If we have a group $G$ and $H \leq G$ with infinite index, show there is an infinite sequence of vertices $(a_i)_{i \in \mathbb{N}}$ in the Cayley graph $\Gamma$ of $G$ (with respect to some not necessarily finite generating set) such that:
- $a_0 \in H$
- $d(a_i, a_{i+1}) = 1$
- $d(h, a_i) \geq i$ for all $h \in H$ and $i$
It is easy to show that if $H$ forms a net for $G$. i.e. there is some $M$ such that for all $g \in G$, $d(g, H) \leq M$, then $H$ is finite index (since all cosets have a representative less than $M$ long). So infinite index means we can find elements arbitrarily far from $H$ (contrapositive).
I am trying to prove this by induction. Base case is where we start with $a_0 = e$, then since $H \neq G$, there is a generator not in $H$, we pick that for $a_1$. Then if we have the sequence up to $a_n$, we know there is some vertex $v$ of $G$ that is further than $n+1$ distance from $H$. I suppose we can choose the $v$ closest to $a_n$ and that should work as a choice for $a_{n+1}$? But how do we know that $v$ is adjacent to $a_n$? How do we prove that we will never have to move closer to $H$ to get to $v$ (I guess this is not a problem if we prove that $v$ is adjacent to $a_n$)?