Subgroup of an abelian group isomorphic to a given quotient group

STATEMENT: Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroups that is isomorphic to $G/H$.

QUESTION: Could someone offer a proof using dual groups. I have found one use the fundamental theorem of finitely generated abelian groups, but I cannot seem to find on using dual groups.

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Let $S^1$ be the circle group. Consider the exact sequence:

$$0\to H\to G \to G/H \to 0$$

Applying the functor $\operatorname{Hom}(-,S^1)$, we get another exact sequence:

$$0\to \operatorname{Hom}(G/H,S^1) \to \operatorname{Hom}(G,S^1) \to \operatorname{Hom}(H,S^1)$$

Now, it so happens that $\operatorname{Hom}(G,S^1) \cong G$ for a finite abelian group $G$—though I don’t know if there’s a good way to prove that without using the fundamental theorem. But if you accept it, then this sequence gives us a (non-canonical) injection $G/H\to G$.