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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 $G$.

Proof:

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. –  Roy Jul 14 '12 at 7:27
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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. –  Roy Jul 14 '12 at 7:42

1 Answer 1

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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
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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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