# Groups with exponentially growing centre

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 $$S$$).

1. 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 $$|S^n \cap gZ| \leq |S^{2n} \cap Z|$$.

1. 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.

1. 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.

• Probably. I'd try Hall's group $G(A)$ of matrices $$\begin{pmatrix}1 & x & z\\0 & t^n & y\\ 0 & 0 & 1\end{pmatrix}$$ with $x,y,z\in A$, $n\in\mathbf{Z}$; here $A$ is the ring generated by $t,t^{-1}$, which leaves several choices, for instance $A=\mathbf{F}_p[t^{\pm}]$ (central extension of the lamplighter) or $A=\mathbf{Z}[1/p]$, $t=p$ (related to Baumslag-Solitar group $\mathrm{BS}(1,p)$). – YCor Sep 11 at 9:18
• Yes, that's simpler. It's not unlike my suggestion, but with the advantage that by taking a different ring than $\mathbf{Z}$ you can take the automorphism $\varphi$ to be diagonal, which makes it easier on the brain. – Sean Eberhard Sep 11 at 9:32
• With $A = \mathbf{F}_p[t^{\pm1}]$ and the generating set consisting of $\varphi^{\pm1}$ times all matrices with entries of degree $0$, I think you have $|S^n Z/Z| \approx p^{2n}$, $|S^n| \approx p^{4n}$, $|S^n \cap Z| \approx p^{2n}$. – Sean Eberhard Sep 11 at 9:35
• My previous comment was hasty, and the situation is actually rather delicate. It turns out that for most points of $S^n$ the coordinate $z$ is determined by $x$ and $y$, basically because of the geometry of the lamplighter: for most points of $S^n$, the lamplighter made an injective walk, and if you only allow $\pm1$ steps then the walk must be unidirectional. I think you can get around this by allowing bigger steps or by walking in $\mathbf{Z}^2$ instead of $\mathbf{Z}$, but I'm still thinking about it... – Sean Eberhard Sep 12 at 13:55
• I have however managed to convince myself that $|S^n \cap Z|$ grows exponentially, so this probably answers question 3 negatively. – Sean Eberhard Sep 12 at 14:00

In fact $$\omega_{G/Z}(SZ/Z) = \omega_G(S)$$, despite the plausible-sounding examples in the comments. The argument is actually embarassingly simple. Let $$S_1$$ be a lift of $$SZ/Z$$ to $$G$$, so that $$|S_1| = |SZ/Z|$$ and $$S \subset S_1 Z_1$$ for some $$Z_1 \subset Z$$. Then $$S^n = S_1^n Z_1^n$$, and $$Z_1^n$$ grows polynomially, so $$\omega_G(S) \leq |S_1| = |SZ/Z|$$. Thus also $$\omega_G(S^n) \leq |S^nZ/Z|$$, but $$\omega_G(S^n) = \omega_G(S)^n$$ and $$|S^nZ/Z|^{1/n} \to \omega_{G/Z}(SZ/Z)$$.
The examples in the comments do however have the property that $$|S^n \cap Z|$$ grows exponentially, but this is not obvious. Let $$G$$ be Hall's group (as suggested by YCor) of matrices of the form $$\left( \begin{array}{ccc} 1 & x & z \\ 0 & t^k & y \\ 0 & 0 & 1 \end{array} \right),$$ where $$k \in \mathbf{Z}$$ and $$x,y,z\in A = \mathbf{F}_p[t^{\pm 1}]$$. Let $$H$$ be the subgroup defined by $$k=0$$ and $$x,y,z\in\mathbf{F}_p$$, let $$\varphi$$ be the diagonal matrix $$(1, t, 1)$$. Let $$S$$ be the symmetric generating set $$H \varphi^{\pm1} H$$. Modulo $$Z$$, this group is just the lamplighter group $$\mathbf{Z}^2 \wr \mathbf{Z}$$, and the generating set $$S$$ represents "change a lamp, take a step, change a lamp". The $$(x, y)$$ coordinates therefore record the lamp configuration at the end of the walk, and the $$z$$ coodinate records second-order information. To be precise, we have $$S^n = \bigcup H^{\varphi^{i_0}} H^{\varphi^{i_1}} \cdots H^{\varphi^{i_n}} \varphi^{-i_n},$$ where the union is over all $$n$$-step walks $$i_0, \dots, i_n$$ in $$\mathbf{Z}$$, by which I mean $$i_0=0$$ and $$|i_t - i_{t-1}| = 1$$ for each $$t > 0$$. We have $$H^{\varphi^{i_0}} H^{\varphi^{i_1}} \cdots H^{\varphi^{i_n}} = \left\{\left( \begin{array}{ccc} 1 & x_0 t^{i_0} + \cdots + x_n t^{i_n} & \sum_{s < t} x_s y_t t^{i_s - i_t} + z_0 + \cdots + z_n \\ 0 & 1 & y_0 t^{-i_0} + \cdots + y_n t^{-i_n} \\ 0 & 0 & 1 \end{array} \right) \colon x_i, y_j, z_k \in \mathbf{F}_p \right\}.$$ Thus the $$z$$ coordinate counts differences $$i_s - i_t$$ between lamp positions where the first lamp was lit before the second lamp. Now consider a walk of the following form: first light an $$x$$ lamp, then walk around lighting $$y$$ lamps however you like, then return to the origin and extinguish the $$x$$ lamp, then go extinguish all the $$y$$ lamps, and finally return to the origin. If you do this then your final $$x$$ and $$y$$ coordinates are zero, since they just record the final lamp configuration, but your $$z$$ coordinate records the configuration of $$y$$ lamps between the two $$x$$ lightings. This argument proves that $$|S^n \cap Z|$$ grows at least as fast as $$p^{n/4}$$.