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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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I tried to fix the spelling, but I don't follow the penultimate sentence. Maybe it would help to give names to things? Are you saying that the quotient is a direct summand of the direct sum? –  Dylan Moreland Jan 18 '12 at 5:44
    
Your use of "the" makes this problematic. While it is true that a quotient of a direct sum of a given family of cyclic $p$-group is a direct sum of some family of cyclic $p$-groups, this family need not be the original one. For instance, $C_2\oplus C_2$ is a quotient of $C_4\oplus C_4$, but the family involved in the latter consists of two copies of $C_4$ and the family involved in the former of two copies of $C_2$. It is also unclear on what grounds you make the jump from "quotient of a direct sum of cyclic $p$-groups" to "direct sum of cyclic $p$-groups". That's the meat of the problem! –  Arturo Magidin Jan 18 '12 at 5:46
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@Gerry: That's only applicable to finitely generated abelian $p$-groups. Of course, it all depends on whether "$p$-groups" means "group whose order is a power of $p$" or "group all of whose elements have order a power of $p$". –  Arturo Magidin Jan 18 '12 at 5:49
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@Ali Gholamian: By "$p$-group" do you mean "finite group whose order is a power of $p$", or do you mean "group all of whose elements have order a power of $p$"? The result may not be true in the latter case if your group is not finitely generated. –  Arturo Magidin Jan 18 '12 at 5:50
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@Arturo, I always take as my default assumption that all groups are finite. OP can always disabuse me of my misperception. –  Gerry Myerson Jan 18 '12 at 8:24
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1 Answer 1

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).

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