sqrt{2})$

$\langle s \rangle$ which fixes $\Bbb Q(\sqrt[4]{2})$

$\langle r^2s \rangle$ which fixes $\Bbb Q(i\sqrt[4]{2})$

$\langle r^2 \rangle$ which fixes $\Bbb Q(i,\sqrt{2})$

$\langle rs \rangle$ which fixes $\Bbb Q((1+i)\sqrt[4]{2})$

$\langle r^3s \rangle$ which fixes $\Bbb Q((1-i)\sqrt[4]{2})$

We know that $D_8$ has order $2^3$ every group with prime order to a power is nilpotent therefore all maximal subgroups are normal and so we can immediately assume that $\langle r \rangle$ $ ,\langle r^2,s \rangle$ $\langle r^2,rs \rangle $ are normal ( we could also have used the fact that any subgroup of index two is normal). We also know that there intersection is normal so $\langle r^2 \rangle$ must be normal .

The rest are all checked by using generating sets E.G. If A is a generating set for G and B is a generating set for H then we must check if $bab^{-1}, b^{-1}ab$ are still in H for all a and b. We see there are no other Normal subgroups

So we have all of our normal subgroups and we're ready to start using the fundamental theorem of Galois Which states that if $k/F$ is Galois and the group of automorphisms which fix K is normal then the Galois group of $k/F$ is $Gal(K/F)\cong G/H$.

So take $\langle r \rangle$ which fixes $\Bbb Q(i)$,

$D^8/\langle r \rangle= \{ r \langle r \rangle\,s\langle r \rangle \}$ as we only need to consider how generators will act on $\langle r \rangle$.

So $D^8/\langle r \rangle= \{ \{r,r^2,r^3,e\} , \{s,sr,sr^2,sr^3 \} \}$.

Now this part I'm really unsure of but I think that we can say because the first set is still just $\langle r \rangle$ and the second set has nothing but elements of order two this must be isomorphic to $C_2$ ( any advice in this aspect would be very much appreciated)

So We've found that the Galois group of $Q(i)$ is $C_2$ as would be expected from complex conjugation.

The other subfields correspond to groups which aren't normal and so they themselves are not normal as extensions and so are not Galois extensions .It doesn't seem like there is a way to find there Galois groups given in the fundamental theorem. I want to say perhaps It's because their Galois groups are trivial but I'm not sure.. I think it could be possible that we adjoin say two roots of a polynomial with 3 roots and then get a Galois group of size two which is not normal but maybe I'm wrong. If there is a way to find Galois groups for non-normal extensions then what is it ?


If $G$ is a finite subgroup of $Aut(K)$ and $H$ a subgroup of $G$ then $K/K^G$ is Galois with Galois group $G$ and $$Aut(K^H/K^G) = \{ gH \in G /H, H = gHg^{-1}\} = N_G(H) / H$$

(the normalizer subgroup)

Proof : $g(K^H) = K^{g H g^{-1}}$ so $g \in Aut(K^H)$ iff $K^H = K^{g H g^{-1}}$ iff $H = g H g^{-1}$.


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