I have a group $G$, then I take a quotient $G/\Bbb Z$ and I seem to get a Prufer 2 group which has $2^n$ elements in the $n$th subgroup. But the original group had way more than $\Bbb Z$ elements in each subgroup. It was more like $\Bbb Z^n$ elements in the nth subgroup.
This leads me to think I've made a mistake. Does this hang together or does it sound iffy?
Obviously $\Bbb Z^n:n\in\Bbb N$ is not a set of $n$ distinct quantities as I imply, but hopefully you will get what I mean with respect to the increasing structure and complexity of the group.
If I understand you correctly, the answer is no. If $G$ is finitely generated, and you have an exact sequence
$$\DeclareMathOperator{rank}{rank} 0 \to \mathbf{Z} \to G \to G/\mathbf{Z} \to 0, $$
then $\rank G = \rank \mathbf{Z} + \rank G/\mathbf{Z}$. To see this, tensor with $\mathbf{Q}$ to get an exact sequence of vector spaces and apply rank-nullity. Note that $\rank M = \dim_{\mathbf Q} \mathbf{Q} \otimes M$.
If the map $f : \mathbf{Z} \to G$ isn't injective, then it factors through $\mathbf{Z}/\ker f \to G$ and $\rank G = \rank G/\mathbf{Z}$ since $\rank \mathbf{Z}/\ker f = 0$.