# Conjugating generators in free groups

Let $$F_n := \langle x_1,\dotsc,x_n\rangle$$ be the free group on $$n$$ generators. Let $$w_1,\dotsc,w_n\in F_n$$ and consider the endomorphism $$\varphi:F_n\to F_n, x_i\mapsto w_ix_iw_i^-$$.

1. I conjecture that $$\varphi$$ is always injective. Is this true?
2. In some cases, $$\varphi$$ is not surjective, e.g. for $$w_1=x_1x_2$$. Can we say in general in which cases $$\varphi$$ is surjective? Maybe if $$w_i$$ does not contain $$x_i$$?
• Am I missing something: isn't conjugation always an automorphism? – Aniruddh Agarwal May 1 '19 at 16:21
• @AniruddhAgarwal: They are conjugating each generator by a different word. – Arturo Magidin May 1 '19 at 16:31
• Yes it's injective, because the image in the abelianization of $F_n$ of $\varphi(F_n)$ is all of $\mathbf{Z}^n$, and hence $\varphi(F_n)$ has rank $\ge n$, and every surjective endomorphism of $F_n$ is injective. – YCor May 1 '19 at 16:49

Proposition 1: Let $$G=A_1 \ast \cdots \ast A_n$$ be a free products and let $$w_1, \ldots, w_n \in G$$ be $$n$$ elements. Then $$H:=\langle w_1A_1w_1, \ldots, w_nA_nw_n^{-1} \rangle = w_1A_1w_1^{-1} \ast \cdots \ast w_nA_nw_n^{-1}.$$

Sketch of proof. Think of $$G$$ as the fundamental group of the graph of groups having a central vertex labelled by $$\{1\}$$ linked to $$n$$ vertices labelled by $$A_1, \ldots, A_n$$ and whose edges are all labelled by $$\{1\}$$. Let $$T$$ denote the corresponding Bass-Serre tree.

Choose $$z_1, \ldots, z_n \in G$$ such that $$H= \langle z_1A_1z_1^{-1}, \ldots, z_nA_nz_n^{-1} \rangle$$ with $$|z_1| + \cdots + |z_n|$$ minimal. For every $$1 \leq i \leq n$$, let $$v_i$$ denote the vertex of $$T$$ stabilised by $$z_i A_i z_i^{-1}$$. Also, let $$v$$ denote the vertex of $$T$$ corresponding to the coset $$\{1\}$$.

I claim that, for every $$i \neq j$$, $$v_i$$ does not belong to $$[v_j,v]$$. Otherwise, take a $$z \in \mathrm{stab}(v_i)$$ sending the first edge of $$[v_i,v_j]$$ to the first edge of $$[v_i,v]$$. Then the length of $$zz_j$$ is smaller than the length of $$z_j$$ because $$z \cdot v_j$$ is closer to $$v$$ than $$v_j$$. This contradicts the choice of the $$z_i$$'s.

As a consequence, there exists a subtree $$S \subset T$$ whose leaves are the $$v_i$$'s. Now a ping-pong argument leads to the desired conclusion. $$\square$$

2. About surjectivity: Given an $$n$$-tuple $$(a_1, \ldots, a_n)$$ of words written over the generators (and their inverses) of $$\mathbb{F}_n$$, consider the following elementary transformations:

1. if $$a_i$$ is not reduced, then reduce it;
2. if $$x_i^{\pm 1}$$ is the last letter of $$a_i$$, remove it;
3. if $$a_jx_j^{\pm 1}$$ is a prefix of $$a_i$$, remove from $$a_i$$ the last letter of this prefix.

Notice that the total length $$|a_1|+ \cdots +|a_n|$$ decreases when an elementary transformation is applied. Call $$(a_1, \ldots, a_n)$$ a reduction of $$(w_1, \ldots, w_n)$$ if $$(a_1, \ldots, a_n)$$ is obtained from $$(w_1, \ldots, w_n)$$ by applying elementary transformations until it is no longer possible.

Proposition 2: Let $$(a_1, \ldots, a_n)$$ a reduction of $$(w_1, \ldots, w_n)$$. Then $$\varphi$$ is surjective if and only if $$a_1= \cdots a_n=1$$.

Sketch of proof. Let $$\Gamma_0$$ be a bouquet of balloons, such that each wire is an oriented path of length $$|w_i|$$ labelled by $$w_i$$ and its corresponding balloon is a single oriented edge labelled by $$x_i$$. Let $$\Gamma_1, \ldots, \Gamma_k$$ be a sequence of graphs such that $$\Gamma_i$$ is obtained from $$\Gamma_{i-1}$$ by applying a Stallings' fold, and such that no such a fold applies to $$\Gamma_k$$. Notice that $$\Gamma_k$$ looks like a bouquet of balloons in which the wires are glued together like a rooted tree. Let $$a_1, \ldots, a_n$$ denote the words labelling the oriented paths from the root to the balloons.

By looking at how a Stallings' fold works, we find that $$(a_1, \ldots, a_n)$$ is a reduction of $$(w_1, \ldots, w_n)$$.

If $$\Gamma$$ is a bouquet of $$n$$ circles, because no Stallings' fold applies to $$\Gamma_k$$, the canonical map $$\Gamma_k \to \Gamma$$ is a local isometry*. As shown by Stallings, it is possible to add edges to $$\Gamma_k$$ in order to get a new graph $$\Psi$$ such that the local isometry $$\Gamma_k \to \Gamma$$ extends to a covering map $$\Psi \to \Gamma$$. Consequently, $$\varphi$$ is surjective if and only if $$\Psi = \Gamma_k$$, which amounts to saying that the tree-like part of $$\Gamma_k$$ is empty; in other words $$a_1= \cdots = a_n=1$$. $$\square$$

Below is an illustration of the method for $$\mathbb{F}_2= \langle a,b \rangle$$, $$w_1 = ab$$ and $$w_2=a$$:

*As a consequence, $$\Gamma_k \to \Gamma$$ is $$\pi_1$$-injective, whence a third argument showing the injectivity of $$\varphi$$.