Let’s define a system of numeration in a group $G$ as an endomorphism $\phi$ of $G$, that satisfies the following three conditions:
1)$\bigcap_{n = 1}^\infty \phi^n(G) = E$
2)$\forall n \in \mathbb{N}$ $\phi^{n + 1}(G) \triangleleft \phi^n(G)$
3)$\forall n, m \in \mathbb{N}$ $\frac{\phi^n(G)}{\phi^{n + 1}(G)} \cong \frac{\phi^m(G)}{\phi^{m + 1}(G)}$
Let’s call a group integer if it «accepts» a system of numeration.
It is not hard to see, that the only finite integer group is the trivial one. It is also not hard to see, that $C_\infty$ is integer (all systems of numeration in $C_\infty$ are of the form $a \mapsto a^n$, where $n \in \mathbb{Z}\setminus \{-1; 0; 1\}$). Also the direct product of integer groups is also integer. From that we can conclude that all free abelian groups are integer.
My question is: Is an integer group always necessary free abelian? And if not, can it be non-abelian?
No and yes. For any group $G$ the group $G^{\mathbb{N}}$ satisfies this property with $\phi$ given by the forward shift
$$\phi(g_1, g_2, g_3, \dots) = (e, g_1, g_2, \dots).$$
We can also take, for example, the $p$-adic numbers $\mathbb{Z}_p$ with $\phi$ given by multiplication by $p$, for any prime $p$.