I had tried some assignments earlier but couldn't ask questions here then due to being ill so, I am asking now. I have done a masters level course in group theory.
If $H$ is a subgroup of a finite abelian group $G$, then $G$ has a subgroup that is isomorphic to $G / H$.
Order of subgroup always divides order of group. Let $|H|=m$ and $|G| =n$. Now there also exists a group of order $n/m$ as $n/m$ divides $n$ and as in abelian groups if $x \mid |G|$ then there always exists a subgroup of order $x$. So, there exists a group of order $|G/H|.$
But How can I prove that it is isomorphic to $G/H$?
I have read theory of finitely generated Abelian Groups but I am at loss of ideas.
