The question is: prove that an infinite finitely generated group $G$ contains an isometric copy of $\mathbb{R}$, i.e., contains a bi-infinite geodesic ($G$ is equipped with the word metric).
I do not even know what I have to prove. It does not make sense to me. The word metric of $G$ assumes values in the natural numbers. How could there be an isometry between a subgraph of the Cayley graph of $G$ and the real line $\mathbb{R}$.
I am really confused.
I found this question here (sheet 6, ex. 1).
On
Here's the general answer. First, show that any connected infinite locally finite graph contains a geodesic ray. Indeed, fix a vertex $x$ and find a vertex $x_n$ at distance $n$ to $x$ and a geodesic segment $S_n$ joining $x$ to $x_n$. By a compactness argument, the sequence $(S_n)$ accumulates on a geodesic ray emanating from $x$.
Now assume in addition that your graph is isometry-transitive (e.g. the Cayley graph of a f.g. group). Then by the previous case for each $n$ there a geodesic segment $T_n$ of length $2n$, which, using homogeneity, can be supposed to be centered at the fixed point $x$. The same compactness argument shows that $(T_n)$ accumulates to a bi-infinite geodesic. This is an isometrically embedding of the Cayley graph of $\mathbf{Z}$.
I'm just going to focus on what you've said you are confused about, namely:
"How could there be an isometry between a subgraph of the Cayley graph of G and the real line $\mathbb{R}$?".
We can extend the word metric on $G$ to a metric on the Cayley graph in a natural way, with each edge being an isometric copy of a unit interval. Under this metric, the Cayley graph of $\mathbb{Z}$ with respect to the generator $1$ is isometric to $\mathbb{R}$.