# systole and residualy finite fundamental group

it is well known that if the fundamental group of a manifold M is residualy finite then for every point p in M and every epsilon positive there is a finite covering such that for a point q in the fiber over p we have systole(q)>epsilon . can anyone tell me how do we prove that ???

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Assume that $M$ is compact. Let $\tilde{M}$ be the universal cover of $M$, and let $G$ be the fundamental group of $M$, so that $G$ acts properly, freely, co-compactly on $\tilde{M}$. Let $x_0$ be any point in $\tilde{M}$. As is well-known, the orbit map $G\rightarrow\tilde{M}:g\mapsto gx_0$ is a quasi-isometry, where $G$ is endowed with word metric associated with some finite generating set. Fix $R>0$, and let $N$ be a normal, finite index subgroup in $G$ which meets the ball $B(e,R)$ in $G$ only at the identity $e$ (here we use residual finiteness of $G$). Then in the finite graph $N\backslash G$, the systole at $e$ will be larger than $R$. This means that the finite cover $N\backslash\tilde{M}$ will have at $x_0$ a systole larger than some fixed affine function of $R$.
Recall the following result by Milnor (Lemma 2 in {\it A note on the fundamental group}, J. Diff. Geom. 2 (1968), 1-7), sometimes called "Fundamental theorem of geometric group theory": Let $X$ be a proper geodesic space and $G$ be a discrete group of isometries of $X$ acting properly co-compactly on $X$. Then $G$ is finitely generated and, when endowed with the word metric, $G$ is quasi-isometric to $X$. – Alain Valette Apr 30 '11 at 12:58