Give a rank $n$ free group $G=\langle a_1,a_2,\dots,a_n\rangle$, let $g_1,g_2,\dots,g_n \in G$, $b_j=g_j^{-1}a_jg_j$ . If $b_1,b_2,\dots,b_n$ can generates the whole $G$, what can we say about $g_1,g_2,\dots,g_n$ ?

I have tried to take $G$ as a fundamental group,and give some topology properties of $g_1,g_2,\dots,g_n$ but failed .

I am not a professional mathematician,may be this question is a trivial, but I still need your help. I will be grateful if any of you can give me some useful advices. Thanks!


I'll make some simplifying assumptions. First, $g_j$ does not begin with $a_j$ nor with $a_j^{-1}$, equivalently the expression $g_j^{-1} a_j g_j$ is reduced. Second, the $g_j$'s are written in order of nonincreasing length.

If $n=2$ then the conclusion is that $g_1 = g_2 a_2^{\pm m}$ for some integer $m \ge 0$.

The method can be used inductively to get conclusions for higher ranks as well, although at the moment I don't have a simple way to state the conclusion.

The proof uses the method of Stallings folds, a common proof technique. Define an "{$a_i$} labelled graph" to be a directed graph each of whose edges is labelled by one of the generators $a_i$, $i=1,...,n$. The "base rose" denoted $R_n$ is the $\{a_i\}$ labelled graph with one loop for each generator $a_i$. Any $\{a_i\}$ labelled graph has a well-defined "induced map" to $R_n$ that preserves direction and labels; in particular any path in an $\{a_i\}$ labelled graph gives a word in the generators and their inverses.

Start by using the expression $g_j^{-1} a_j g_j$ to construct an $\{a_i\}$ labelled graph with one loop edge labelled $a_j$ and base point $p_j$, and with an arc from $p_j$ to opposite endpoint $q_j$ so that the word of that arc is $g_j$. Next, identify all of the vertices $q_j$ to a single base vertex $Q$. Let $\Gamma_0$ be the resulting graph. By construction the induced map $\Gamma_0 \to R_n$ is a local embedding at every point except $Q$.

Now start folding: if there is a vertex $v$ at which the induced map is not a local embedding (initially the only possibility is $v=Q$), then at $v$ there are two incident edges $e,e'$ with identical label and orientation pointing in the same direction with respect to $v$; identify those two edges to produce $\Gamma_1$; this is "folding $e,e'$". Repeat. You obtain a sequence of quotient maps $\Gamma_0 \to \Gamma_1 \to \ldots \to \Gamma_K \to R_n$, stopping only when the map is a local embedding at each point. From your hypothesis one concludes that final quotient map $\Gamma_K \to R_n$ is a graph isomorphism and that every quotient map along the sequence is a homotopy equivalence.

Now let me describe the case $n=2$. Since $g_2$ is not longer than $g_1$ the first stage of the folding process starting $\Gamma_0$ will identify all of $g_2$ with an initial segment of $g_1$; let this stage end at $\Gamma_k$. If $k=K$ we're done with $m=0$. If not, the only possibility for continuing to fold is that the remaining terminal segment of $g_1$ itself has an initial segment which is a power of $a_2$, and the next stage of the folding process starting with $\Gamma_k$ is to identify that segment by wrapping it around $a_2$; let this second stage of the process end at $\Gamma_l$. If $g_1$ has still not yet been used up, i.e. if there is any remaining unfolded terminal segment of $g_1$, then that segment has endpoints at $p_1$ and $p_2$, the base points of the $a_1$ and $a_2$ loops, forming an "eyeglass" graph that is locally embedded in $R_2$ and will not have any further identifications under the folding process; but this contradicts the fact that that $\Gamma_K = R_2$. Therefore $g_1$ has indeed been entirely used up and we are done.

  • $\begingroup$ First thanks for your attention.I think the case $n=2$ is an easy case since we can assume $g_1=1$ and by induction we can easily get that $g_2$ must be of the form $a_1^k$,but it seems this methods will failed in case $n>2$ $\endgroup$ – Grub Oct 28 '13 at 1:45

Letting $a_i$ and $b_i$ be as in your question, observe that the homomorphism $F_n \rightarrow F_n$ that takes $a_i$ to $b_i$ for all $1 \leq i \leq n$ is an isomorphism if and only if the $b_i$ generate $F_n$ (this uses the classical fact that a set of $n$ elements of $F_n$ that generate $F_n$ must be a free generating set). This suggests examining the resulting subgroup of $\text{Aut}(F_n)$.

Define $\Gamma_n$ to be the subgroup of $\text{Aut}(F_n)$ consisting of automorphisms $f$ such that $f(a_i)$ is conjugate to $a_i$ for all $1 \leq i \leq n$; this group is known as the pure symmetric automorphism group. Your question basically boils down to asking about the structure of $\Gamma_n$. Here there is quite a bit known. For instance, it is a classical theorem of Nielsen that $\Gamma_2$ is exactly the set of inner automorphisms of $F_2$, which is equivalent to what Lee did in his answer. Things are more complicated for higher $n$ (I don't think there is an easy description of the possible $g_i$ in your question for $n \geq 3$ ), but a large amount is known about $\Gamma_n$. For instance, the paper

Humphries, Stephen P., On weakly distinguished bases and free generating sets of free groups. Quart. J. Math. Oxford Ser. (2) 36 (1985), no. 142, 215–219.

proves that it is generated by the elements $c_{ij}$ for distinct $1 \leq i,j \leq n$ defined by $$c_{ij}(a_k) = \begin{cases} a_k & \text{if $k \neq i$},\\ a_j a_i a_j^{-1} & \text{if $k = i$} \end{cases}.$$

Also, McCool found a relatively simple presentation for it in terms of these generators in

McCool, J., On basis-conjugating automorphisms of free groups. Canad. J. Math. 38 (1986), no. 6, 1525–1529.

There is also a large literature that comes from the theorem of Dahm that says that $\Gamma_n$ is isomorphic to the pure braid group of a set of $n$ unlinked circles in $\mathbb{R}^3$; see

Goldsmith, Deborah L. The theory of motion groups. Michigan Math. J. 28 (1981), no. 1, 3–17.

for a proof of this (Dahm never published his result) and also see

Brendle, Tara E.; Hatcher, Allen, Configuration spaces of rings and wickets. Comment. Math. Helv. 88 (2013), no. 1, 131–162.

for many interesting generalizations of this. One important result that comes out of this fact is a computation of the integral cohomology ring of $\Gamma_n$ in

Jensen, Craig; McCammond, Jon; Meier, John, The integral cohomology of the group of loops.
Geom. Topol. 10 (2006), 759–784.

  • $\begingroup$ Thanks for your professional answer. As you said,the pure symmetric automorphism group is too complex to see it clearly.Then if we consider the following subgroup of it: Let $a_{n+1}=(a_1a_2...a_n)^{-1}$, $H=\{h\in Aut(G)|h\ send\ a_i\ to\ a\ element\ conjugates\ to\ a_i,i=1,2,...,n+1\}$ .What can we say about this subgroup?If we take G as the fundamental group of $S^2-P$ where $P$ contains $n+1$ points,Let $a_1,a_2,...,a_{n+1}$ be a "well-choosed" simple loop which surrond points in P respectly, Then I guess that each element in $H$ is induced by a orientation-preserving homeomorphism. $\endgroup$ – Grub Oct 28 '13 at 2:13
  • 1
    $\begingroup$ @Grub : You are correct that this subgroup is the mapping class group of a punctured sphere. This follows from the version of the Dehn-Nielsen-Baer theorem for punctured surfaces; see, e.g. Theorem 8.8 in Farb-Margalit's "Primer on mapping class groups". But be warned that the mapping class group of a punctured sphere is still a pretty rich and complicated object. $\endgroup$ – Andy Putman Oct 28 '13 at 2:34

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