Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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.]

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.