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

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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.

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 = \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.