Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$.
My approach: Let's try to prove by induction on $o(G)$.
If $o(G)=1$ then the result is obvious. Suppose it is true for all groups with order less than $n$.
Suppose that $o(G)=n$ and $H\neq \{e\},G$ because otherwise the result is trivial.
Let $x\in H$ s.t. $x\neq e$ and consider subgroup generated by $x$, i.e. $\langle x\rangle$ then $H/\langle x\rangle\leq G/ \langle x\rangle$. Note that $(G/\langle x\rangle)/(H/\langle x\rangle)\cong G/H$. Since $o(G/\langle x\rangle)<o(G)=n$ then by induction hypothesis there exists subgroup $\bar{K}\leq G/\langle x\rangle$ such that $\bar{K}\cong (G/\langle x\rangle)/(H/\langle x\rangle)$ $\Rightarrow$ $\bar{K}\cong G/H$.
Then I was trying to find subgroup of $G$ which is isomorphic to $G/H$ but no results. I've considered the surjective mapping $\pi: G\to G/\langle x\rangle$ which does give any useful results.
I would be very thankful if anyone can show how to complete my approach?
