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Let $G$ be an abelian group. Prove that $H =\{g \in G \;|\; |g| < \infty\}$ is a subgroup of $G$. Given an explicit example where this set is not a subgroup when $G$ is non-abelian.

I am confused with the notation of $g$. Since it has cardinality, is it a set? a group? Hence I am having trouble showing its inverse is in $H$.

Thank you~

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    $\begingroup$ If $g$ is an element of $G$, then $|g|$ is the least integer such that $|g|$ copies of $g$ multiplied together gives the identity. If no such integer exists, $|g|$ is set equal to infinity. $\endgroup$ – KReiser Sep 3 '13 at 16:44
  • $\begingroup$ @KReiser: I believe your comment is an answer. If you think so too, feel free to leave it as an answer so it can be accepted. $\endgroup$ – RghtHndSd Sep 3 '13 at 16:45
  • $\begingroup$ possible duplicate of Torsion Subgroup (Just a set) for an abelian (non abelian) group. $\endgroup$ – Seirios Sep 3 '13 at 16:57
  • $\begingroup$ The notation $|g|$ for an element refers to the size of the subgroup that it generates. $$|g|=|\lbrace 1,g,g^{-1},g^2,g^{-2},...\rbrace |$$ If the group is finite then the size will be the least integer $n$ such that $g^n=1$ $\endgroup$ – MyUserIsThis Sep 3 '13 at 17:20
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    $\begingroup$ @Seirios: The question here is about notation, not about how to solve the problem. This is not a duplicate. $\endgroup$ – Jim Sep 3 '13 at 17:35
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On the suggestion of rghthndsd, I'm turning my comment into an answer:

If $g$ is an element of $G$, then $|g|$ is the least integer such that $|g|$ copies of $g$ multiplied together gives the identity. If no such integer exists, $|g|$ is set equal to infinity.

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  • $\begingroup$ Thank you KReiser. May I request for an counter example? I was thinking of reflections, but then I find it is not a group...? $\endgroup$ – Tumbleweed Sep 3 '13 at 18:25
  • $\begingroup$ Well, from the question you know that such a group must be both infinite and non-abelian- do you know any examples of groups that satisfy these properties? Do you know constructions that can build new groups, like semi-direct product or free product? $\endgroup$ – KReiser Sep 3 '13 at 18:49
  • $\begingroup$ @Tumbleweed See this previous question. (Actually, you can do it with reflections of the plane. Take two perpendicular lines, then reflecting along these each of these lines independently has order two, while if you do one then the other you get infinite order.) $\endgroup$ – user1729 Sep 4 '13 at 7:47

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