Torsion and Torsion-Free Quotient Group The book by Fraleigh says that if $G$ is a torsion group, then so is $G/H$ for every normal subgroup $H$ in $G$. It also says if $T$ is the torsion normal subgroup of an abelian group $G$, then then $G/T$ is torsion-free.
I cannot reconcile both these facts:


*

*$G$ torsion implies $G/H$ torsion

*$H$ torsion implies $G/H$ torsion-free ($G$ abelian)


If $G$ is torsion, then shouldn't $H$ be torsion too? and that means there is a contradiction, I am hopeless
Proof 1)
Because $G$ is a torsion group, we know that $x^m = e$ in $G$ for some positive integer $m$. Computing $(xH)^m$ in $G/H$ using the representative $x$, we have $(xH)^m = x^mH = eH = H$, so $xH$ is of finite order. Because $xH$ can be any element of $G/H$, we see that $G/H$ is a torsion group.                       
Proof 2)
Suppose that $xT$ is of finite order in $G/T$; in particular, suppose that $(xT )^m = T$. Then $x^m \in T$. Because $T$ is a torsion group, we must have $(x^m)^r = x^{mr} = e$ in $G$ for some positive integer $r$. Thus $x$ is of finite order in $G$, so that $x \in T$. This means that $xT = T$. Thus the only element of finite order in $G/T$ is the identity $T$, so $G/T$ is a torsion-free group.
It is probably very obvious, and I apologize if it is, I just cannot wrap my head around it with exams coming up.
 A: If $T$ is the torsion subgroup of an abelian group $G$, then $G/T$ is indeed torsion-free.
Indeed, if $xT$ is torsion in $G/T$, then $eT=(xT)^m=x^mT$ for some $m>0$. This means $x^m\in T$, so $(x^m)^n=e$ for some $n>0$. Therefore $x^{mn}=e$ and $x\in T$, so $xT=eT$ and there is no nontrivial torsion element in $T$.
If $G$ is already torsion, then $T=G$ and $G/T$ is the trivial group, which is torsion-free because it has no nontrivial torsion element (having no element different from the identity).
The trivial group $\{e\}$ is indeed both torsion and torsion-free. There's no contradiction.
A: The proof for the second statement is wrong.
you proved that every element of finite order in $G/T$ comes from an element of finite order.
I believe this is a counter example let $G=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ and let $T=\mathbb{Z}/2\mathbb{Z}\times\{0\}$. Clearly $T$ is a normal torsion subgroup of $G$. But $G/T\cong \mathbb{Z}/2\mathbb{Z}$ which is not torsion-free.
Note that there is a difference between a torsion group and the torsion subgroup.
Definition (Torsion group): Let $G$ be a group, we say that $G$ is torsion if every element in $G$ is of finite order.
Definition (the Torsion subgroup): Let $G$ be a group, the torsion subgroup $T$ of $G$ is the subgroup of $G$ containing all elements of finite order. (i.e $T$ is the maximal torsion subgroup of $G$).
And I'm pretty sure that what is actually proved in the book is the following
Theorem: Let $G$ be a group and let $T$ be its torsion subgroup. Then $G/T$ is torsion-free.
