# Torsion group is a subgroup.

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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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. – KReiser Sep 3 '13 at 16:44
@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. – RghtHndSd Sep 3 '13 at 16:45
possible duplicate of Torsion Subgroup (Just a set) for an abelian (non abelian) group. – Seirios Sep 3 '13 at 16:57
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 $g^n=1$ – MyUserIsThis Sep 3 '13 at 17:20
@Seirios: The question here is about notation, not about how to solve the problem. This is not a duplicate. – Jim Sep 3 '13 at 17:35

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.