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

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Revision as of 20:51, 28 April 2009 by Nomeata (talk | contribs) (→‎Proof: Removing redundant generating and minor changes to wording.)

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−1y−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 Aut(F). We use this generating set to show that H is automorph-conjugate in F via fact (1):

  • Replacing x by its inverse: τx([x,y])=x−1yxy−1=x−1yxy−1⋅x−1x=x−1[y,x]x=x−1[x,y]−1x∈x−1Hx.
  • Swapping x and y: σ([x,y])=[y,x]=[x,y]−1∈H.
  • Replacing x by xy: η([x,y])=xyyy−1x−1y−1=xyx−1y−1=[x,y]∈H.