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Let $x,y$ be generators for the free group $F_2$. It's known that $Aut(F_2)$, and hence $Out(F_2)$ preserves the conjugacy class of the subgroup $\langle[x,y]\rangle$ generated by $[x,y]$ (This conjugacy class is in some contexts called the nielsen invariant)

On the other hand, if we view $x,y$ as topological generators of $\hat{F_2}$ (hence fixing an embedding $F_2\hookrightarrow\hat{F_2}$), then $Out(\hat{F_2})$ does not have the same property of preserving the conjugacy class of $\langle [x,y]\rangle$ (this follows from the fact that $\hat{F_2}$ has the strong lifting property). By the strong lifting property I mean that for any finite group $G$ and two surjections $\varphi,\psi : \hat{F_2}\rightarrow G$, there is an $\alpha\in Aut(\hat{F_2})$ such that $\psi = \varphi\circ\alpha$. If $Aut(\hat{F_2})$ were to preserve the Nielsen invariant, then the image of the conjugacy class of $\langle [x,y]\rangle$ in $G$ would be the same for any surjection $\hat{F_2}\rightarrow G$, which is to say that all commutators of generating pairs of finite groups generate conjugate subgroups. This is easily verified to be false - the smallest example is the alternating group $A_5$ - it has two generating pair with commutators of order 3 and 5 respectively.

My question is - What is the stabilizer of the conjugacy class of $\langle[x,y]\rangle$ in $Out(\hat{F_2})$?

The stabilizer should be a closed subgroup, lets call it $S$ - could it be $\overline{GL_2(\mathbb{Z})}\cong\overline{Out(F_2)}\cong \widehat{Out(F_2)}\subset Out(\hat{F_2})$?

If not, can we describe the difference $S/\overline{GL_2(\mathbb{Z})}$ somehow?

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  • $\begingroup$ Could you say a little more about the strong lifting property? $\endgroup$ – HJRW Sep 5 '15 at 20:43
  • $\begingroup$ @HJRW I've explained it a bit more as an edit. $\endgroup$ – Will Chen Sep 8 '15 at 16:11
  • $\begingroup$ @HJRW Also I should have said $\hat{F_2}$ has the strong lifting property (not $Aut(\hat{F_2})$). I've edited to correct that. $\endgroup$ – Will Chen Sep 8 '15 at 16:38

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