$Question$
Let $G$ be an abelian group and let $H$ be a subgroup consisting of all elements of $G$ that have finite order .Prove that every non-identity element in $ G/H $ has infinite order
$ Attempt$
Suppose $G/H$ has a non- identity element of finite order .
Let $g_i=n$ which implies ${g_i}^nH = H$ which implies ${g_i}^n\in H$ which implies $g_i\in H$. Since $g_i$ is an arbitrary element, this implies $H= G$, a contradiction.
Is this proof correct ? If not can someone please give a hint or suggestion on how to move forward.
Best Answer
You need no contradiction, but you're missing a couple of points.
Let $gH$ be an element of finite order in $G/H$. Therefore there exists $n>0$ with $(gH)^n=H$, that is, $g^nH=H$. Hence $g^n\in H$, so there is $m>0$ with $(g^n)^m=1$. As a consequence $g^{nm}=1$, so $g$ has finite order, hence $g\in H$ and $gH=H$.
Therefore the only element of finite order of $G/H$ is the identity $H$.
The points you are missing: $g^n\in H$ doesn't mean by itself that $g\in H$; there is no assumption that $H\ne G$.