An element $g\in G$ in a group $G$ is called infinitely divisible if $b=y^n$ for infinitely many different $n\in {\Bbb Z}$. It is not hard to find a finite CWcomplex (or even a compact manifold) with a fundamental group containing an infinitely divisible element. For example, consider a group generated by $x$ and $b$ with a relation $xbx^{1}=b^2$. Then $b$ is infinitely divisible. However, in a hyperbolic manifold every element can be represented by a unique shortest geodesic, which implies that infinite divisibility does not occur. Now, suppose that $G$ is a finitely generated Gromov hyperbolic group (a posteriori, it is finitely presented, as Gromov proved). It seems that it cannot contain infinitely divisible elements of infinite order. I would be very grateful for any reference to this statement.

2$\begingroup$ Torsion elements are infinitely divisible. So you're asking about elements of infinite order. $\endgroup$– YCorSep 22 at 19:38

6$\begingroup$ Let $g$ be of infinite order, with arbitrary large roots $g_n$. Each $g_n$ commutes with $g$. But the centralizer of $g$ is finitebycyclic, i.e., surjects onto $\mathbf{Z}$ with finite kernel. Projecting, we get a contradiction. $\endgroup$– YCorSep 22 at 19:40

$\begingroup$ Thanks! Yes, I was asking about an ifinite order element. I will make a correction $\endgroup$– Misha VerbitskySep 22 at 20:02
1 Answer
YCor has answered your question in the comments.
But here is another proof anyway. The "asymptotic translation lengths" of elements (in a fixed word hyperbolic group) are uniformly rational. Also, the asymptotic translation length of an element is zero if and only if the element has finite order. Thus if an element is infinitely divisable, so is its asymptotic translation length, which thus must be zero, hence the element is torsion.
Bowditch's paper Tight geodesics in the curve complex says that this result is contained in Gromov's long paper Hyperbolic groups, and "an elegant proof can be found in [D]" (which is Delzant's paper Sousgroupes distingues et quotients des groupes hyperoliques). I could not find the result in either. But Bowditch is very reliable, so I probably missed it...

2$\begingroup$ You can find also find a proof in Bridson and Haefliger. $\endgroup$– HJRWSep 23 at 6:12