# Infinitely divisible elements in Gromov hyperbolic groups

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 CW-complex (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.

• Torsion elements are infinitely divisible. So you're asking about elements of infinite order.
– YCor
Sep 22 at 19:38
• Let $g$ be of infinite order, with arbitrary large roots $g_n$. Each $g_n$ commutes with $g$. But the centralizer of $g$ is finite-by-cyclic, i.e., surjects onto $\mathbf{Z}$ with finite kernel. Projecting, we get a contradiction.
– YCor
Sep 22 at 19:40
• Thanks! Yes, I was asking about an ifinite order element. I will make a correction Sep 22 at 20:02