# $\operatorname{Gal}({\mathbb Q(i,\sqrt[4]{2})}/{\mathbb Q})$

I'm studying $\operatorname{Gal}({\mathbb Q(i,\sqrt[4]{2})}\Big/{\mathbb Q})$; I find it has order $8$. Any non-trivial subgroup will have order $2$ or $4$. Subgroups of order $2$ are easy to find, but how can I find quickly the subgroups or order $4$ ?

In a more general manner, is there a way to know the number of subgroups of order $k$ in a group of order $n$ ? (if $k\mid n$ of course). I know the second Sylow theorem which says that the number $n_p$ of $p$-Sylows of a group $G$, where $\operatorname{Card}(G)=p^ns$ must divide $s$, and be $\equiv1 \space \pmod{p}$.

Are there any further results that help to know the number of subgroups of order $k$, when $k$ is not prime ?

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No, there is no general way to know that, though there are partial results here and there. For example, if the group is abelian finite then it has at least one subgroup of order each divisor or the group's order, and the group's cyclic iff it has one unique such subgroup (of each divisor). – DonAntonio Dec 31 '13 at 16:44
Aside: while it does have 8 elements, it is not isomorphic to $(\mathbb{Z} / 2 \mathbb{Z})^3$ -- one way to see this is that this group does not have any subgroups of order 4, but the Galois group does! – Hurkyl Dec 31 '13 at 16:49
Thanks. Then it must be isomorphic either to $\mathbb Z/8\mathbb Z$, or to $\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$. How can I know ? – Asinus Dec 31 '13 at 16:54
@Asinus: It might be a nonAbelian group too -- my memory is a but fuzzy, but I think (without doing any computation) it is that it's actually the dihedral group with 8 elements. The only way I know for sure is by tedious calculation: e.g. find a presentation for the Galois group. – Hurkyl Dec 31 '13 at 16:57
It can't be the former because there are two different subgroups of order $4$, corresponding to $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{2})$. – Michalis Dec 31 '13 at 16:57

The group is generated by an element $a$ which takes $i$ to $-i$ while fixing $\root4\of2$, and an element $b$ which fixes $i$ and takes $\root4\of2$ to $i\root4\of2$. $a^2=1$, $b^4=1$. See if you can prove $ab=b^{-1}a$. This will enable you to list all the elements of the group, and find all of its subgroups.