Let $F$ be a free group of finite rank, and $p, b \in F$, where $b$ is a root element (i.e. not a proper power). I have a case where $p^{n_k} = V_{n_k}^{-1}b^{-1} V_{n_k} \cdot U_{n_k}^{-1}b U_{n_k}$, for some $n_k \in \mathbb{Z}$ ...i.e. some powers of $p$ are products of two conjugates of $b$ and $b^{-1}$.

What can be said about $p$ and $b$ ?

Some immediate implications: By Karras-Magnus-Solitair, since $b$ is root, one-relator group $ < F \ | \ b >$ is torsion-free, so if a proper power of $p$ is in $ncl(b)$ then $p\in ncl(b)$.

Also, by going to an abelianization of $F$, it is clear that $p\in \[ F, F \]$.

I was hoping that $p$ is conjugate to $b$...by Magnus, if we can show that $b$ is also in the the normal closure of $p$ then that would be the case.

I also have a somewhat related general question. Say, if we have $b$ as normal root of $p$ i.e. $p \in ncl(b)$, so that $p = \displaystyle{\prod_{i=1}^{n}} T_i^{-1} b^{\epsilon _i} T_i$, where $\epsilon _i = \pm 1$. Clearly the above product is not unique. Is there a notion of associating to $p$ a minimal integer $n_p \geq 1$ so that $p$ can be decomposed to the product $p = \displaystyle{\prod_{i=1}^{n_p}} T_i^{-1} b^{\epsilon _i} T_i$? For example if $n_p=1$ then $p$ and $b$ are conjugate.

Apologies if I missed something obvious or the question(s) doesn't make sense.


  • 1
    $\begingroup$ So, 1/2 is a universal lower bound for stable commutator length in free groups. Doesn't it follow that your $n_k\leq 2$? If so, then the possibilities are very restricted, by a theorem of Lyndon. I'll try to write more details later, if I have time. $\endgroup$
    – HJRW
    Mar 24, 2013 at 8:31
  • $\begingroup$ Thanks Henry. Let me go one step back and consider a case that led to an assumption of $b$ being root. If we drop that assumption we have that $b=\tilde{b}^k$ for some $k\geq 2$. Suppose that there is a proper power of $p$ that can be represented as $p^m= V^{-1} b^{-k}V\cdot U^{-1} b^{k}U$ for some $U$ and $V$. We then have that $$ p^m = \tilde{V}^{-k} \tilde{U}^{k} ,$$ where $\tilde{V} = V^{-1} b^{-1}V$,$\tilde{U} = U^{-1} b U$, and $|m|, k \geq 2$. Then by Schutzenberger-Lyndon we have that $\tilde{U}, \tilde{V}, p$ commute. So that $V = b^l U$ and $p=1$. $\endgroup$ Mar 24, 2013 at 9:51
  • $\begingroup$ Not sure how to edit comments, but $V^{-1} b^{-k} V$ in the above comment should be read as $V^{-1} \tilde{b}^{-k} V$ etc.... Anyhow, so now the question is that $b$ is root, and some powers of $p$ can be represented as $V^{-1} b^{-1} V\cdot U^{-1} b U $ form. For example, $p^m = V^{-1} b^{-1} V\cdot U^{-1} b U$ and $p^n = X^{-1} b^{-1} X\cdot Y^{-1} b Y$ for some $m, n \in \mathbb{Z}$ and $U, V, X, Y \in F$. What could be inferred about $p$ and $b$? $\endgroup$ Mar 24, 2013 at 10:14

1 Answer 1


This is just to flesh out the details of my comment above. I think we can show that $n_k=1$ or $p=1$.

After conjugating (and simplifying notation slightly), your equation easily becomes


(for $w=vu^{-1}$). The stable commutator length of $p$ is equal to the infimum of $cl(p^n)/n$ over all $n>0$, so in your case we have

$\mathrm{scl}(p)\leq 1/n$.

But a theorem of Duncan--Howie asserts that scl in a free group is always at least 1/2, so $n\leq 2$. (This is all explained in Danny Calegari's book scl.) Suppose therefore than $n=2$.

In this case, your equation can be realized by a map of the fundamental group of the surface $\Sigma$ of Euler characteristic -1, which has presentation $\langle p,w,b|p^2=[w,b]\rangle$.

A theorem of Lyndon asserts that every such map factors through $H_1(\Sigma)$ (though Lyndon didn't say it like that, of course). Because $p$ is torsion in $H_1(\Sigma)$ and free groups are torsion-free, it follows that $p=1$.

I'll add some proper references when I have more time.

  • $\begingroup$ oh...i didn't notice that the left side could be represented as a commutator. damn. In that case we have that $$p^n = V^{-1}b^{-1}V \cdot U^{-1}bU = \[ U^{-1}V, U^{-1}bU \] $$, but then $n=1$ or $p=1$. This is Schutzenberger again (if $x,y,z\in F$ and $\[ x, y\] = z^n$ for $n\geq 2$ then $z=1$. $\endgroup$ Mar 24, 2013 at 10:27
  • $\begingroup$ sorry, not sure how to edit comments... those $\[$ are commutator brackets. Thanks a lot!! $\endgroup$ Mar 24, 2013 at 10:29

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