Finite index subgroup is quasi-isometric to original group

Question

I have been attempting the following problem for a while:

Let $G$ be a group generated by some finite set $X$. If $H \underset{f.i.}{\leq} G$ is a subgroup of finite index, then $H$ is quasi-isometric to $G$.

The notes I'm following suggest this can be done without anything too fancy. I am allowed to use the fact that $H$ is finitely generated (this was proved earlier). The metric we are using here is the word metric; $d_G(g_1,g_2) = \min \{ n\in \mathbb{N} \mid g_2=g_1x_1\dots x_n, \text{ where } x_i\in X\}$. And in $H$ the metric is $d_H(h_1,h_2) = \min \{ n\in \mathbb{N} \mid h_2=h_1y_1\dots y_n, \text{ where } y_i\in Y \}$ for some generating set $Y$ of $H$.

The main trouble I am having is to get one of the bounds. My attempt went something like this: Take the inclusion map $H \to G$. We must show there is a $\lambda \geq 1, \epsilon > 0$ such that $$ \frac{1}{\lambda} d_H(h_1,h_2) -\epsilon \leq d_G(h_1,h_2) \leq \lambda d_H(h_1,h_2) + \epsilon $$ for all $h_1,h_2\in H$. Furthermore there must be some $C\geq0$ such that for all $g\in G$ there is a $h\in H$ such that $d_G(g,h)\leq C$.

This last property I can show using the finite index. I can also find an upper bound on $d_G(h_1,h_2)$ by writing the generators of $H$ in terms of the generators of $G$. However, I have had no luck getting any kind of lower bound. Any advice would be much appreciated.

(Some people have also suggested trying a different map, like one from $G\to H$ based on a transversal, but I have had no luck with this either).

Answer 1

The most elegant proof is probably using the Milnor-Svarc lemma: let $H$ be a group acting geometrically (i.e., properly discontinuously and coboundedly by isometries) on a proper geodesic metric space $X$; then $H$ is finitely generated and q.i. to $X$. Given $H$ of finite index in $G$, you can let $H$ act on a Cayley graph of $G$ extending the action of $H$ on $G$ by left multiplication, and check that this action is geometric.

If you want to just prove it from the definitions, here is a way to simplify a lot the calculations for your last estimate: it is a general fact that if $H$ is of finite index in $G$, then there exists a subgroup $N$ of finite index in $H$ that is normal in $G$. So if you prove the theorem assuming normality then you are done: $N$ is q.i. to $H$ and $G$, and so $H$ and $G$ are q.i. Normality allows you to choose a transversal which is both left and right, and all calculations simplify.

Answer 2

Here's a brief account of how to do this.

Suppose you have $h_1,h_2 \in H$. Regarded as elements of $G$, you then construct a shortest expression $$h_1^{-1} h_2 = x_1...x_n, \text{where $x_i \in X$} $$ where $n = d_G(h_1,h_2)$.

You say that you know the "Furthermore..." part. So, from that you conclude that there's a finite set $D \subset G$, independent of $h_1$ and $h_2$, such that for each $i=1,...,n$ the element $x_1 .... x_i$ differs from some element of $H$ by a single element of $D$ which I'll denote $d_i$, and so $$x_1 ... x_i d_i \in H $$ Now you can write $$(*) \qquad h_1^{-1} h_2 = (x_1 d_1) (d_1^{-1} x_2 d_2) (d_2^{-1} x_3 d_3) .... (d_{n-2}^{-1} x_{n-1} d_{n-1}) (d_{n-1}^{-1} x_n) $$ (under the extra assumption that $Y \subset X$, you can visualize this equation nicely in the picture of the Cayley graph of $H$ embedded in the Cayley graph of $G$).

By construction (and induction) each parenthesized expression is in $H$. Furthermore, and this is the key, there are only finitely many elements of $H$ that have one of the forms $d^{-1} x d'$, or $d^{-1} x$, or $x d'$, as $x$ varies over $X$ and $d,d'$ vary over $D$. So we can choose a word in $Y$ for each of those finitely many elements, and we set the number $\lambda$ to be the maximum word length of those choices. The expression (*) then gives the inequality $$d_H(h_1,h_2) \le \lambda n = \lambda d_G(h_1,h_2) $$ so the left inequality is true with $\epsilon=0$.