Every abelian $p$-group is an image of some direct sum of cyclic $p$-groups 
Every abelian $p$-group is an image of some direct sum of cyclic $p$-groups.

This is exercise 4.3.6 on page 113 of Derek J.S. Robinson's A Course in the Theory of Groups (Second Edition).
I don't know why it is true.
To my knowledge, a Prüfer $p$-group is a $p$-group. But how can it be an image of some direct sum of cyclic $p$-groups?
Many thanks!
 A: Let $G$ be an abelian $p$-group. For each $a\in G$, let $\mathrm{o}_p(a)=k$ if and only if the order of $a$ is $p^k$. Then $\mathbb{Z}/p^{\mathrm{o}_p(a)}\mathbb{Z}$ maps to $G$ by the map that sends the generator to $a$.
Thus, we have a family of abelian cyclic $p$-groups,
$$\bigl\{ \mathbb{Z}/p^{\mathrm{o}_p(a)}\mathbb{Z}\}_{a\in G},$$
and for each group in the family, we have a map to $G$. By the universal property of the direct sum, these maps induce a group homomorphism
$$\bigoplus_{a\in G} \mathbb{Z}/p^{\mathrm{o}_p(a)}\mathbb{Z} \to G.$$
The map is onto, because for every $a\in G$ the generator of the $a$th group in the family maps to $a$. Therefore, $G$ is a quotient of the direct sum of the family.
(Of course, this direct sum is extremely wasteful: if $a$ has order $p$, then we get one direct summand for $a$, one for $a^2$, one for $a^3$, and so on, when "really", we only need one direct summand that maps its generator to $a$; you can easily come up with a "nicer," less wasteful family, and you may want to try it. Or, you can try to figure out what the "induced homomorphism" from the direct sum to $G$ will be, explicitly). 
