This is probably a basic algebraic (or even set-theoretic) matter. I am reading "Groups, Graphs and Trees" by J. Meier.
It's about Lemma 11.30 which is left as an exercise to the reader. Before I quote the lemma, I need to quote the following prescription:
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 $(A\cup B)\setminus (A\cap B)$.
I have issues seeing why the given lemma is true. Now let's take the finitely generated two-ended group $\mathbb{Z}$ with the generating set $\{1\}$ and consider the respective Cayley graph $\operatorname{Cay}(\mathbb{Z},\{1\})$.
Let's take $C = \mathcal{B}(0,2) =\{-2,1,0,1,2\}$ as the finite subgraph, that is the ball centered at the vertex corresponding to $0$ with radius $2$.
Then we get the two unbounded connected components
$$D_+ = \{3,4,5,...\}$$ and $$D_- = \{...-5,-4,-3\}$$
Now take the subset $E \subset D_+$ as $$E = \{5,6,7,8,...\}$$
Let's say i choose any positive integer $g \ge 1$, for example $g=2$, then clearly $2E \subset E$. But then we have $$E\Delta 2E = (E\cup 2E)\setminus (E\cap 2E) = E\setminus gE = \{5,6,7,...\}\setminus \{10,12,14,...\} = \{3,4,11,13,15...\}$$ so obviously $$\vert E\Delta 2E \vert = \infty.$$
However, the complement yields $$(E\Delta 2E)^\complement = (E\setminus 2E)^\complement = D_-\cup \mathcal{B}(0,2) \cup \{5,6,7,8,10,12,...\}$$ which is clearly also infinite
$$\vert (E\Delta 2E)^\complement \vert = \infty$$
So my question is: Why is lemma 11.30 true? What am i doing wrong? I am clearly missing something or doing something fundamentally wrong in my argument.