 Solution

A map $f:X\to Y$ is a quasi-isometry if there are $\lambda \ge 1$ and $\epsilon ,R\ge 0$ such thatfor all $x,x^{\prime }\in X$, and every point of $Y$ lies within distance $R$ of $f(X)$. The Milnor–Švarc lemma says that a group acting properly discontinuously, cocompactly and isometrically on a proper geodesic metric space is finitely generated, and each orbit map from a word metric is a quasi-isometry.

Now let $S,T$ be finite generating sets of $G$. PutReplacing each letter in an $S$-word by a $T$-word and conversely givesThus the identity map is a bilipschitz equivalence, hence a quasi-isometry. All finite generating sets of a finitely generated group give quasi-isometric Cayley graphs.