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Let $X$ be a complete, simply connected, Riemannian manifold of negative curvature. Let $\Gamma$ be an almost nilpotent group of non-elliptic isometries containing a hyperbolic isometry $\gamma$. Then $\Gamma$ fixes a unique geodesic $\ell$ and is infinite cyclic.

It should follow from the fact that hyperbolic isometries fix a unique geodesic on each point of which they attain the minimum translation distance $d_\gamma=\min_{x\in X}d(x,\gamma(x))$.

I can prove the first part of the statement for nilpotent groups, because they have non-trivial center and if two isometries $\gamma_1,\gamma_2$ commute and one fixes a geodesic $\ell=\gamma_1(\ell)$, then also the other fixes it.

On the other hand I have some problem showing that $\Gamma$ is infinite cyclic. I was looking for counterexamples and came up with this: let $\gamma_1$ and $\gamma_{\sqrt{2}}$ two isometries of $X$ which translate the same geodesic $\ell$ of $1$ and $\sqrt{2}$. Their action on $\ell$ is the same of $\mathbb{Z}^2$. I know that there are no counterexamples, so I suspect that these two isometries generate a non-nilpotent group, and I would be able to detect it if I'd look at their action outside $\ell$.

So here are the questions: How can I prove the statement? Why the counterexample is in fact not a counterexample?

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    $\begingroup$ You are missing the discreteness assumption. $\endgroup$ – Moishe Kohan Jan 9 at 21:38
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In order to close this question. Yes, your example is a counter-example to the claim you are trying to prove. The key condition you are missing is that the group $\Gamma$ is discrete. Under this assumption the claim indeed holds. One proves this in three steps:

  1. Suppose that $g_1, g_2$ are hyperbolic isometries which share exactly one fixed point $p$ in the sphere at infinity of $X$. Then $[g_1, g_2]$ is a parabolic isometry $f$ of $X$ fixing $p$.

  2. If $g$ is a hyperbolic isometry of $X$ and $f$ is a parabolic isometry of $X$ such that $g, f$ have a common fixed point $p$ in the sphere at infinity of $X$, then either the sequence $$ f_n:=g^n f g^{-n} $$ or $$ f_{n}=g^{-n} f g^{n} $$ satisfies the property $$ d(x, f_n(x))\to 0 $$ for (any) $x$ on the axis of $g$. Thus, a discrete subgroup of isometries cannot contain two hyperbolic elements which share exactly one fixed points, cf. this question.

Step 3. If $\Gamma$ is almost nilpotent and contains non (nontrivial) elliptic elements then $\Gamma$ either has exactly one fixed point $p$ in the ideal boundary of $X$ or preserves a 2-point subset $\{p, q\}$ in this ideal boundary of $X$.

a. Steps 1 and 2 show that the first case is impossible.

b. In the second case, $\Gamma$ will preserve the unique geodesic $c\subset X$ connecting $p, q$. Then, restricting $\Gamma$ to this geodesic we obtain an isomorphism of $\Gamma$ to a discrete isometry group of the real line. From this, you conclude that $\Gamma$ is cyclic (since it contains no elliptic elements).

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