# Free groups are CT-groups [closed]

A group $$G$$ is called CT-group if being commutative elements is transitive relation on $$G\setminus\{1\}$$ i.e. if $$1 \neq x,y,z\in G$$ and $$[x,y]=1, [y,z]=1$$ then $$[x,z]=1$$.

I encountered the fact that free groups are CT-group, but every proof to this is very long and work for much general case (I know its true also for hyperbolic torsion free groups, which is nice but overkill for me).

Any one knows a self contained proof for this? The papers I saw uses arguments from algebraic topology, which are way over my knowledge, I only need the case of free groups.

• The centralizer of an element in a free group is cyclic.... – Steve D Mar 27 at 12:21
• There's an easy self-contained proof using basics on tree automorphisms (namely that an automorphism of a nonempty tree without fixed point preserves a unique axis). Now everything depends on what's "way over your knowledge". – YCor Mar 27 at 12:26
• OMG @SteveD, this is so simple, I don't know how I missed it, thanks! – Barak Ohana Mar 27 at 12:48
• @YCor could you send reference to this proof, it sound interesting (I do my thesis is geometric group theory, this is exactly my type of things). – Barak Ohana Mar 27 at 12:50
• @SteveD: The statement that centralizers are cyclic is as easy (or as complicated) as the statement that free groups are CT or as the statement that a freely acting automorphism of a tree has a unique axis. – user6976 Mar 29 at 0:17

(1) A free group admits a free action on a tree. (Namely its Cayley graph with respect to a free generating subset).

(2) If a group acts freely on a nonempty tree, then it's a CT-group.

(Indeed, if $$u,v$$ are non-identity elements, they are loxodromic since the action is free, if they commute, they have the same axis, and conversely if they have the same axis then $$[u,v]$$ is identity on this axis and hence is identity.)

• (This argument doesn't take for granted that centralizers of nontrivial elements are cyclic, and actually proves it.) – YCor Mar 27 at 14:38

For a completely self-contained proof see Theorem 1.8.28 in my book https://math.vanderbilt.edu/sapirmv/book/b2.pdf. The idea is that you first prove that if $$uv=vu$$ in the free semigroup then $$u,v$$ are powers of some other word $$w$$ which is trivially proved by induction (Theorem 1.2.9). And then prove a similar fact for free group by looking at possible cancelation in $$uv$$ or $$vu$$. Case 1 is when $$u$$ is not cyclically reduced and case 2 is when $$u$$ is cyclically reduced.

• Just wanted to thank you for the pointer to your great book, which is about an interesting topic that I didn't know much about beforehand. – anomaly Mar 27 at 20:49
• I think that what's missing here is the fact the converse fact that is we have two elements $u$ and $v$ of free group which some power of them coincide then they are both the power some element, and thus commuting. – Barak Ohana Mar 28 at 19:36
• Yes, but it is proved the same way. Suppose $p^m=q^n$ for some reduced words $p, q$ . If $p,q$ are cyclically reduced then the equality holds in the free semigroup and one can use, for example, Theorem 1.2.28 (Fine-Wilf) from my book. If, say, $p=ap_1a^{-1}$ then $p_1^m=a^{-1}q^na$. Since $p_1$ cannot start with $a^{-1}$ or end with $a$, $q$ must start with $a$ and end with $a^{-1}$, $q=aq_1a^{-1}$. Then $q_1^n=p_1^m$ where $p_1, q_1$ are shorter than $p, q$. – user6976 Mar 28 at 21:05
• Actually in the semigroup case you do not even need Fine-Wilf. Suppose $p^m=q^n$. Consider the graph consisting of $m|p|$ vertices forming a cycle with edges labelled by letters of $p^m$. That graph admits a rotation $\alpha$ of order $m$. Since $p^m=q^n$ this graph also admits rotation $\beta$ of order $n$. But the group of all rotations of any cyclic labelled graph is cyclic. Therefore there exists a rotation $\gamma$ such that $\alpha$ and $\beta$ are powers of $\gamma$. The same proof works for the description of commuting words in a free semigroup. It also can be found in my book. – user6976 Mar 28 at 23:41