$\newcommand{\ord}{\text{ord}}$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.