A metabelian quotient of a free group - MathOverflow

A metabelian quotient of a free group

2011-12-07T03:36:13Z

I don't know much about free groups (excepted the very basics), and the following question may be trivial, although it isn't to me. Let $F$ be a free group with $n$ generators $x_1,\dots,x_n$. Consider the 'augmentation' map $a:F \rightarrow \mathbb{Z}$ that sends $x_i$ to $1$ for $i=1,\dots,n$, and let $A = \ker f$. So $A$ is a free group as a subgroup of the free groups $F$, of infinite rank I presume. Is it easy to describe a free family of generators of $A$? This would surely allow to answer my question, which is:

Can we describe the action of $\mathbb{Z}$ on $A^{ab}$ ?

Here, $A^{ab} = A /D(A)$ is the abelianization of $A$ (with $D(A)$ its derived subgroup), and the action in question is the one given by the short exacts sequence: $$1 \rightarrow A/D(A) \rightarrow F/D(A) \rightarrow \mathbb{Z} \rightarrow 1$$ The group $F/D(A)$ is the metabelian group of the title. Of course, describing the action is essentially equivalent to describing the group $F/D(A)$, since the extension splits.

(The question feels elementary to me, but at the same time I feel helpless to solve it, because I don't know how to recognize when a family of elements in a free group is free.)

Now the truth is that the real question I need is when $F$ is a free pro-$p$ group with $n$ generators instead of a free group, $a$ is the continuous map from $F$ to $\mathbb{Z}_p$ that sends the generators to $1$, and $D(A)$ is the closed derived subgroup. But I believe (perhaps naively) that the solution of the discrete problems will easily give a solution of its pro-$p$ analog, and that the discrete problem is more natural.

This question in turn comes from my trying to understand the structure of the maximal metabelian quotient of some pro-$p$ Galois group of number fields with prescribed ramification. In some cases such a group is the $p$-adic $F/D(A)$ considered in this question.

Answer by Mark Sapir for A metabelian quotient of a free group

2011-12-07T03:46:57Z

To find generators of $A$, use the Nielsen-Schreier method. It is very easy in that case: http://en.wikipedia.org/wiki/Nielsen%E2%80%93Schreier_theorem

Answer by Cam McLeman for A metabelian quotient of a free group

2011-12-07T04:28:22Z

Do you mean the augmentation map on the group ring $\mathbb{Z}[F]$ (and in the pro-$p$ case, the completed group ring $\mathbb{Z}_p[[F]]$?) I ask only because this augmentation map comes up frequently and significantly in the study of large number-theoretic Galois groups.

Assuming this is the case (and apologies for misinterpreting if not -- hopefully the answer will still be of some use to you), there is a tremendous amount of machinery set up for dealing exactly with questions of this sort -- probably the best starting place is the phrase "pro-p Fox Differential Calculus." (And so, indeed, your intuition that solving the discrete problem turns out to provide the correct pro-$p$ analog is correct. It was Iwasawa who carefully established the fundamental analogy here. In fact, thanks to the topology of $\mathbb{Z}_p$, in some ways the pro-p Fox calculus is nicer than the discrete version.) In particular, if you filter the group ring $\mathbb{Z}_p[[F]]$ by powers of the augmentation ideal (the group-ring version of your $A$), you land upon the sequence of "dimension subgroups" of F.

These subgroups have shown up repeatedly in the analysis of pro-$p$-groups arising in the study of large Galois groups arising from restricted ramification questions (as appears to be the case for you). A couple of the highlights of the theory are the work of Vogel and Morishita interpreting number-theoretic analogs of the a priori knot-theoretic notion of Milnor invariants, refined versions of Golod-Shafarevich-type inequalities, and perhaps most relevant for your question, work of Arrigoni (e.g., "On Schur $\sigma$-groups") which I think explicitly answers questions of your type. For a more fundamental reference, see Koch's "Galois theory of $p$-extensions.")

Sorry to be mostly hand-wavey -- I'm away from good references at the moment.