# 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 an exercise on page 113 of Derek J.S. Robinson's A course in the Theory of Groups.

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!

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The Prufer $p$-group is the image of $\oplus_{i=1}^{\infty}(\mathbb{Z}/p^i\mathbb{Z})$, modulo the subgroup that identifies $p+p^{k+1}\mathbb{Z}$ in the $(k+1)$st summand with $1+p^{k}\mathbb{Z}$ in the $k$th summand. Think about constructing it by first taking a cyclic group of order $p$, then adjoining a $p$th root of the generator; then adjoining a $p$-th root of the $p$th root of the generator, etc. – Arturo Magidin Sep 14 '11 at 17:21

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).