Galois Theory.Subgroups of Galois Group

Question

How to List the subgroups of the Galois group in general.Im not interested in a specific Example but to make it easier. Supposes the galois group $G=Gal[Q(v,i):Q]$ $$v= \sqrt[4]{2}$$ I know how to find the possible Automorphisms (elements of the Galois group using the fact that is $a$ is a root of an irreducible polynomial its $(σa)$ is also a root (σ) is an automorphism) but how do i list its subgroups.And how to know that i have listed them all and im not missing one?

Answer 1

if we know the Galois group, this means we know its structure, in this case it is a problem of group theory and not of Galois theory, in this case we can, as indicated in the commentary, to proceed by recognizing growing small subgroups while the group is not already indicated in a classification list!!!. But it seems of interest to you to use Galois theory for a Galois extension $ K / \Bbb{Q} $ of degree $ n $; in this case, $K = \Bbb {Q} (\alpha)$, a relevant information is that the group $Gal (K / \Bbb{Q})$ is a group of order $n$ and is isomorphic to a transitive subgroup of $ S_{n}$. So

1) it is essential to have a primitive element $ \alpha $ of $K$ and know all its conjugates (the roots of $ irr (\alpha, K / \Bbb{Q}) $, say $ \alpha = \alpha_1, \alpha_2 \cdot\cdot\cdot \alpha_n$.

2) Learn the different divisors $ m $ of $ n $, to avoid trivial subgroups keep only $ m \in \{2, \cdot\cdot\cdot, n-1 \} $.

3) form the different product $P= (X- \alpha_1) (\prod(X -\alpha_{i_1}) \cdot\cdot\cdot (X -\alpha_{i_r}))$ with $ i_1 + \cdot\cdot\cdot+ i_r = m-1$ and $i_j\neq i_l)$ search the field generated by $\Bbb{Q }$ and the coefficients of the polynomial $P$ in 3)

4) The Galois correspondence determines the subgroups as required.

For the given example: a primitive element is $ ^4\sqrt{2} + i$, whose conjugates are in number $ 8$. Then the Galois group is of order 8, as we may see as compositor of the real extension $\Bbb{Q} (^4\sqrt{2})$ with $\Bbb{Q}(i)$. Moreover, he identifies with a transitive subgroup of $ S_4 $,that is a 2-Sylow of $S_4$, hence a dihedral group generated by a 4-cycle and a suitable transposition. In this stage we can lock to the classification list (Group Theory) and determined the sub groups of the dihedral group $D_4 = \{e, a, a^2, a^3,b, ab, ab^2, ab^3 \} $, there are:

sub-groups of order 4: $ \{e, a, a^2, a^3 \}, \{e, a^2, b, ba^2 \} , \{e, a^2, ba, ba^3 \} $ of index 2, so normal in $D_4$.

sub-groups of order 2: $ \{e, a^2 \}, \{e, b \}, \{e, ba\},\{e, ba^2\}, \{e,ba^3 \} $

others of order 1 and 8 are obvious.

so then if we need to determine the intermediate extensions, the Galois correspondence is applied after choosing well isomorphism between Galois group and D_4.

Returning to the determination of these subgroup by Galois theory, ie by means of the intermediate extensions of $K / \Bbb{Q}$. by the method described above.

$\alpha = ^4\sqrt{2} + i $, start by $(\alpha-i)^4=2$ to arrive at $irr(\alpha,\Bbb{Q})=X^8-16X^6+28X^2+1$, i know the roots: $^4\sqrt{2} + i$ , $^4\sqrt{2} - i$,$-^4\sqrt{2} + i$ and $-^4\sqrt{2} + i$, $i^4\sqrt{2} + i$ , $i^4\sqrt{2} - i$,$-i^4\sqrt{2} + i$ and $-i^4\sqrt{2} + i$, we form all product of two factors s.t (X-\alpha) divide the product $(X-\alpha) (X-^4\sqrt{2} + i) (X-\alpha) (X+^4\sqrt{2} - i) (X-\alpha) (X+^4\sqrt{2} + i) (X-\alpha) (iX-^4\sqrt{2} - i) (X-\alpha) (iX-^4\sqrt{2} + i) (X-\alpha) (X+i^4\sqrt{2} - i) (X-\alpha) (X+i^4\sqrt{2} + i)$ and we form all product of four factors s.t (X-\alpha) divide the product

$(X-(\sqrt[4]{2}+i))(X-(\sqrt[4]{2}-i))=X^2-2X\sqrt[4]{2}+\sqrt{2}+1$

$(X-(\sqrt[4]{2}+i))(X-(-\sqrt[4]{2}-i))=X^2-\sqrt{2}-2i\sqrt[4]{2}+1$

$(X-(\sqrt[4]{2}+i))(X-(-\sqrt[4]{2}+i))=X^2-2iX-\sqrt{2}-1$

$(X-(\sqrt[4]{2}+i))(X-(i\sqrt[4]{2}-i))=X^2-iX\sqrt[4]{2}-X\sqrt[4]{2}+i% \sqrt{2}-i\sqrt[4]{2}-\sqrt[4]{2}+1$

$(X-(\sqrt[4]{2}+i))(X-(i\sqrt[4]{2}+i))=X^2-iX\sqrt[4]{2}-2iX-X\sqrt[4]{2}+i% \sqrt{2}+i\sqrt[4]{2}-\sqrt[4]{2}-1$

$(X-(\sqrt[4]{2}+i))(X-(-i\sqrt[4]{2}+i))=X^2+iX\sqrt[4]{2}-2iX-X\sqrt[4]{2}% -i\sqrt{2}+i\sqrt[4]{2}+\sqrt[4]{2}-1$

$(X-(\sqrt[4]{2}+i))(X-(-i\sqrt[4]{2}-i))=X^2+iX\sqrt[4]{2}-X\sqrt[4]{2}-i% \sqrt{2}-i\sqrt[4]{2}+\sqrt[4]{2}+1$

we get the field generayed with this polynoms....

This is the first time I do this kind of calculation, it is tiring but we can always try.

Note that can be helpful is that the third polynomial generates the Galois extension $\Bbb{Q}(i,\sqrt{2})$ which corresponds to the normal sub group $\{e,a^2\}$ of $D_4$. sorry for my English and Good luck