Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$? I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states

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

Thinking about this for a bit, the only reasonable approach I could think of was to construct some surjective homomorphism $\phi\colon G\to K$ for $K\leq G$, and $\ker\phi=H$, and then just use the isomorphism theorems to get the result.
After a while of trying, I've failed to come up with a good map, since $H$ seems so arbitrary. I'm curious, how can one construct the desired homomorphism? This is just the approach I thought of, if there's a better one, I wouldn't mind seeing that either/instead. Thank you.
 A: Since a finite abelian group is the direct sum of its $p$-parts, it suffices to establish the result when $G$ is a finite abelian $p$-group.
If $G=C_{p^{a_1}} \oplus\cdots\oplus C_{p^{a_k}}$, with $1\leq a_1\leq\cdots \leq a_k$, and let $Q$ be a quotient of $G$. Then $Q$ is a finite abelian $p$-group that is generated by $k$-elements (the images of the generators of $G$), and so when we express it as a direct sum of cyclic $p$-groups, it will have at most $k$ direct summands,
$$Q \cong C_{p^{b_1}}\oplus\cdots\oplus C_{p^{b_m}},$$
$1\leq b_1\leq \cdots\leq b_m$, $m\leq k$. 
Now, $b_m\leq a_k$, because every element of $G$ is of order dividing $p^{a_k}$, hence the same is true for $Q$. So $C_{p^{a_k}}$ has a subgroup of order $p^{b_m}$.
Likewise, $b_{m-1}\leq a_{k-1}$ (count the number of elements of order greater than $p^{a_{k-1}}$ in $G$; an element of order greater than $p^{a_{k-1}}$ in $Q$ must be an image of one of these). So you can find a subgroup of $C_{p^{a_{k-1}}}$ of order $p^{b_{m-1}}$.
Continue this way until you get all the cyclic summands you need out of the cyclic summands of $G$ to construct a subgroup isomorphic to $Q$.
A: I have some notes on (mostly finite) abelian groups for an undergraduate audience here.
The fact that if $G$ is abelian every subgroup is normal appears on page 1.
The result you are asking about is Theorem 19 on page 8 of my notes.  Beware that although a complete proof is in the notes, it takes a little while to get there...the point is that this uses, in addition to the basic character theory of finite abelian groups, the fact that a finite abelian group is non-canonically isomorphic to its character group, which in turn uses the main structure theorem for finite abelian groups.
Added: It is possible to dispense with the character theory (although to my taste this is a nice, clean way to phrase it), but it does not seem possible to avoid the structure theorem, which is a famous, and famously nontrivial, result.  Note that in particular Arturo's nice answer does not use character theory but does use the structure theorem...twice.
