Let x,y be generators for the free group F2. It’s known that Aut(F2), and hence Out(F2) preserves the conjugacy class of the subgroup ⟨[x,y]⟩ 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 ^F2 (hence fixing an embedding F2↪^F2), then Out(^F2) does not have the same property of preserving the conjugacy class of ⟨[x,y]⟩ (this follows from the fact that ^F2 has the

strong lifting property). By the strong lifting property I mean that for any finite group G and two surjections φ,ψ:^F2→G, there is an α∈Aut(^F2) such that ψ=φ∘α. If Aut(^F2) were to preserve the Nielsen invariant, then the image of the conjugacy class of ⟨[x,y]⟩ in G would be the same for any surjection ^F2→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 A5 – 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 ⟨[x,y]⟩ in Out(^F2)?

The stabilizer should be a closed subgroup, lets call it S – could it be ¯GL2(Z)≅¯Out(F2)≅^Out(F2)⊂Out(^F2)?

If not, can we describe the difference S/¯GL2(Z) somehow?

**Answer**

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