I am reading this text in combinatorial group theory. In this text, the definition of a free basis is defined:
Let $S$ be a subset of a group $F$ then $S$ is a free basis if any set map $\varphi: S \to G$ from $S$ to a group $G$ can be extended to a unique homomorphism $\theta: F \to G$, i.e. $\varphi(s) = \theta(s)$ for all $s \in S$.
Let $C$ be the infinite cyclic group generated by $a$, then both $\{a\}$ and $\{a^{-1}\}$ af free bases of $C$. The text then states that these are the only free bases, but I can not figure out why this is. I tried with $G$ the group of order $2$ and mapping $a^k$ onto the non-identity element ($k$ not equal to $\pm 1$), but could not figure out a contradiction. Any help is appreciated.
You have almost the right idea. Write $C_k$ for the cyclic group of order $k$, which has identity $0$ and generator $g$, and let $S$ be the underlying set of the infinite cyclic group $C$ which contains generator $a$.
Let $k$ be fixed, and consider the set map $\phi: \{a^k\} \to C_k$, given by $\phi: a^k \mapsto 0$. Then we can extend this to the trivial homomorphism $\theta: x \mapsto 0$ for all $x \in C$. On the other hand, we can also extend it to the surjective $\psi: a^i \mapsto (i \ \mathrm{mod}\ k)$.
These are two different homomorphisms since $i \not = \pm 1$, so our set $\{a^k\}$ is not a free basis.