# Automorphism of the free group

Let $\mathbb{F}_2$ be the free group of rank $2$ with generators $a$ and $b$. I would like to build an automorphism $\varphi$ of $\mathbb{F}_2$ such that :

1) $\varphi([a,b]) = [a,b]$

2) $\overline{\varphi}$ the induced application on the abelianization of $\mathbb{F}_2 \simeq \mathbb{Z}^2$ is the identity.

For any $n \in \mathbb{Z}$ the conjugation by $[a,b]^n$ does the trick. I was wondering if any such $\varphi$ must be of this form.

(This question is related to the study of elements of the Torrelli group of a closed surface which acts only on a embeded $1$-holed torus).

• These are the only automorphisms: They have to be inner and centralizer of the commutator element is generated by this element. Dec 3, 2013 at 16:58
• Could you detail a bit I'm not sure I understand your proof. Dec 3, 2013 at 17:02

The main reference is "Primer on mapping class group" by Farb and Margalit. The point is that for once punctured torus the abelianization of the fundamental group defined an isomorphism of the mapping class group to $GL(2,Z)$. Thus, under your assumption the automorphism has to be inner, conjugation by some element $g$ of the free group. Now, use the assumption that it sends $[a,b]$ to itself. This means that $g$ centralizes the commutator. Lastly, the commutator is represented by a simple loop, thus, it generates a maximal cyclic subgroup.
• I'm not sure I get what you want to say. If you read carefully Farb and Margalit(p.87, I guess there is only one edition), the mapping class group of the once punctured torus is isomorphic to the central extension of $GL(2,\mathbb{Z})$. Maybe you make the distinction whether the torus is thought with or without boundary. Nevertheless, I would be glad to have further explaination on this proof. Dec 4, 2013 at 10:06