Torsion group is a subgroup.

Question

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~

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 03 '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 03 '13 at 16:45
possible duplicate of Torsion Subgroup (Just a set) for an abelian (non abelian) group. — Seirios, Sep 03 '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 03 '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 03 '13 at 17:35
Another common notation for order is $o(g)$. That is, $o(g)=|g|$. — user1729, Sep 03 '13 at 19:33

score 4 · Accepted Answer · answered Sep 03 '13 at 18:10

4

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.

answered Sep 03 '13 at 18:10

KReiser

74,746

Thank you KReiser. May I request for an counter example? I was thinking of reflections, but then I find it is not a group...? – Tumbleweed Sep 03 '13 at 18:25
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? – KReiser Sep 03 '13 at 18:49
@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.) – user1729 Sep 04 '13 at 07:47

Torsion group is a subgroup.

1 Answers1