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Generally, for a prime $p$, Prüfer $p$ group is defined to be direct limit of the system of groups $\{\mathbb{Z}/p^n\mathbb{Z}\colon n\in \mathbb{N}\}$ with the homomorphisms $\mathbb{Z}/p^n\mathbb{Z} \rightarrow \mathbb{Z}/p^{n+1}\mathbb{Z}$ induced by multiplication by $p$. But on Wikipedia, it is defined with some "characterization":

  • It is the unique infinite $p$ group, in which every element has $p$ $p$'th roots.

  • It is the unique infinite $p$ group which is locally cyclic (every finite se of elements of group generates a cyclic group.)

How do we prove the "uniqueness" in these characterizations using the definition which is given in terms of direct limits?

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    see also http://math.stackexchange.com/questions/22979/characterizations-of-the-p-prufer-group2011-06-20
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    I think you misunderstand something. To prove there is a unique blah, you prove that there is at least one blah (the direct limit for instance) and any two blah are isomorphic. The second part has nothing to do with the direct limit.2011-06-20
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    @Jack: I don't think there is a substantial problem. One can prove that the direct limit construction satisfies the given property (your first clause), and then try to prove the uniqueness clause along the lines you suggest, but by appealing to the universal property of the direct limit somehow. I read the question as asking that.2011-06-20
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    You can also define it to be the group of complex numbers which when raised to some $p$-power-th power yield 1.2011-06-20
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    @Arturo, oh, like every group which satisfies it must be a direct limit, or something? Yeah, OK, that actually seems natural.2011-06-20
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    @Jack: Or: "if $G$ is a group that is an infinite $p$-group and every element has $p$ $p$-th roots, then we can define maps from the directed system.... and by the universal property we get map from the direct limit..." etc.2011-06-20

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