Let $G$ be a finite abelian group and $H$ be a subgroup of $G$. Then there is an epimorphism from $G$ to $H$. Let $G$ be a finite abelian group and $H$ be a subgroup of $G$. Then there is an epimorphism from $G$ to $H$.
(That is, an epimorphism from a finite abelian group to its subgroup.)
 A: By Exercise II.5.8 in Hungerford's Algebra,
$G\cong \prod_{i=1}^{n}P_i$ is the direct product of its Sylow $p$-subgroups.
By Exerise 24.56 in Gallian's Contermporary Abstract Algebra,
$H\cong \prod_{i=1}^{n}(H\cap P_i)$ is the direct product of the Sylow $p$-subgroups of $H$.
So it is sufficient to consider the case $G$ is a $p$-group.
By this,
we can assume $G\cong \Bbb{Z}_{p^{r_1}}\oplus\Bbb{Z}_{p^{r_2}}\oplus\cdots \oplus \Bbb{Z}_{p^{r_s}}$
and $H\cong \Bbb{Z}_{p^{t_1}}\oplus\Bbb{Z}_{p^{t_2}}\oplus\cdots \oplus \Bbb{Z}_{p^{t_u}}$,
where $s\geq u$ and $r_i\geq t_i$ for each $i=1, 2, ..., u$.
Recall that for $\Bbb{Z}_n$,
if $d\mid n$,
then consider the onto natural homomorphism $\theta:\Bbb{Z}\to \Bbb{Z}_d$.
Note that $\ker{\theta}=\langle d\rangle$ and $\langle n\rangle\subseteq \ker{\theta}$.
By Hungerford, p.43, thm.I.5.6,
there exists an onto homomorphism $f:\Bbb{Z}_n\cong \Bbb{Z}/\langle n\rangle\to \Bbb{Z}_d$.
Apply this argument to each $i=1, 2, ..., s$.
That is,
there is an onto homomorphism $f_i$ from $\Bbb{Z}_{p^{r_i}}$ to $\Bbb{Z}_{p^{t_i}}$ for each $i=1, 2, ..., s$.
($f_i$ maps $\Bbb{Z}_{p^{r_i}}$ to trivial group if $i>u$.)
Then $(f_1, f_2, ..., f_s)$ is an onto homomorphism from $G$ to $H$.
A: It follows from the structure theorem of finitely generated abelian groups that a finite abelian group $G$ is (not canonically) isomorphic to its character group $\def\ch#1{\operatorname{Hom}(#1,\mathbb{Q}/\mathbb{Z})}\ch{G}$. Moreover, the functor $\ch{{-}}$ maps epimorphisms into monomorphisms and conversely.
So if you have $H$ a subgroup of $G$, consider the epimorphism $\ch{G}\to\ch{H}$ obtained from the inclusion and use the isomorphisms $G\cong\ch{G}$ and $H\cong\ch{H}$.
