# 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 $[x,y] = xyx^{-1}y^{-1}$: $H = \langle [x,y] \rangle$.

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

## Proof

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

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

Proof: By fact (2), the elementary Nielsen automorphisms of $F$ generate $\operatorname{Aut}(F)$. 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: $\tau_x([x,y]) = [x^{-1},y] = x^{-1}yxy^{-1} = x^{-1}yxy^{-1} \cdot x^{-1}x = x^{-1}[y,x]x = x^{-1}[x,y]^{-1}x \in x^{-1}Hx$.
• Replacing $y$ by its inverse: $\tau_y([x,y]) = [x,y^{-1}] = xy^{-1}x^{-1}y = y^{-1}[x,y]^{-1}y in y^{-1}Hy$.
• Swapping $x$ and $y$: $\sigma([x,y]) = [y,x] = [x,y]^{-1}\in H$.
• Replacing $x$ by $xy$: $\eta([x,y]) = xyyy^{-1}x^{-1}y^{-1} = xyx^{-1}y^{-1} = [x,y] \in H$.