If G is a finite Abelian group, and H is a subgroup of G, How can I prove that G/H is isomorphic to another subgroup N of G?. If G is an abelian finite group, and H is a subgroup of G, How can I prove that G/H is isomorphic to another subgroup N of G?.
I'll write abelian groups additively.

You can use the character group: if $G$ is a finite abelian group, then its character group is $\hat{G}=\operatorname{Hom}(G,\mathbb{Q}/\mathbb{Z})$.

From the structure theorem of abelian groups, it follows that $G$ is the direct sum of cyclic $p$-groups. If $C$ is a finite cyclic $p$-group, then $\hat{C}\cong C$, hence there is a (non canonical) isomorphism $\varphi\colon G\to\hat{G}$.

If $H$ is a subgroup of $G$, consider $H'=\varphi(H)$, so $\hat{G}/H'\cong G/H$. Now prove that, defining
$$
N=\{x\in G:\chi(x)=0,\text{ for all }\chi\in H'\},
$$
you have $N\cong \hat{G}/H'$.
HINT:

Because $G$ is abelian, every subgroup will be normal. Therefore, it makes sense to consider $G/H$ for a subgroup $H$.

Then, use the first isomorphism theorem such that $H = ker(f)$