a long product of elements in hyperbolic group is not a proper power Let $G$ be a hyperbolic group, i.e., there exist $\delta>0$ and a finite generating set $S$ of $G$ such that the Cayley graph $X$ of $G$ relative to $S$ is a $\delta$-hyperbolic space. Assume also that $G$ is non-elementary (i.e., does not contain a finite-index cyclic subgroup).
In general, given an element $g\in G$, we say that $g$ is a proper power (in $G$) if there exists $h\in G$ and an integer $n>1$ such that $h^n = g$. For example, every torsion element is a proper power.
It is a basic fact that every hyperbolic element of $G$ is a power of an (hyperbolic) element which is not a proper power.
My question is as follows. Does there exists a word $w(X,Y)$ in $F(X,Y)$ (the free group on the set of two elements $\{X,Y\}$) such that for every non-commuting elements $r_1,r_2\in G$, the substitution $w(r_1,r_2)$ is not a proper-power in $G$? What if we allow to $w(r_1,r_2)$ to be torsion (i.e. asking for a word $w$ such that [if $w(r_1,r_2)$ is hyperbolic then not proper power])?
If somebody has an answer with further restrictions (e.g. torsion-free), then, feel free to add them.
Thank You!
 A: I'll make my comments into an answer. In short, if we fix the word and vary the group then the answer is "no", while if we fix the group to be an arbitrary torsion-free hyperbolic group then the answer is "yes".
Firstly, if we fix the word and vary the group then the answer is "no". (For references I've given a paper of McCool and Schupp, which has pretty accessible proofs. Lyndon and Schupp's book "Combinatorial group theory" also contains the same results, but I think without proofs and also I don't have it in front of me so can't reference it properly.)
Lemma 1.
For all words $W\in F(\mathbf{x})$ there exists a hyperbolic group $G_W$ such that the word $W$ is a proper power in $G_W$.
Proof. Fix $W$ and take $G_W:=\langle \mathbf{x}\mid W^n\rangle$ for some $n>1$. Then the word $W$ defines an element of order precisely $n$ in $G_W$ [1, Theorem 2], and is a proper power. Moreover, this presentation is a Dehn presentation [1, Theorem 4], and hence $G_W$ is hyperbolic. QED
Secondly, if we fix the group and vary the word then the answer is dependent on the group. The answer is "no" for finite groups (which are all hyperbolic).
Lemma 2.
If $G$ is a finite group then there exists no word $W\in F(a, b)$ such that for every non-commuting elements $r_1, r_2\in G$ the substitution $W(r_1, r_2)$ is not a proper power in $G_W$.
Proof. As $G$ is finite, every element is a proper power. QED
Next, the answer is "yes" for torsion-free hyperbolic groups.
Lemma 3.
If $G$ is a torsion-free hyperbolic group then there exists a word $W\in F(a, b)$ such that for every non-commuting elements $r_1, r_2\in G$ the substitution $W(r_1, r_2)$ is not a proper power in $G$.
In fact, for all $m\geq4$ the word $W$ can be chosen to be of length $\geq m$.
Proof. Firstly, assume that $G$ is free. Lyndon and Schützenberger proved that in a free group, if $x^i = y^jz^k$ with $i, j, k\geq2$ then the elements pairwise commute [2]. Therefore, taking $j, k\geq2$ such that $j+k=m$, we have that the word $W:=y^jz^k$ is never a proper power, as required.
Next, let $G$ be an arbitrary torsion-free hyperbolic group. Then there exists some number $N\in\mathbb{N}$ such that for all $a, b, c$ pairwise non-commuting elements of $G$ and numbers $p, q, r>N$ the subgroup $\langle a^p, b^q, c^r\rangle$ is free on the given basis (this is standard). Therefore, take both $j$ and $k$ to be $\max(m ,N)$, and suppose $x^i=y^jz^k$. If $i\geq N$ then the result holds by the above paragraph. Otherwise, take powers to $N$ to get $x^{Ni}=(y^jz^k)^N$, and note that this identity holds in the subgroup $\langle x^N, y^j, z^k\rangle$. Hence, the subgroup $\langle x^N, y^j, z^k\rangle$ is not free, and so either $[x, y]=1$, $[y, z]=1$ or $[x, z]=1$. Applying the fact that $x^i=y^jz^k$ also holds, and that we are in a torsion-free hyperbolic group so centralisers of elements are cyclic, we have that $x$, $y$ and $z$ pairwise commute, as required.
QED
Note that there exists an infinite family of hyperbolic groups for which the equation $x^iy^jz^k=1$ has non-commuting solutions [3] (but here $i, j, k<N$).
[1] McCool, James, and Paul E. Schupp. "On one relator groups and HNN extensions." Journal of the Australian Mathematical Society 16.2 (1973): 249-256.
[2] Lyndon, Roger C., and Marcel-Paul Schützenberger. "The equation $ a^ M= b^ Nc^ P $ in a free group." The Michigan Mathematical Journal 9.4 (1962): 289-298.
[3] Brady, N., Ciobanu, L., Martino, A. and O Rourke, S., "The equation $x^{p} y^{q}= z^{r}$ and groups that act freely on $\Lambda$-trees." Transactions of the American Mathematical Society 361.1 (2009): 223-236.
