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.

  • 1
    $\begingroup$ 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 $\mathrm{BS}(1,p)$). $\endgroup$
    – YCor
    Commented Sep 11, 2019 at 9:18
  • $\begingroup$ 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. $\endgroup$ Commented Sep 11, 2019 at 9:32
  • 1
    $\begingroup$ 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 $|S^n Z/Z| \approx p^{2n}$, $|S^n| \approx p^{4n}$, $|S^n \cap Z| \approx p^{2n}$. $\endgroup$ Commented Sep 11, 2019 at 9:35
  • $\begingroup$ 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... $\endgroup$ Commented Sep 12, 2019 at 13:55
  • $\begingroup$ I have however managed to convince myself that $|S^n \cap Z|$ grows exponentially, so this probably answers question 3 negatively. $\endgroup$ Commented Sep 12, 2019 at 14:00

1 Answer 1


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 $p^{n/4}$.

So, Q1: no. Q3: no. Q2: still mysterious.


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