Let $F$ be a finitely generated free group, $H_1, \dots, H_n$ finitely generated subgroups of infinite index in $F$, and $\epsilon > 0$. Must there be an epimorphism to a finite group $\phi \colon F \to G$ such that $$\frac{|\phi(H_1 \cdots H_n)|}{|G|} < \epsilon ?$$ Where $H_1 \cdots H_n = \{h_1 \cdots h_n : h_1 \in H_1, \dots, h_n \in H_n\}$.

Equivalently: Is the Haar measure of the closure of $H_1 \cdots H_n$ in the profinite completion of $F$ equals $0$ ?

This question asks for a generalization of Is a free group a product of f.g subgroups of infinite index?


The answer is yes.

Let's prove this by induction on $n$. If $n=1$, then by M. Hall's theorem there is a finite index subgroup $K \leqslant F$ such that $H_1 \subseteq K$ and $|F:K|>1/\epsilon$. Choose any finite index normal subgroup $N \lhd F$, which is contained in $K$, let $G=F/N$ and let $\phi:F \to G$ be the natural epimorphism. Note that $|G:\phi(K)|=|F:K|$ as $N\subseteq K$. Then $$\frac{|\phi(H_1)|}{|G|} \le \frac{|\phi(K)|}{|G|}=\frac{1}{|G:\phi(K)|}= \frac{1}{|F:K|}<\epsilon .$$

Now, suppose that $n\ge 2$ and the statement has been proved for smaller $n$. By Theorem 1.1 in http://arxiv.org/abs/1308.3192, there is a finitely generated subgroup $H$ of infinite index in $F$, such that $H_{n-1} \subseteq H$ and $|H_n: (H_n \cap H)|=k<\infty$.

Thus $H_n \subseteq \bigcup_{j=1}^k H f_j$ for some $f_1,\dots, f_k \in F$. By the induction hypothesis, we can find an epimorphism $\phi$ from $F$ onto a finite group $G$ such that $$\frac{|\phi(H_1\dots H_{n-2}H)|}{|G|}< \frac\epsilon k.$$ Since $H_1 \dots H_n \subseteq \bigcup_{j=1}^k H_1 \dots H_{n-2}Hf_j$, we have \begin{multline*}\frac{|\phi(H_1\dots H_{n-2}H_{n-1}H_n)|}{|G|}\le \sum_{j=1}^k \frac{|\phi(H_1\dots H_{n-2}H)\phi(f_j)|}{|G|} \\ =\sum_{j=1}^k \frac{|\phi(H_1\dots H_{n-2}H)|}{|G|}<k \, \frac\epsilon k =\epsilon,\end{multline*} as required.

  • $\begingroup$ I asked this question in order to find a new proof for the theorem that you have used. That is, I know how to prove Theorem 1.1 in arxiv.org/abs/1308.3192 using my question (I only need $n = 2$), so I am really looking for a different proof (I knew this proof for $n=2$). $\endgroup$
    – Pablo
    Dec 22 '14 at 16:37
  • 2
    $\begingroup$ For $n=2$ you don't need the full strength of the theorem from Olshanskii's paper. You can use an easier fact that there is a f.g., subgroup $H$, of infinite index in $F$, containing finite index subgroups of $H_1$ and $H_2$. This is not difficult to prove geometrically (see, for example, Thm. 1 in [Gitik, Rita, "Ping-pong on negatively curved groups". J. Algebra 217 (1999), no. 1, 65–72.] for a more general statement for quasiconvex subgroups in hyperbolic groups). $\endgroup$ Dec 22 '14 at 16:54
  • $\begingroup$ Now I am interested in the hyperbolic case as a result of your comment and this interesting paper. So, do you know whether these amalgamated products constructed in the paper can be assured to have infinite index? I am ready to assume that the ambient hyperbolic group is residually finite or QCERF... $\endgroup$
    – Pablo
    Jan 18 '15 at 17:46
  • $\begingroup$ Yes, with amalgamated products together with QCERF it should be easy to ensure that they have infinite index. $\endgroup$ Jan 18 '15 at 19:49

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