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. 


Thanks!



A question on normal closures of elements in free groups.
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


$p^n=[w,b]$


(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.


