Subgroups of finite abelian groups. For every subgroup $H$ of a finite abelian group $G,$ there exists a subgroup $N$ of $G$ such that $G/N \cong H.$ I need to prove this or give a counter example.
I am aware of isomorphism theorems and classification of abelian groups, direct products etc. 
 A: Yes, this is true. 
Theorem: Let $G$ be a finite abelian group. The following two statements hold.
(A) Each subgroup of $G$ is isomorphic to a quotient group of $G$, 
(B) Each quotient group of $G$ is isomorphic to a subgroup of $G$.
For a proof see [this MSE question](
Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?, which proves $(B)$; but also the same idea works for proving $(A)$. A further reference for the proofs is 
L. Fuchs, Abelian Groups. Oxford 1960, page $53$.
A: the group $G$ is Abelian finite , by the structure theorem  $G$ is
a direct product of cyclic sub groups $G_i, i=1,...n$  where
$|G_1|$ is the exponent of $ G$ and $|G_{i+1}|$ devise $|G_{i}|$
for all $i=1,...,n-1$ also $H$ is a direct product of cyclic sub
groups $H_i, i=1,...m$ where $|H_1|$ is the exponent of $ H$ and
$m\leq n$, therefor the exponent of $ H$ devise the exponent of
$G$. as $G$ is also a direct product of his $p$-Sylow, we get a
prove for $G$ p-group and in the end we established the general
case.
So we can suppose $n=m$ (because the  exponent propriety ensure
that we can suppose $H_i$ is in $G_i$ for all $i$), which if necessary  to
complete by the trivial subgroup. in this case we tack in each
$G_i$ a subgroup $N_i$ of order $[G_i:H_i]=$ and form $N$ as the
direct product of $N_i$ in $G$, so we obtain $G/N\simeq H$.
in general case is the same argument,decompose $G$ and $H$ in a
direct p-Slow, and   in each p-Sylow of $G$, tack the appropriate
$N_p$ and form $N$ as direct product of his p-Sylow $N_p$
