# Isometry of balls in residually finite groups

A finitely generated group is called residually finite if there exists a sequence of normal subgroups of finite index whose intersection is trivial. I am to prove the following: "Let $$G$$ be a residually finite group and let $$N_{i} \subseteq G$$ be a sequence of finite index subgroups with trivial intersection such that $$N_{i} \subseteq N_{j}$$ whenever $$J\leq i$$. Then for every $$R > 0$$ there exists $$i_{0} \in \Bbb N$$ such that $$B_{G}(e,R)$$ is isometric to $$B_{G/N_{i}}(e,R)$$ for all $$i \geq i_{0}$$". While working with some concrete examples, I guess that we have to choose (and we can) $$i_{0}$$ such that $$|G/N_{i_{0}}| > 2R$$ for given $$R > 0$$. Now if we take an element of length $$l$$, it has to be preserved in the ball of the quotient group. I do not see how I may proceed further. More precisely, suppose $$g \in B_{G}(e,R)$$ is expressed as $$s_{1}^{k_{1}}s_{2}^{k_{2}}.....s_{m}^{k_{m}}$$, where $$s_{i}$$ are generators and $$\Sigma |k_{i}| = l$$. The element $$gN_{i_{0}}$$ in $$B_{G/N_{i_{0}}}(e,R)$$ will become $$s_{1}^{k_{1}}s_{2}^{k_{2}}.....s_{m}^{k_{m}}N_{i_{0}}$$. It may so happen that some of the generators are elements of $$N_{i_{0}}$$, but under the given conditions, I have to show that the word length is preserved.

We have $$d_{G/N_i}(aN_i,bN_i)=\inf\{d_{G}(a,bx)|x\in N_i\}$$ as a word $$w$$ in the generators satisfies $$waN_i=bN_i$$ if and only if $$wa=bx$$ for some $$x\in N_i$$.

We may choose $$i_0$$ sufficiently large that $$B_G(e,4R)\cap N_{i_0}=\{e\},$$ as balls contain finitely many elements.

Then for $$i\geq i_0$$, if $$a,b\in B_G(e,R)$$ and $$wa=bx$$ for some $$x\in N_i$$, then $$x=b^{-1}wa$$, so if $$x\neq e$$ we have the length of $$w$$ must be at least $$4R-2R=2R$$.

Thus for $$x\in N_i \backslash \{e\}$$: $$d_G(a,b)\leq 2R\leq d_G(a,bx).$$

Thus $$d_{G/N_i}(aN_i,bN_i)=\inf\{d_{G}(a,bx)|x\in N_i\}=d_G(a,b).$$

We may conclude that the map sending $$a\mapsto aN_i$$ is an injective isometry $$B_G(e,R)\to B_{G/N_i}(e,R)$$.

Finally note this map is surjective, as if $$d_{G/N_i}(e,aN_i), then $$wa=x$$ for some $$w$$ a word of length at most $$r$$ in the generators and $$x\in N_i$$. Then we have $$w(ax^{-1})=e$$ and $$ax^{-1}\mapsto aN_i$$, with $$ax^{-1}\in B_G(e,R)$$.

• "We may choose $i_0$ sufficiently large that $B_G(e,4R)∩N_{i_0} = \{e\},$ as balls contain finitely many elements" - Can you expand this a little? I sort of get the geometric intuition but am not able to prove this. Commented May 9 at 15:50
• The ball $B_G(e,4R)$ contains finitely many elements $g_1,\cdots,g_n$. As the intersection of the $N_i$ is empty, for each $k=1,\cdots,n$ we have some $i_k$ such that $g_k\notin N_{i_k}$. Then just let $i_0$ be the maximum of $i_1,\cdots,i_n$.
– tkf
Commented May 9 at 16:26