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I wish to show that there does not exists a subgroup $H$ of $\mathbb{Z}$ where $|H|=10$.

This is what I have so far:

  • Assume by contradiction that there does exist a subgroup $H$ of $\mathbb{Z}$ where $|H|=10$.

  • Since $|H|=10$, a finite number, it must have a greatest element — call it $n$.

  • If $n \ne 0$, by closure, $n+n=2n$ must be in $H$, which contradicts the assumption that $n$ was the greatest element.

  • If $n=0$, take the least element, called $m$. By closure, $m+m=2m$ must be in $H$. Since $m <0, 2m$ must be less than $m$, contradicting the assumption that $m$ was the least element.

  • Thus, the order of $H$ cannot be finite, and thus cannot be $10$.

Is there a simpler way to show this? I get the feeling that my logic is wrong, but I can't explain why.

ShashComandur
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    All nonzero subgroups of $\Bbb Z$ are isomorphic to $\Bbb Z$, so have infinitely many elements. – Dietrich Burde Dec 10 '23 at 16:35
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    Actually, the proof at the duplicate for the general case has the same idea as your proof. But it is a little bit more concise and excludes all finite subgroups (except $0$) by showing that all subgroups are of the form $n\Bbb Z$. This is very useful for all other exercises concerning subgroups of $\Bbb Z$. – Dietrich Burde Dec 10 '23 at 16:44
  • Of course the classification of the subgroups of $\mathbb{Z}$ is a very standard fact, but this exercise can also be solved more directly. $H$ must have some nonzero element $a\in\mathbb{Z}$. But then $a,2a,3a,4a,...$ are all also elements of $H$, and so there must be infinitely many of them. – Mark Dec 10 '23 at 16:53

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