Does the abelian group $\mathbb{Z}[\frac{1}{2}]$ have uncountably many subgroups?
Uncountably many subgroups of an abelian group
7
$\begingroup$
group-theory
abelian-groups
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3@m.k.: It means that we take the subring of the rationals $\mathbb{Q}$ generated by $\mathbb{Z}$ and the element $\frac{1}{2}$. – 2012-02-22
1 Answers
6
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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0Dear @Jack: you are absolutely right and I have now edited my answer. Thanks$a$lot for spotting and correcting my former erroneous statement. – 2012-02-22