In order to make a smaller example for my question Galois group of the field of all constructible complex numbers, I am posing this new question. I know already, that E is a galois extension of $\mathbb{Q}$ of dimension 16, it being a galois extension of the splitting field $L$ of the polynomial $X^4-2$ of dimension 2, since $E=L(\sqrt{3})$.

Now the candidates (groups of order 16) are limited and I also know, that the following groups must be isomorphic to normal subgroups:

  1. $\mathbb{Z}_2$ since $Gal(E/L)=\mathbb{Z}_2$
  2. $(\mathbb{Z}_2)^3$ since $Gal(E/\mathbb{Q}(\sqrt{2})=(\mathbb{Z}_2)^3$
  3. $D_4$ since $Gal(E/\mathbb{Q}(\sqrt{3})=D_4$
  4. $\mathbb{Z}_4$ since $Gal(E/\mathbb{Q}(i, \sqrt{3})=\mathbb{Z}_4$

If this is correct, it narrows down the choice of groups of order 16 to only one: $D_4 \times \mathbb{Z}_2$. So first question: are these all correct? Second question: is there an easier way to compute this group, which has now escaped me?

  • 1
    $\begingroup$ "E is a galois extension of $\mathbb{Q}$ of dimension 16, it being a galois extension of the splitting field L" is not quite correct, as being a galois extension is not transitive. You need $3\in\mathbb{Q}$ here. $\endgroup$ – j.p. May 4 '15 at 15:22
  • $\begingroup$ I'd guess that most people would look at the quotients $Gal(K/\mathbb{Q})$ instead of looking at $Gal(E/K)$ like you. I doubt that this would make the calculation easier, but it fits better to your goal of understanding the projective limit. $\endgroup$ – j.p. May 4 '15 at 16:04

Condition 1 is satisfied by all groups of order $16$, so that is unhelpful. In fact there is only one isomorphism type of groups $G$ satisfying 2,3,4.

Let $N$ be the elementary abelian subgroup of order 8, and let $t \in G \setminus N$. Since, by 3, $G$ is nonabelian, we can choose the generators $u,v,w$ of $N$ such that $t^{-1}ut=v$, $t^{-1}vt=u$, and $t^{-1}wt=w$.

To complete the definition, we need to specify $t^2$ as an element of $N$. Since this must centralize $t$, the possibilities for $t^2$ are $1,uv,w,uvw$. In fact $t^2=1$ and $t^2=uv$ give isomorphic groups $D_8 \times C_2$ (if $t^2=uv$ then $(tu)^2=1$, so replace $t$ by $tu$). Also, since $w$ and $uvw$ are interchangable, $t^2=w$ and $t^2=uvw$ give isomorphic groups, so there are just two isomorphic types of groups satisfying 2 and 3.

Now the first option, $D_8 \times C_2$, has the normal subgroup subgroup $\langle tu \rangle \cong C_4$, so that is a candidate. In fact, I checked by tedious calculation that the second option has no normal subgroups isomorphic to $C_4$, so we must have $G \cong D_8 \times C_2$.

Alternatively observe that, since $L \cap {\mathbb Q}(\sqrt{3})={\mathbb Q}$, the normal subgroups $C_2$ and $D_8$ in 1 and 3 have trivial intersection, so $G$ is their direct product.


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