if every countable subgroup of $G$ is $h$-embedded in $G$.

Thus property ${{\mathbf \sharp}}$ is, in the terminology of [@hernmaca03], the dual property of pseudocompactness.

The relation between property ${{\mathbf \sharp}}$ and Question \[Qcomp\] is clear from the following Lemma. Although a combination of Propositions 3.4 and 4.4 of [@hernmaca03] would provide an indirect proof, we offer a direct proof for the reader’s convenience.

\[nocompact\] Let $(G,{\mathcal{T}_{\mbox{\tiny $H$}}})$ denote a totally bounded group with property ${{\mathbf \sharp}}$. Then $(G,{\mathcal{T}_{\mbox{\tiny $H$}}})$ has no infinite compact subsets.

We first see that all countable subgroups of $G$ are ${\mathcal{T}_{\mbox{\tiny $H$}}}$-closed. Suppose otherwise that $x\in \cl_{(G,{\mathcal{T}_{\mbox{\tiny $H$}}})}N\setminus N$ with $N$ a countable subgroup of $G$. The subgroup $\widetilde{N}={\langle\,}N\cup
\{x\}{\,\rangle}$ is also countable and, by hypothesis, inherits its maximal totally bounded group topology from $(G,{\mathcal{T}_{\mbox{\tiny $H$}}})$. Since subgroups are necessarily closed in that topology, it follows that $N$ is closed in $\widetilde{N}$, which goes against $x\in \widetilde{N}\setminus N$.

Now suppose $K$ is an infinite compact subset of $G$ and let $S\subset K$ be a countable subset of $K$. Define $\widetilde{G}={\langle\,}S{\,\rangle}$ and denote by $\widetilde{G}$ and $\overline{(\widetilde{G},{\mathcal{T}_{\mbox{\tiny $H$}}})}$ the completions of $\widetilde{G}^\sharp$ and $(\widetilde{G},{\mathcal{T}_{\mbox{\tiny $H$}}})$ respectively. Since ${\langle\,}S{\,\rangle}$ is $h$-embedded the identity function $j\colon \widetilde{G}^\sharp \to 