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Let $G$ be an abelian group, and let $H\leq G$. Prove that if $G/H$ is torsion free, then $H$ contains the torsion group of $G$.


Let $x\neq1$ be an element in the torsion group. Thus there exist $k\in \mathbb{N} $ with $x^k=1$.

Now we look at $(xH)^k = x^kH = H $

$G/H$ is torsion free, so $xH$ must be equal to $H$ and therefore $x\in H $, as we wanted.

My question is that: In this proof I didn't really use the fact that $G$ is abelian. I could assume only that $H\lhd G$ and the proof still stands. Is it true?

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You might want to add to the actual question the assumption that $G$ is finite, not just leave it in the tags. – Asaf Karagila Jul 14 '12 at 7:22
@Asaf, It was a mistake, $G$ could be inifinite. – catch22 Jul 14 '12 at 7:27
If $G$ is not abelian, then the torsion elements might not form a subgroup. For example if $G = \langle x,y\ |\ x^2 = y^2 = e\rangle$, then $x$ and $y$ are torsion elements, but $xy$ is not. – David Wheeler Jul 14 '12 at 7:32
@David, Thank you. – catch22 Jul 14 '12 at 7:42
up vote 1 down vote accepted

As David Wheeler says in the comments, we cannot generally speak of the torsion subgroup of groups that are not abelian, because the torsion points need not be closed under multiplication; an obvious example being the infinite dihedral group. On the other hand the proof you've written does demonstrate that all torsion points are contained in $H$.

It is possible for non-abelian groups to have torsion subgroups, of course. If $G$ is finite then the torsion points are just all of $G$, for instance, and Wikipedia states that the torsion subset of any nilpotent group forms a normal subgroup (so this would include infinite groups). The orders of elements do not behave predictably in nonabelian settings, however: whereas in abelian groups we have that the order of $ab$ is the $\rm lcm$ of the orders of $a$ and $b$, we can actually pick any integers $r,s,t$ and there will be a group $G$ with elements $a,b,ab$ of those orders respectively (this is Thm 1.64 in Milne's freely available group theory course notes), specifically the $\rm PSL_2$s of some finite fields.

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No, $2mnr|(q-1)$ and $|\mathbb{F}_q^{\times}|=q-1$, as in the notes, are correct. As defined in the notes, $q$ is a power of $p$, i.e., $q=p^a$ for some integer $a$. It is not the exponent. – user36021 Jul 17 '12 at 23:53
There is no typo in the notes (at least not in regard to that). $\mathbb{F}_q$ is generally used to denote the field of $q = p^k$ elements for some fixed $p$ and $k$. – Brandon Carter Jul 18 '12 at 0:23
@BrandonCarter You are correct, for some reason I read "some power of it, say $q$," to mean $p^q$, even though I've seen $q=p^r$ so many times... very silly of me. – anon Jul 18 '12 at 0:26
One may add, since you mention nilpotency, that in the nilpotent case one can give bounds on the order of $ab$ in terms of the orders of $a$ and $b$ and the nilpotency class. – Arturo Magidin Jul 18 '12 at 1:35
@anon: Sorry for misunderstanding what the situation was with the off-answer. (Yes, I know "anon" is short of anonymous). Since comments don't come with icons, at first I thought the comment you posted was by the OP; then I realized it was by you, but rather than overwrite my impression I simply appended the new information to it... – Arturo Magidin Jul 18 '12 at 3:35

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