Given a finitely generated group $G$ and a finite subset $S$, let $\omega_G(S) = \inf_{n\geq1} |S^n|^{1/n}$ (this is the exponential growth rate of $S$).

Is there a finitely generated group $G$ with centre $Z$ and symmetric generating set $S$ such that $\omega_{G/Z}(SZ/Z) < \omega_G(S)$?

Here is my idea for a construction. Let $H$ be the discrete Heisenberg group, and let $\varphi$ be an automorphism of $H$ that acts on $H^\text{ab} \cong \mathbf{Z}^2$ as a matrix with eigenvalue $\lambda > 1$, and let $G = H \rtimes_\varphi \mathbf{Z}$. Let $S$ be the union of the standard generating sets for $H$ and $\mathbf{Z}$. Note that $\varphi$ preserves the centre of $H$, so $Z(G) = Z(H) \cong \mathbf{Z}$, and $G/Z(G) \cong \mathbf{Z}^2 \rtimes_\varphi \mathbf{Z}$. Since $\lambda > 1$ we have exponential growth in $G/Z(G)$, and my intuition is that there is enough extra space in $H$ that we should have even greater growth in $G$. The details elude me, however, and maybe somebody has a reference or a simpler example.

Assuming that the answer to question 1 is "yes", it follows that $|S^n \cap Z|$ grows exponentially. Indeed since $S^n$ is growing much faster than $S^n Z/Z$, the max fibre-size $\max_g |S^n \cap gZ|$ must grow exponentially, and $|S^n \cap gZ| \leq |S^{2n} \cap Z|$.

How fast can $|S^n \cap Z|$ grow compared to $|S^n|$? In other words, what is the supremum of $\inf_{n\geq1} |S^n \cap Z|^{1/n} / \omega_G(S)$ over all $G$ and $S$? Is it $1$?

At the moment I cannot even convince myself that you could not have even $|S^n \cap Z| \gg |S^n|$ for all $n$.

Finally, a technical question.

Do we always have $|S^n \cap Z|^{1/n} \to \omega_G(S) / \omega_{GZ/Z}(SZ/Z)$?

By the argument before quetion 2, the exponential growth rate of $|S^n \cap Z|$ is at least $(\omega_G(S) / \omega_{G/Z}(SZ/Z))^{1/2}$. It's plausible but not obvious that the largest fibre over $Z$ is $S^n \cap Z$ itself, so it's plausible that we can remove the square-root. The other inequality is more suspect, because we might have lots of small fibres.