If $G$ is a finite abelian group, then we have a decomposition $$G\cong \prod_{p} G(p)$$ where $G(p)$ is the $p$-Sylow subgroup of $G$. This product makes sense as for all but finitely many primes $p$, we have $G_p=\{0\}$. This is proven by showing that the cardinality of $G$ and $\prod_{p} G(p)$ agree. If we now assume that $P$ is a profinite abelian group, there still exists the notion of a $p$-Sylow subgroup $P(p)$ which is now a pro-$p$-group. I'm curious if there exists an isomorphism $$P\cong \prod_{p} P(p).$$
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4$\begingroup$ One can verify this by computing indices as 'supernatural numbers', in the sense of Serre's "Galois cohomology"; but probably easier is to consider the maps $P \to P(p)$ given by $g \mapsto \lim_{N \to \infty} g^{p^N - 1}$, where $N \to \infty$ in the divisibility order. (I might also consider calling your profinite group something other than $P$, which looks like a pro-$p$ group ….) $\endgroup$– LSpiceCommented Jul 31, 2020 at 16:40
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$\begingroup$ Oh, by the way there is a very natural proof for finite groups, which consists in proving that the natural map $\prod_p G(p)\to G$ is a group isomorphism (rather than computing cardinalities). $\endgroup$– YCorCommented Jul 31, 2020 at 22:28
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This is Proposition 2.3.8 of Ribes and Zaleskii - Profinite groups (second edition). (I originally gave references specifically for the finer structure of profinite Abelian groups, but assuming finite generation, in Section 4.3 of the same book.)
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$\begingroup$ I'm particularly interested in the non-finitely generated case. Does anyone know more about that case? $\endgroup$ Commented Jul 31, 2020 at 17:01
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2$\begingroup$ It turns out those Section 4.3 references were mainly about the finer structure of pro-$p$ groups. I modified now to cite Proposition 2.3.8, which is about pronilpotent groups in general (no finite generation hypotheses). Sorry for the confusion! $\endgroup$– LSpiceCommented Jul 31, 2020 at 17:04