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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1What does the notation $\mathbb{Z}[\frac{1}{2}]$ mean? – 2012-02-22
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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
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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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1You 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). – 2012-02-22
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0About Georges's answer and Jack's comment: It suffices to show that the proper subgroups are cyclic. – 2012-02-22
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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