# Avoiding countable subgroups of general uncountable groups

The following problem is a general form of another problem (motivation is available there). Initially, the problems were posted together, but the first one is solved below, a solution that does not apply to the more focused version of the problem, which is the needed scenario.

Problem 1. Let $G$ be an uncountable group, $H$ a countable subgroup of $G$, and $g\in G\smallsetminus H$. Is there necessarily an element $x\in G\smallsetminus H$ such that $g\notin\langle H\cup \{x\}\rangle$?

Counterexample for Problem 1: According to this answer Saharon Shelah constructed a "Jónsson group" of order $\aleph_1,$ i.e., an uncountable group $G$ in which every proper subgroup is countable. Choose an element $g\in G\setminus\{e\}.$ Let $H$ be a maximal subgroup of $G$ not containing $g.$ Since $H$ is a proper subgroup of $G,$ $H$ is countable. By the maximality of $H,$ there is no element $x\in G\setminus H$ with $g\notin\langle H\cup\{x\}\rangle.$
Fix an odd prime $p$. Let $A=\mathbf{F}_p((t))$ be the ring of Laurent series over the field $\mathbf{F}_p$ on $p$ elements; we only view it as an abstract group, and let $A_0$ be the set of series $\sum a_nt^n$ in $A$ with $a_0=0$. Let $B$ be the non-unital subring $t^{-1}\mathbf{F}_p[t^{-1}]$. Define a symplectic form on $A_0$ by $\langle t^n,t^m\rangle=0$ if $n+m=0$ and $\langle t^n,t^{-n}\rangle=1$ for all $n\ge 0$, by extending it by "formal" linearity (namely $$\langle\sum a_n t^n,\sum b_nt^m\rangle=\sum_{n>0}a_nb_{-n}-a_{-n}b_n\qquad\qquad ).$$ That it is non-degenerate is straightforward. Observe that $B$ is a maximal isotropic subspace.
Define a (nilpotent) uncountable group $G$ as the set of pairs $(x,t)$ with $x\in A_0$, $t\in\mathbf{F}_p$, and group law $(x,t)(x',t')=(x+x',t+t'+\langle x,x'\rangle)$. Then $B$ is a countable subgroup and every group properly containing $B$ contains $(0,1)\notin B$.
• NB: Shelah's example has cardinal $\aleph_1$ regardless of CH (as any Jonsson group). This example has cardinal $2^{\aleph_0}$ and works replacing $t\mathbf{F}_p[[t]]$ with any subgroup containing $t\mathbf{F}_p[t]$, so we can get any cardinal in $[\aleph_1,2^{\aleph_0}]$ (which is of course an improvement only under the negation of CH). – YCor Jun 12 '15 at 13:32