TITLE: Uncountably many subgroups of an abelian group
QUESTION [7 upvotes]: Does the abelian group $\mathbb{Z}[\frac{1}{2}]$ have uncountably many subgroups?

REPLY [6 votes]: No. The strict subgroups are of the form $a\cdot 2^m\mathbb Z \; (m\in \mathbb Z \;,\; a\in 2\mathbb N+1)$.      
[Core of proof: look for  elements in the subgroup with smallest possible power of two ($=m $) . If there are  some  take the one with least positive possible odd $a$. Else the subgroup is not strict.]