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I am interested in the growth rate of this type of group: $G=\mathbb{Z}^2\rtimes_{\sigma} \mathbb{Z}$, where $\sigma(a)=\begin{pmatrix}x&y\\z&w\end{pmatrix}\in SL_2(\mathbb{Z})$, where $a$ is the generator on the right copy of $\mathbb{Z}$ and the action is just by matrix multiplication.

Here are two examples:

For $\sigma(a)=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, this gives us the discrete Heisenberg group $H_3$, which is nilpotent, and hence by Gromov's theorem, it has polynomial growth rate(see here).

When $\sigma(a)=\begin{pmatrix}2&1\\1&1\end{pmatrix}$, it was mentioned here this group has exponential growth rate.

So my first question is:

1, Could anyone give me a reference to show the link between whether the above group $G$ has polynomial growth rate or not and the property, say eignvalue, of the matrix $\sigma(a)$?

Note that the above $G$ is a polycyclic-by-finite group, my question is:

2, Could anyone give me a polycyclic-by-finite group not of the type of $G$ with exponential growth rate?

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  • $\begingroup$ The theorem that $H_3$ and other (virtually) nilpotent groups have polynomial growth is a theorem of Milnor, with the exact degree of polynomial growth computed by Bass. Gromov's theorem is the converse: every group of polynomial growth is virtually nilpotent. $\endgroup$
    – Lee Mosher
    Commented Aug 11, 2013 at 16:02
  • $\begingroup$ @Lee, in your answer, you mentioned Milnor's paper, I checked it, but it is still not clear to me how to relate the nilpotentness of $G$ to the property of $\sigma(a)$, could you give more hints? $\endgroup$
    – Jiang
    Commented Aug 11, 2013 at 22:14
  • $\begingroup$ @Lee, especially, is the lemma 1 in Milnor's paper useful in our situation? $\endgroup$
    – Jiang
    Commented Aug 11, 2013 at 22:17
  • $\begingroup$ @Jiang: For your group to be (virtually) nilpotent, $\sigma(a)$ must fix a point (otherwise the center of $G$ would be trivial). You can also check this is a sufficient condition (quotient by the fixed subgroup, and check it is (virtually) abelian). $\endgroup$
    – Steve D
    Commented Aug 13, 2013 at 2:47

1 Answer 1

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To answer your 1st question see the paper of Milnor, "Growth of finitely generated solvable groups."

To answer your 2nd question, simply generalize your second example to higher dimensions, e.g. take $\mathbb{Z}^3\rtimes_{\sigma} \mathbb{Z}$ where $\sigma \in SL_3(\mathbb{Z})$ has an eigenvalue not on the unit circle.

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  • $\begingroup$ thanks! For the type of $G$, I mean the type of $\mathbb{Z}^d\rtimes \mathbb{Z}$, maybe I should state it clearly next time. $\endgroup$
    – Jiang
    Commented Aug 11, 2013 at 16:14
  • $\begingroup$ Would one of type ${\mathbb Z}^d \lhd {\mathbb Z}^2$ suit? $\endgroup$
    – Derek Holt
    Commented Aug 11, 2013 at 17:36
  • $\begingroup$ @DerekHolt, what does $\vartriangleleft$ mean? $\endgroup$
    – Jiang
    Commented Aug 11, 2013 at 19:28
  • $\begingroup$ I typed the wrong symbol. I meant ${\mathbb Z}^d \rtimes {\mathbb Z}^k$ with $k>1$. You can construct examples like that from algebraic number fields $K$, where you let the torsion-free part of the group of units of $K$ act on the additive group of the integers of $K$, and the action is given by multiplication in $K$. $\endgroup$
    – Derek Holt
    Commented Aug 11, 2013 at 20:15

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