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Let $F_n$ be a free group of finite rank $n$ and let $V$ be a verbal subgroup of $F_n.$ In a "nice" situation one could expect that

$(*):$ the group $\mathrm{Aut}_V(F_n) \le \mathrm{Aut}(F_n)$ whose elements fix each element of $F_n$ mod $V$ acts transitively on the set of all primitive elements from the coset $xV$ of a primitive element $x$ of $F_n.$

I have two questions related to the situation above.

1) Is the property $(*)$ true for the verbal subgroups $V=V([x_1,\ldots,x_m])=\gamma_m(F_n),$ i.e. for terms of the lower cenral series of $F_n?$ (Asked this question on math.stackexchange, got no answers).

Most likely, the answer to 1) is affirmative, and might be obtained from the classic description of the generators of the group $\mathrm{IA}(F_n)$ of IA-automorphisms of $F_n$ due to Magnus. I just don't see how, and being not versed in German I am not able to read the original Magnus' proof. Any comments on the plan etc. of Magmus' proof are welcome.

2) Are there examples of verbal subgroups $V$ for which the property $(*)$ is not true?

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(This started out as a comment, but it grew too long)

My guess is that this is a very hard problem, though I'd be happy to be proven wrong. But I can say something about Magnus's theorem on generators for $IA_n$. I know of three proofs of this result.

  1. Magnus's original proof. This is based on the short exact sequence $$1 \longrightarrow IA_n \longrightarrow Aut(F_n) \longrightarrow GL(n,\mathbb{Z}) \longrightarrow 1.$$ Magnus constructed a finite presentation $\langle S | R \rangle$ for $GL(n,\mathbb{Z})$ (you can find a modern exposition of this in Milnor's book on algebraic k-theory). The elements $S$ can be lifted in a natural way to a generating set $\tilde{S}$ for $Aut(F_n)$. This implies that the lifts $\tilde{R}$ of the relations are normal generators for $IA_n$. Magnus then proved with a brute-force computation that the subgroup generated by his finite generating set was normal and contained $\tilde{R}$, which implies that his finite set is a generating set.

  2. In the paper M. Bestvina, K.-U. Bux and D. Margalit, Dimension of the Torelli group for Out(Fn), Invent. Math. 170 (2007), no. 1, 1–32 there is a topological proof (using the action of $IA_n$ on auter space) that Magnus's normal generating set is a normal generating set. This paper also contains a brief description of the calculation needed to go from this to the finite generating set.

  3. In my paper "The complex of partial bases for F_n and finite generation of the Torelli subgroup of Aut(F_n)" with Matt Day we gave yet another topological proof that Magnus's normal generating set is a normal generating set using a space analogous to the curve complex. We also have a somewhat more complete sketch of the calculation needed to go from this to the finite generating set.

EDIT : I forgot to point out that in the paper described in the third item above, we also prove that the desired result holds for $V$ equal to the commutator subgroup. I'm pretty sure that this is the only special case of the desired result that appears in the literature.

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@Andy Putman: Thank you very much indeed. –  Taff Mar 2 '12 at 20:20
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