# Avoiding countable subgroups of a group homeomorphic to the Cantor space

Update: Further work with Adam (who answers below) and Piotr led to a rather satisfactory result about the problem that motivated the problem below, see our recent paper The Haar Measure Problem. In particular, we answer there a problem mentioned in the discussion below.

The following question is motivated by the paper [Brian, Mislove, Every compact group can have a non-measurable subgroup]. A positive solution to a variation of the following problem implies a positive solution to the problem studied there, i.e., that every infinite compact group has a subgroup that is not Haar measurable. A more general form of this question was answered in the negative there.

Every infinite metric profinite group is, as a topological space, homeomorphic to the Cantor space [Hofmann, Morris, The Structure of Compact Groups, Theorem 10.40].

Problem. Let $G$ be an infinite metric profinite group (or, more generally, one homeomorphic to the Cantor space), $H$ a countable subgroup of $G$, and $g\in G\smallsetminus H$. Is there necessarily an element $x\in G\smallsetminus H$ such that $g\notin\langle H\cup \{x\}\rangle$?

Discussion. I find it intriguing that in this scenario, the problem is about a topological group structure on the Cantor space. A rather well-understood object from the topological point of view. In particular, this fact is used in the Brian--Mislove paper cited above to observe that the answer to the big question in the first paragraph is positive when there is a non-null set of reals of cardinality smaller than the continuum ($\mathrm{non}(\mathcal{N})<\mathfrak{c}$). Thus, it remains to consider the case $\mathrm{non}(\mathcal{N})=\mathfrak{c}$ and a transfinite induction can be used to construct the needed nonmeasurable subgroup. There, in each step, the intermediate subgroup is Lebesgue null. In our problem, we model this by "countable". Even assuming the Continuum Hypothesis, I do not know the answer to Problem 2 (or the big problem).

Comment: I have separated this question from the quetion there to make it possible to declare the other question as solved.

• Two remarks: 1) for a topological group, "homeomorphic to Cantor space" is equivalent to "metrizable profinite". 2) A "metric group" is a group endowed with a metric, not just a metrizable group. – YCor Jun 25 '15 at 18:45

Since $G$ is profinite, we may write it as $G=\lim_{n\in\mathbb{N}} G_n$, where $G_n$ are finite and the projections $\pi_n:G\to G_n$ are onto. Since $H$ is countable, we may write it as $H=\bigcup_{n\in\mathbb{N}} H_n$, where $H_n\subseteq H_{n+1}$ are finite, $H_0$ nonempty. Let $\bar H_n = H_n\cup H^{-1}_n\cup\{e\}$ where $X^{-1}=\{x^{-1}\mid x\in X\}$. Let $X^r=\{x_1x_2\ldots x_r\mid x_i\in X\}$.
We construct inductively a sequence $A_{k}\subseteq G_{n_k}$ as follows. Pick any $h\in H_0$ and choose $n_0$ so that $\pi_{n_0}(h)\neq \pi_{n_0}(g)$. Let $A_0=\{\pi_{n_0}(h)\}\subseteq G_{n_0}$. Suppose $n_k$ and $A_k\subseteq G_{n_k}$ are defined. For every $a\in A_k$ we choose two points $h_a,h'_a\in H$ such that $\pi_{n_k}(h_a)=\pi_{n_k}(h'_a)=a$. This is always possible since $H$ is a group and the kernel of $\pi_{n_k}$ restricted to $H$ is infinite. Find $m_k$ such that every $h_a$ and $h'_a$ belongs to $H_{m_k}$. We may assume that the sequence $m_k$ is increasing. Then choose $n_{k+1}>n_k$ so that $\pi_{n_{k+1}}(g)\notin\pi_{n_{k+1}}(\bar H^{k}_{m_k})$. This is possible since $\bar H^{k}_{m_k}\subseteq H$ is finite and $g\notin H$.
Define $A_{k+1}=\{\pi_{n_{k+1}}(h_a)\mid a\in A_k\}\cup\{\pi_{n_{k+1}}(h'_a)\mid a\in A_k\}\subseteq G_{n_{k+1}}$. Let $A=\lim_{k\in\mathbb{N}} A_{n_k}$. Then the cardinality of $A$ is the continuum and we are done once we prove the following.
Claim: For every $x\in A$ we have $g\notin\langle H,x\rangle$.
Suppose that, to the contrary, $g\in\langle H,x\rangle$ for some $x\in A$. Then we have elements $h_1,h_2,\ldots,h_s\in H$ and a word $w$ such that $w(x,h_1,h_2,\ldots,h_s)=g$. Let $r$ be the length of this word. There exists $k>r$ such that $h_1,h_2,\ldots,h_s\in H_{m_k}$. Then by construction of $A$ we have $$\pi_{n_{k+1}}(x)\in\pi_{n_{k+1}}(A)=A_{k+1}\subseteq\pi_{n_{k+1}}(H_{m_k})$$ hence $$\pi_{n_{k+1}}(g)\notin\pi_{n_{k+1}}(\bar H^{k}_{m_k})\ni\pi_{n_{k+1}}(w(x,h_1,h_2,\ldots,h_s))$$ which is a contradiction.