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I have started reading about subgroup growth and, to my surprise, I haven't found a reference to whether direct products preserve subgroup growth.

Recall that, given a finitely generated group $G$, the function $s_n(G)$ is given by $$s_n(G)=\#\{\text{subgroups of } G\text{ of index }\leq n\}.$$ We say that $G$ has subgroup growth type $f$ for some function $f$ if there are $a$, $b>0$ such that \begin{align} s_n(G)&\leq f(n)^a \quad & &\text{for } n \text{ large enough},\\ f(n)&\leq s_n(G)^b \quad & &\text{for infinitely many } n. \end{align}

What I have found is the following: Let $G$ be a finitely generated group, let $N$ be a normal subgroup and let $Q=G/N$. Then Proposition 1.3.2 in Subgroup growth by A. Lubotzky and D. Segal states \begin{align} s_n(G)&\leq s_n(Q)s_n(N)n^{\text{rk}(Q)},\\ s_n(G)&\leq s_n(Q)s_n(N)c^n,\qquad \text{where}\ c=3^{d(Q)/3}. \end{align} Of course, these inequalities can be applied to a direct product $G\times G$ by taking $N=G\times 1$.

This seems to suggest that there should exist a group $G$ of intermediate growth and infinite rank such that $G\times G$ has strictly faster subgroup growth than $G$; more precisely, given any $a>0$, $$s_n(G\times G)>s_n(G)^a \quad \text{for infinitely many } n.$$ Does such an example exist?

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  • $\begingroup$ Including a definition of “subgroup growth” would be a good reminder here. $\endgroup$ Commented Sep 9, 2022 at 9:06
  • $\begingroup$ I added more details. Thanks for your suggestion. $\endgroup$
    – user44172
    Commented Sep 9, 2022 at 9:45

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Subgroup growth of direct product is quite difficult in general. However, it is easier for pro-$p$ groups. You might like to see a recent paper: Y. Barnea and J.-C. Schlage-Puchta, Branch groups, orbit growth, and subgroup growth types for pro-p group, Forum Math. Pi 8 (2020). We compute there the subgroup growth of the Grigorchuk group (and more complicated stuff). The big obstacle was exactly dealing with direct sums and we were able to solve it because we did think about it as a pro-$p$ group rather than in general.

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