$eH \in G/H$ is the only element with finite order $\newcommand{\ord}{\text{ord}}$This is a question in my book: > $G$ is an infinite Abelian group and $H=\\{ a \in G \mid \ord(a) < \infty \\}$ is a normal subgroup of $G$, show that $eH$ is the only element that has an finite order. And this is my attempt: $eH=H$ and $aH=H$ for all $a \in G$ with $\ord(a) < \infty$, because then $a \in H$. And we know $eH$ has $\ord(eH)=1$ so that is the only element with a finite order. If we take $a \in G$ and $a \notin H$, then $\ord(aH)$ is infinite. I hope this is correct!

If $aH$ has finite order in $G/H$, then $(aH)^n=H$, for some $n$. This means that $a^n\in H$. By definition of $H$, there is $m$ such that $(a^n)^m=1$. Hence $a^{mn}=1$ and so $a\in H$ and $aH=H=eH$.

Two remarks:

* $H$ is called the _torsion subgroup_ of $G$ and denoted $tor(G)$. The result says that $G/tor(G)$ is torsion-free.

* The hypothesis that $G$ is infinite is not needed. The hypothesis that $G$ is abelian is crucial because it ensures that $H$ is a subgroup (and normal). If $G$ is not abelian, $H$ will only be a normal _subset_.