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Given a group $G$, the subgroup growth functions are given by

$a_n(G) :=$ the number of subgroups of $G$ of index exactly $n$.

$s_n(G) :=$ the number of subgroups of $G$ of index at most $n$.

The latter is clearly increasing. I wonder when the sequence $a_n(G)$ enjoys similar behavior.

To be specific, say a sequence $\{ x_n \}_{n=1}^\infty$ is roughly increasing if there exists $C > 1$ such that $x_{C n} \geq x_n$ for all natural numbers $n$.

Question: For what groups $G$ is $a_n(G)$ roughly increasing? In particular, is $a_n(G)$ roughly increasing for $G = SL_3(\mathbb{Z})$?

Note that the sequence $a_n(G)$ is roughly increasing for any finitely generated nilpotent $G$ that is infinite. On the other hand, the sequence $a_n(G)$ is never roughly increasing for finite $G$.

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  • $\begingroup$ Does this not follow from the rather precise estimates for $a_n$ obtained by Lubotzky--Nikolov (mathscinet.ams.org/mathscinet-getitem?mr=2155033) for arithmetic groups with CSP? (I'm asking before doing the computation to see if you tried yourself). $\endgroup$
    – Jean Raimbault
    Aug 27, 2018 at 8:55
  • $\begingroup$ Thank you for your reply and for thinking about my question. The main challenge to applying their work is that they only control the growth of $\log(s_n)$. $\endgroup$
    – Khalid Bou-Rabee
    Aug 27, 2018 at 10:30
  • $\begingroup$ You're right, I assumed they dealt with $a_n$ but I was mistaken. Do you know if their proof or that of Goldfeld--Pyber--Lubotzky at least implies that $a_n \ge 1$ for sufficiently large $n$? $\endgroup$
    – Jean Raimbault
    Aug 27, 2018 at 11:31
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    $\begingroup$ I believe the answer to Jean Raimbault's question is no, since there seems to be infinitely many $n$ where $a_n(G) = 0$ for $G = SL_3(\mathbb{Z})$. For this, apply Guralnick's theorem from his J. Algebra paper "Subgroups of Prime Power Index in a Simple Group" to show that for sufficiently large p and k, $a_{p^k}(G) = 0$. Note this doesn't show that $a_n(G)$ is not roughly increasing. $\endgroup$
    – Khalid Bou-Rabee
    Aug 27, 2018 at 12:44
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    $\begingroup$ My gut feeling is that $a_n(SL_3(\mathbb{Z}))$ ought to be roughly increasing. If one takes $C$ to be an index in which a lot of finite-index subgroups of index $C$ in $SL_3(\mathbb{Z})$ exist, then intersecting these with subgroups of index $n$, I would expect "most" to have maximal index, i.e. $Cn$. $\endgroup$
    – Ian Agol
    Aug 29, 2018 at 5:51

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