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Is there a structure theorem for finitely generated profinite abelian group like a structure theorem of f.g. abelian group?

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  • $\begingroup$ The version as stated is clearly false as $\mathbf Z_p$ is not of this form. But I think that every finitely generated profinite abelian group is a product of its pro-$p$ Sylow subgroups, and each of those is isomorphic to $\mathbf Z_p^n \times H$ for $H$ a finite abelian $p$-group. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 22, 2022 at 15:21
  • $\begingroup$ Thanks for clarification. $\endgroup$
    – Sunny
    Commented Dec 22, 2022 at 15:24
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    $\begingroup$ Yes. A f.g. profinite abelian group is the same as a profinite abelian group $G$ that is a quotient of $\hat{\mathbf{Z}}^d$ for some $d$. By Pontryagin duality, this is the same as a locally finite abelian group that is isomorphic to a subgroup of $(\mathbf{Q}/\mathbf{Z})^d$ for some $d$. The latter is easy to classify: this is a direct sum of copies of Prüfer and its quotients, such that each prime appears $\le d$ times. So the classification is: $\prod_{p\text{ prime}}A_p$ with $A_p$ is a product of $\le d$ quotients of the $p$-adic group $\mathbf{Z}_p$ (with $d$ independent of $p$). $\endgroup$
    – YCor
    Commented Dec 22, 2022 at 15:27
  • $\begingroup$ A reference is Ribes–Zalesskii's Profinite groups, §4.3. (Although it seems that they use Pontryagin duality between profinite abelian groups and torsion abelian groups in the proof, which should be unnecessary.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 22, 2022 at 15:30
  • $\begingroup$ @R.vanDobbendeBruyn yes, it's not really necessary, it's just a way for me to feel safer (performing the argument for the discrete dual) but definitely one can do everything at the profinite side. $\endgroup$
    – YCor
    Commented Dec 22, 2022 at 15:48

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