Groups with finitely many roots Let $G$ be a finitely generated infinite group and suppose that $g \in G$ is of infinite order. We say that $x \in G$ is a primitive root of $g$ (denote $x \in \sqrt[G]{g}$) in $G$ if $x^n = g$ and $n \in \mathbb{N}$ is maximal possible.
Is there some condition on $G$ to ensure that primitive root (not necessarily unique) exist for every $g \in G$?
I guess different wording would be for which groups does the equation $x^n = g$ have solution only for finitely many $n \in \mathbb{N}$?
 A: There are indeed several important (and rich) classes of groups which satisfy this condition, drawn from the field of geometric group theory.


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*hyperbolic groups (including fundamental groups of compact hyperbolic manifolds)

*$\text{CAT}(0)$-groups (including fundamental groups of compact Riemannian manifolds having sectional curvatures $\le 0$)

*automatic groups
In each of these cases, there is a certain amount of topology/geometry behind the proof that each element has a primitive root. 
Here is a brief example of such a proof. Suppose $M$ is a compact Riemannian manifold of sectional curvature $\le 0$ (for example a flat torus with fundamental group $\mathbb{Z} \oplus \mathbb{Z}$, or a closed hyperbolic surface of genus $\ge 2$). Every element of $\pi_1(M)$ has a "length" which can be thought of either as the translation length of that element under the deck transformation action on the universal covering space, or as the length of a closed geodesic loop to which a based loop representing that element is freely homotopic. There are two geometric properties of "length", proved using the curvature hypothesis, from which it follows that primitive roots exist:


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*There is a constant $L$ such that $0 < L \le \text{length}(g)$ for all $g \in \pi_1(M)$

*$\text{length}(g^n) = n \cdot \text{length}(g)$ for all $g \in \pi_1(M)$ and $n \in \mathbb{Z}$.

