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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

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    @A$r$turo, I alway$s$ take as my default assum$p$tion that all groups are finite. OP can always disabuse me of my misperception.2012-01-18

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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).