# Conjugation in the Free Group

1. Jul 12, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
Note: I did not get this problem from a textbook.

Let denote the (nonabelian) free group on the generators , and let be arbitrary. My question is, does there exist a such that , besides (the identity); is such an equation in the free group possible? Obviously this equation would imply they commutate, but that won't necessarily be a contradiction. It appears that I am effectively asking whether the centralizer of a given element in is nontrivial. I can't determine the answer to this question; perhaps someone would be so kind as to guide me towards it

2. Relevant equations

3. The attempt at a solution

2. Jul 12, 2016

### andrewkirk

There are infinitely many other elements being $h^k$ for any integer $k$, as $h^{k}(h)h^{-k}=h$.

Last edited: Jul 12, 2016
3. Jul 14, 2016

### andrewkirk

Further I am fairly confident I can prove that the set of elements that commute with $h$ is:

$$\{g^k\ :\ g\in G\wedge k\in\mathbb Z\wedge \exists m\in\mathbb N(h=g^m)\}$$
That is, powers of any element of which $h$ is a power.

But I shan't bother to write an indication of how the proof would go unless evidence appears against the fact that this thread seems to have been abandoned by the OP (or unless somebody else shows interest).

Last edited: Jul 14, 2016
4. Jul 15, 2016

### Bashyboy

In other words, you claim that the centralizer of $h$ is $\{g^k\ :\ g \in G,~ k \in \mathbb{Z},~\exists m \in \mathbb{N} (h=g^m)\}$?

I have another question, which seems to be related to your second post, andrewkirk. According to what I have been reading, the centralizer of a nonidentity $h$ in the free group is an infinite cyclic, i.e., a subgroup generated by one element; and evidently all subgroups of the free group are themselves free groups (wiki). Would that not make the centralizer $C_{F_n}(h)$ a free group on one generator, and therefore an abelian group, as the free group on one generator is always abelian? If so, that is rather interesting that $C_{F_n}(h)$ is always an abelian subgroup for all nonidentity $h \in F_n$.

Last edited: Jul 15, 2016
5. Jul 15, 2016

### andrewkirk

Yes, it would make the centraliser a free group on one generator.

6. Jul 16, 2016

### micromass

Staff Emeritus
This is not evident at all! It's a pretty deep result.