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

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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

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up vote 6 down vote accepted

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.]

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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
About Georges's answer and Jack's comment: It suffices to show that the proper subgroups are cyclic. –  Pierre-Yves Gaillard Feb 22 '12 at 18:29
Dear @Jack: you are absolutely right and I have now edited my answer. Thanks a lot for spotting and correcting my former erroneous statement. –  Georges Elencwajg Feb 22 '12 at 20:10

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