Groups with two ends: showing that either $E\Delta gE$ is finite or $(E\Delta gE)^\complement$ is finite. 
Let $G$ be a finitely generated group with $e(G) = 2$, and let
$\Gamma$ be a Cayley graph of $G$. There is then a finite subgraph $C$
such that $\Gamma \setminus C$ has exactly two connected, unbounded
components. By adding all the finite components of $\Gamma \setminus
 C$ to $C$, we may assume that $\Gamma \setminus C$ consists of exactly
two connected components, each of which is unbounded. Choose one of
these two complements, and let $E \subset G$ consist of those elements
of $G$ that correspond to the vertices in that component. Notice that
we may apply elements of $G$ to $E$, forming subsets $$gE = \{g \cdot
 h \mid h \in E\}.$$ Similarly we can left multiply $E^\complement$: $$gE^\complement
 = \{g\cdot k \mid k \not\in E\}.$$

Now the lemma:

Lemma 11.30. Let $G$ and $E$ be as above, and let $g \in G$. Then, because $G$ is two-ended, either $E\Delta gE$ is finite or $(E\Delta
 gE)^\complement$ is finite.

Note:

*

*$A\Delta B$ denotes the symmetric difference $(A\cup B)\setminus (A\cap B)$.


*$e(G)$ denotes the number of ends of the group $G$.
I already asked a question about this lemma here: Geometric Group Theory, Meier lemma 11.30 about a two-ended groups $G$
This time, i'd love to ask for an idea on how to prove this lemma. The reason why i'm a bit struggling on this is because i am really uncertain if i am not missing any scenario on how $gE$ could possibly look like.
If we imagine $G$ to be the group $(\mathbb{Z},+)$ it's quite simple. $gE$ becomes the set of elements $\{g+h \mid h \in E\}$ which i can picture as some sort of translation.
But i can't find a way to generalize this to all possible scenarios where the group operation can be anything else.
The key observation is certainly that the group is two ended, but i haven't figured out how that implies the given claim.
Any help is appreciated.
 A: Let's write
$$\Gamma = E' \cup C \cup E
$$
where $E,E'$ are the two components of $\Gamma-C$, both unbounded.
We also have
$$\Gamma = gE \cup gC \cup gE'
$$
where $gE,gE'$ are the two components of $\Gamma - gC$, also both unbounded.
Now we need a case analysis. The group $G$ acts on the set consisting of its two ends. Thus each individual group element acts as a permutation of those two ends.
There are now two cases to consider, depending on whether or not $g$ fixes or swaps the two ends.
Case 1: $g$ fixes the the two ends. Thus $E$ and $gE$ represent the same end and so $E \Delta gE$ is bounded.
Case 2: $g$ swaps the two ends.
Thus $E$ and $gE'$ represent the same end, so $E \Delta gE'$ is bounded, and so $E \Delta gE^\complement = E \Delta (gE' \cup gC)$ is bounded.
Also $E'$ and $gE$ represent the same end, and so similarly $E^\complement \Delta gE$ is bounded.
Also $E'$ and $gE'$ represent different ends, so $E' \cap gE'$ is bounded, and so $E^\complement \cap gE^\complement = (E' \cup G) \cap (gE' \cup gC)$ is bounded.
Putting these together it follows that
$$(E \Delta gE)^\complement = (E \cap gE^\complement) \cup (E^\complement \cap gE^\complement) \cup (E^\complement \cap gE)
$$
is bounded.
A: The group is virtually cyclic and infinite. Therefore in the Cayley graph there is a line (more precisely, a quasi-geodesic)  $l$ such that the whole Cayley graph is in a $\delta$-neighborhood of $l$ for some $\delta$ This implies your statement (just as for the group of integers).
