TITLE: Is this true? Every abelian $p$-group is isomorphic to some direct sum of cyclic $p$-groups.
QUESTION [5 upvotes]: Every abelian $p$-group is isomorphic to a direct sum of cyclic $p$-groups.

We have that every abelian $p$-group is an image of some direct sum of cyclic $p$- groups. Therefore, every abelian $p$-group is a quotient of the direct sum of the family of cyclic $p$-groups. Now, the quotient of the direct sum of the family of cyclic $p$-groups is direct sum of the family of cyclic p-groups (I am not sure this is correct). Hence every abelian $p$-group is isomorphic to some direct sum of cyclic $p$-groups.

REPLY [7 votes]: The Prüfer $p$-group $\mathbb Z(p^\infty)$ is a counterexample.  Here is one among many ways to see this.  For every $a\in\mathbb Z(p^\infty)$ there exists $b\in \mathbb Z(p^\infty)$ such that $p\cdot b=a$.  This is not true in any direct sum of cyclic $p$-groups (unless all of the summands are trivial).