I tried to find out whether the following good-looking statement about groups is true. "An infinite group $G$ always has a countably infinite subgroup."
I faced this question while pondering upon the existence of a countably additive, (left) $G$-invariant measure on all of $\mathscr{P}(G)$. If the statement is true, then I believe that the Vitali technique of showing the existence of a non-measurable subset of $\mathbb{R}$ can be imitated to establish that such a measure on $\mathscr{P}(G)$ cannot exist. But I am stuck with this group-theoretic obstacle.
To summarize, does an infinite group always has a countably infinite subgroup? Or, does there exist an infinite group, having no countably infinite subgroups? Please enlighten me.
Addendum : By a subgroup, I mean to include the case of the group itself. So when $G$ is countable, the question is not very interesting.
Consider $p^n$-th roots of unity in $\mathbb{C}$ for a fixed prime $p$, but for all $n\in\mathbb{N}$. This is an infinite group, but a proper subgroup of it is finite and cyclic.
Think this in following way: let $G_n=\{ z\in\mathbb{C} : z^{p^n}=1\}$; it is cyclic group of order $p^n$, and we have a chain $$1\subseteq G_1 \subseteq G_2 \subseteq \cdots;$$ the union of these subgroups (inside $\mathbb{C}^{\times}$) is again a subgroup of $\mathbb{C}^{\times}$. You may then think from this why a proper subgroup of it should be finite.