I want to show that any bijective quasiisometry $f:G\to H$ (whith $G$ and $H$ groups, finitely genered by S and T respectely) is a biLipshitz equivalence.
I know this:
There are constants $\lambda\geq 1$ and $C\geq 0$ such that $f$ is a $(\lambda, C)-$quasiisometry, ie: for any $a,b\in G$: $$\frac {1}{\lambda }d_S(a,b)-C\leq d_T(f(a), f(b))\leq \lambda d_S(a,b)+C$$
My idea: clearly this is true for any $\lambda^\prime\geq \lambda$ and $C^\prime \geq C$, I suposse that i can take $\lambda^\prime$ big enough such that $f$ is a $(\lambda^\prime,0)-$quasiisometry, then $f$ a surjective biLipshitz and thus $f$ is a biLipshitz equivalence.
But I don't know how to prove it.
Your suggestions would be greatly appreciated. :)