Let $G$ be a finitely generated group. Can $G$ be embedded as a finite-index subgroup of a 2-generated group? 100-generated?
I strongly doubt it but I don't know a counterexample.
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2$\begingroup$ For 2-generated embeddings, a counter-example is the (2,3,7)-Coxeter group. For 100, I think you should take a direct product of 101 pairwise nonisomorphic triangle Coxeter groups. $\endgroup$– Moishe KohanCommented Jul 27 at 14:50
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$\begingroup$ @MoisheKohan Interesting, thanks! Does this also rule out quasi-isometry with a 2-gen group? $\endgroup$– Sean EberhardCommented Jul 27 at 14:59
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$\begingroup$ I am unsure how to prove a qi version, but I know how to prove a "virtually isomorphic" ("commensurable") version. It is a bit more complicated variation on the direct product argument. $\endgroup$– Moishe KohanCommented Jul 27 at 16:22
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$\begingroup$ It may be too restrictive a case to be useful, but any group of Euler characteristic $-1$, e.g. a 3-generator 1-relator group, cannot be embedded with finite index as a proper subgroup of any torsion-free group. $\endgroup$– ADLCommented Jul 27 at 16:45
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$\begingroup$ A relevant example could be the free product $G_n:= \mathbb{Z}^2 \ast \mathbb{Z}^3 \ast \cdots \ast \mathbb{Z}^n$. A group $H$ quasi-isometric to $G_n$ decomposes as a graph of groups whose edge-groups are finite and whose vertex-groups are virtually $\mathbb{Z}^k$ for $2 \leq k \leq n$ (at least one factor for each $k$). A naive guess is that, for $n >> m$, $H$ is not $m$-generated. But there is something to prove here. $\endgroup$– AGenevoisCommented Jul 27 at 18:21
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2 Answers
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$\DeclareMathOperator\PSL{PSL}$Let $G_1,...,G_m$ be pairwise non-isomorphic maximal lattices in isometry groups of real-hyperbolic spaces $\mathbb H^{k_i}$ (possibly of different dimensions $k_i\ge 3$) with nontrivial abelianizations. (Note that every lattice is contained in a maximal lattice as a finite index subgroup.) For instance, we can take some pairwise noncommensurable arithmetic subgroups of the simplest type which do not preserve orientation: All such lattices have nontrivial homomorphisms to $\mathbb Z_2$. Take $$ G=G_1\times ... \times G_m. $$ Suppose that we have an inclusion $G\to H$ as a finite index subgroup. Then, by Mostow Rigidity, the group $H$ acts isometrically (possibly, with finite kernel) on the product of hyperbolic spaces $$ X=\mathbb H^{k_1}\times ... \times \mathbb H^{k_m} $$ extending the product action of $G$. In view of maximality of each $G_i$ (and the assumption that these groups are pairwise nonisomorphic), this action of $H$ cannot permute the factors and, thus, again, by maximality, the action has to be equal to the action of $G_i$ on the corresponding factor in the product. Thus, $H$ is an extension of $G$ by a finite subgroup $K$, the kernel of the action of $H$ on $X$. The group $G$ and, hence, $H$, has a surjective homomorphism to a product of $m$ nontrivial abelian groups (abelianizations of the groups $G_i$). Thus, the rank of $H$ (the minimal number of generators) is at least $m$. The same argument works when we allow some $k_i=2$ as long as the corresponding group $G_i$ is a triangle reflection group.
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$\begingroup$ When you say "by Mostow rigidity": I know this as a consequence of QI rigidity (it is explicitly stated in one of the Kleiner-Leeb papers), but do not see immediately why it follows from Mostow rigidity. $\endgroup$– YCorCommented Jul 27 at 21:41
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1$\begingroup$ @YCor: Kleiner-Leeb would be an overkill, I use the fact that the automorphism group of a lattice is a bigger lattice, which follows from Mostow (one first passes to a finite index subgroup in $G_i$ to ensure normality). $\endgroup$ Commented Jul 27 at 23:04
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$\begingroup$ Thanks, I see. So QI rigidity allows to encompass $k_i=2$. $\endgroup$– YCorCommented Jul 28 at 9:19
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$\begingroup$ I suspect the dimension-two case only needs Nielsen realisation. $\endgroup$– HJRWCommented Jul 28 at 21:30
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Here is a counterexample, coming from the theory of hyperbolic 2-orbifold groups.
Consider the fundamental group $\pi_1(S_2)$ of a closed, oriented surface of genus $2$ (its minimum number of generators is $4$).
Consider a group embedding $\pi_1(S_2) \hookrightarrow \Gamma$ whose image has finite index. There's a theorem saying that $\Gamma$ has a finite normal subgroup $N$ whose quotient group $\Gamma/N$, like $\pi_1(S_2)$, is the fundamental group of some closed hyperbolic $2$-orbifold. Since $\pi_1(S_2)$ is torsion free, it follows that the composition $\pi_1(S_2) \mapsto \Gamma \mapsto \Gamma/N$ is an embedding, and the image of this embedding clearly has finite index.
But from the classification of closed hyperbolic 2-orbifolds, none of their fundamental groups is 2-generated.
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2$\begingroup$ No, plenty are 2-generated (van Dyck groups). $\endgroup$ Commented Jul 27 at 14:36
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2$\begingroup$ Ah, shoot, I mis-thought this. I thought about the triangle groups but not the van Dyck groups. $\endgroup$ Commented Jul 27 at 14:49
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1$\begingroup$ @MoisheKohan How do you prove that the 237 reflection group is not 2-generated? $\endgroup$– YCorCommented Jul 27 at 15:42
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1$\begingroup$ (From the abelianization, the 246 reflection group is not generated by less than 3 elements; if it's maximal, we're done.) $\endgroup$– YCorCommented Jul 27 at 15:47
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4$\begingroup$ @MoisheKohan The 237 Coxeter grou is 2-generated (see theorem A) link.springer.com/content/pdf/10.1023/A:1005032526329.pdf $\endgroup$– Ian AgolCommented Jul 27 at 18:24