Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In other words, $uv = vu$ in $F_n$ if and only if $u=w^m$ and $v=w^n$ for some $w\in F_n$.

I would like to prove this without making use of Nielsen-Schreier (every subgroup of a free group is free). We can always find reduced representations $u=t_1^{\epsilon_1}\cdots t_k^{\epsilon_k}$ and $v=s_1^{\eta_1}\cdots s_l^{\eta_l}$ and the statement $uv = vu$ transforms into $$t_1^{\epsilon_1}\cdots t_k^{\epsilon_k}\cdot s_1^{\eta_1}\cdots s_l^{\eta_l}\cdot t_k^{-\epsilon_k}\cdots t_1^{-\epsilon_1}\cdot s_l^{-\eta_l}\cdots s_1^{-\eta_1}\sim 0$$ where $\sim$ denotes the equivalence relation coming from setting $x\cdot x^{-1}\sim 0$. However, the arbitrariness of $u$ and $v$ makes it hard to go on from here.

share|cite|improve this question
What about induction on the lengths of reduced representation, then concentrating only on the first letters..? – Berci Oct 14 '12 at 11:47
First deal with the two special cases: 1) there is no cancellation when forming the products $uv$ and $vu$; and 2) one of $u,v$ cancels completely when forming these products. If neither of these happens, then $u,v$ have the form $au'a^{-1}$, $av'a^{-1}$, where $a$ is a generator (or its inverse), and $u'$, $v'$ are shorter than $u$ and commute, so you can use induction. – Derek Holt Oct 14 '12 at 16:32

I answer the question just to remove it from the list of unanswered questions.

Of course, the simplest solution is probably to use Nielsen-Schreier theorem: if $a$ and $b$ commute, then $\langle a,b \rangle$ is an abelian subgroup; but the only abelian free group is $\mathbb{Z}$, so $\langle a,b \rangle$ is cyclic.

Otherwise, the result can be shown thanks to a combinatorial argument from the classical normal form in free groups. The proof is made by induction on $\mathrm{lg}(a)+\mathrm{lg}(b)$, and the argument follows the hint given by Derek Holt; the same proof can be found in Johnson's book, Presentations of Groups. The case where $\mathrm{lg}(a)$ or $\mathrm{lg}(b)$ belongs to $\{0,1\}$ is obvious, so let us suppose that $\mathrm{lg}(a), \mathrm{lg}(b) \geq 2$.

First, we write $a$ and $b$ as reduced words on a free basis $$\left\{ \begin{array}{l} a= x_1 \cdots x_m \\ b= y_1 \cdots y_n \end{array} \right., \ n \leq m.$$

Then, we have the reduced product $$x_1 \cdots x_{m-r} y_{r+1} \cdots y_n =ab =ba = y_1 \cdots y_{n-s} x_{s+1} \cdots x_m.$$

Notice that $$m+n-2r-1= \mathrm{lg}(ab)= \mathrm{lg}(ba)= m+n-2s-1$$ implies $r=s$. Moreover, $0 \leq r \leq n$.

Case 1: $r=0$, there is no cancellation in the product. Then $y_i=x_i$ for $1 \leq i \leq n$ hence

$$ a = \underset{=y_1 \cdots y_n=b}{\underbrace{ \left( x_1 \cdots x_n \right) }} \cdot \underset{:=u}{\underbrace{ \left( x_{n+1} \cdots x_m \right) }} = bu.$$

Notice that $u$ and $b$ commute: $$bu=a=b^{-1}ab= ub.$$ Therefore, the induction hypothesis applies, and it is sufficient to conclude.

Case 2: $r=n$, the number of cancellations is maximal. Then $y_i=x_{m-i+1}^{-1}$ for $1 \leq i \leq n$ hence

$$a^{-1} = \underset{=y_1 \cdots y_n=b}{\underbrace{ \left( x_m^{-1} \cdots x_{m-n+1}^{-1} \right) }} \cdot \underset{:=u}{\underbrace{ \left( x_{m-n+2}^{-1} \cdots x_m^{-1} \right) }} = bu.$$

You conclude that $b$ and $u$ commute and you apply the induction hypothesis.

Case 3: $0 < r <n$. Then $x_1=y_1$, $y_1=x_m^{-1}$, $y_n=x_m$ and $x_1=y_m^{-1}$. Let $z:=x_1$, $a'= x_2 \cdots x_{m-1}$ and $b'= y_2 \cdots y_{n-1}$. Then $a=za'z^{-1}$ and $b=zb'z^{-1}$. Finally, you may apply the induction hypothesis to $a'$ and $b'$, and conclude.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.