largest subgroup of Out(^F2)Out(\hat{F_2}) which preserves the Nielsen invariant

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?

Answer

Attribution
Source : Link , Question Author : Will Chen , Answer Author : Community