# Is it known if the absolute Galois group is “divisible”?

The definitions of a divisible group that I have seen all seem to assume abelian is an a priori property of the group. My question is as to whether or not it is known that--given a non-torsion element $\tau\in\mathbf{G}$* and $n\in\mathbb{N}$, do we know if $\exists \tau'\in\mathbf{G}_\mathbb{Q}$ such that $\tau=\tau'^n$?

The motivation is that--if this never happens--then the fields $\overline{\mathbb{Q}}^\tau$ should be sort of "maximal" subfields of $\overline{\mathbb{Q}}$ since fixing by the generator would be equivalent to fixed by any power of it.

Edit: anon has noted that quotients of divisible groups are divisible, so we can lay to rest that $\mathbf{G}_\mathbb{Q}$ is not divisible. Can we tell if the opposite is true? I.e. the answer to "can I find such a $\tau '$ given $n$" is "no", but how about the somewhat interesting and related question: If I have a cyclic subgroup of the absolute Galois group, $C=\langle\tau\rangle$ can we find a maximal (with respect to inclusion), cyclic group containing $C$? Of course this is equivalent to a minimal field in a chain, so it seems like if that is so something interesting must be going on.

The other related question which deals with the original spirit of the problem is: are there $\tau$ such that $\langle \tau^n\rangle$ fixes $k\subseteq\overline{\mathbb{Q}}$ then $\langle\tau\rangle$ fixes $k$? i.e. is $k=\overline{\mathbb{Q}}^\tau$ equivalent to $k=\overline{\mathbb{Q}}^{\tau^n}$ possible for some $\tau$ non-torsion? If so can such elements be characterized?

The immediate observation is that, by the FTGT, if we identify $\langle\tau\rangle\cong\mathbb{Z}$ and write $[n]$ for the index $n$ subgroup, that--in the topology of $\mathbf{G}_\mathbb{Q}$--necessarily $\overline{[n]}=\overline{[1]}$. I.e. the monothetic group $\overline{[1]}$ has every element is a generator. If $\overline{[1]}$ were connected, of course this would be trivial to check since the set of generators has Haar measure 1 in this case, and a finite index subgroup has positive Haar measure, and so contains a generator, but the topology here is totally disconnected so it is not easy to see one way or the other.

*My TeX was not rendering properly when I wrote $\mathbf{G}_\mathbb{Q}$ here, so I just included it as a sidenote rather than have it look like a mess.

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Quotients of divisible groups are divisible, which is certainly not true of the absolute Galois group of Q. – anon Nov 28 '12 at 18:39
That's a fantastically simple reason. Thank you, anon. I've edited the original question to be a bit stronger now that the easy part is revealed. – Adam Hughes Nov 28 '12 at 18:48

There are the following results:

• Let $K = \bar{K}$ be algebraically closed and $\sigma \in \mathrm{Aut}(K)$. Then every finite extension of $K^\sigma$ is cyclic.
• Let $x \in K^{sep} \setminus K$. Let $L/K$ be a subfield of $K^{sep}$ maximal with respect to the property of not containing $x$. Then $G_L = \mathbf{Z}/2$ or $\mathbf{Z}_p$.

And $1 \in \mathbf{Z}_p$ is not $p$-divisible.

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Here is a result of Haran that might be relevant : For almost all $\sigma$ in the absolute Galois group of $\mathbb{Q}$ the fixed field $\bar{\mathbb{Q}}^\sigma$ of $\sigma$ in $\bar{\mathbb{Q}}$ has no proper cofinite subextensions. (I.e. $L\subsetneq \bar{\mathbb{Q}}^\sigma$ implies $[\bar{\mathbb{Q}}^\sigma:L]=\infty$.)

Here almost all is in the sense of the Haar measure on the absolute Galois group.

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