# Uncountably many subgroups of an abelian group

Does the abelian group $\mathbb{Z}[\frac{1}{2}]$ have uncountably many subgroups?

-
What does the notation $\mathbb{Z}[\frac{1}{2}]$ mean? – Mikko Korhonen Feb 22 '12 at 17:27
@m.k.: It means that we take the subring of the rationals $\mathbb{Q}$ generated by $\mathbb{Z}$ and the element $\frac{1}{2}$. – Pete L. Clark Feb 22 '12 at 17:37

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.]
You need to combine the two descriptions, $\frac{a}{2^m}\mathbb{Z}$ for non-negative odd numbers $a$ and integers $m$ are all distinct proper subgroups. In other words, 1/2 is not in the subgroup { (3n)/2 : n in Z } which is closed under addition. (negative m give the subgroups of Z). – Jack Schmidt Feb 22 '12 at 18:17