Subgroup generated by commutator of generators of free group on two generators is automorph-conjugate

This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup satisfying a particular subgroup property (namely, Automorph-conjugate subgroup (?)) in a particular group or type of group .

Statement

Let $F$ be a free group on two generators, with $x,y$ being the generators. Let $H$ be the subgroup of $F$ generated by the commutator $[x,y]=xy{x}^{-1}{y}^{-1}$:

$H=⟨[x,y]⟩$.

Then, $H$ is an automorph-conjugate subgroup of $F$.

Facts used

1. Automorph-conjugate iff conjugate to image under a generating set of automorphism group
2. Elementary Nielsen automorphisms generate the automorphism group of a finitely generated free group

Proof

Given: $F$ is a free group with freely generating set $\{x,y\}$. $H=⟨[x,y]⟩$.

To prove: $H$ is automorph-conjugate in $F$.

Proof: By fact (2), the elementary Nielsen automorphisms of $F$ generate $\mathrm{A}\mathrm{u}\mathrm{t}(F)$. We use a modified version of this generating set to show that $H$ is automorph-conjugate in $F$ via fact (1):

• Replacing $x$ by its inverse: ${\tau }_x([x,y])=[x^{-1},y]=x^{-1}yx{y}^{-1}=x^{-1}yx{y}^{-1}\cdot x^{-1}x=x^{-1}[y,x]x=x^{-1}[x,y{]}^{-1}x\in x^{-1}Hx$.
• Replacing $y$ by its inverse: ${\tau }_y([x,y])=[x,y^{-1}]=x{y}^{-1}{x}^{-1}y=y^{-1}[x,y{]}^{-1}yin{y}^{-1}Hy$.
• Swapping $x$ and $y$: $\sigma ([x,y])=[y,x]=[x,y{]}^{-1}\in H$.
• Replacing $x$ by $xy$: $\eta ([x,y])=xyy{y}^{-1}{x}^{-1}{y}^{-1}=xy{x}^{-1}{y}^{-1}=[x,y]\in H$.