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This lecture notes from John Jones https://www2.warwick.ac.uk/fac/sci/maths/people/staff/vincent/cohomology.pdf state that abelian groups are a product of cyclic groups (page 9).

We know that this is true if the group is finite or finitely generated. But does it hold in general for any abelian group?

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    $\begingroup$ No, only when the group is finitely generated. Take $(\mathbb{R},+)$ as a counterexample. $\endgroup$ – Francis Begbie Jan 11 '16 at 5:24
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No, most non-finitely generated abelian groups are not products of cyclic groups. For instance, a product of infinitely many nontrivial cyclic groups is uncountable (since any product of infinitely many sets with more than one element is uncountable). So any countable abelian group that is not finitely generated cannot be a product of cyclic groups. A simple example of such a group is $\mathbb{Q}$.

The proof that $\mathbb{Q}$ is injective in those notes is totally wrong (it's not even correct if you assume $P$ and $Q$ are finitely generated; it seems to be implicitly assuming $P$ is a direct summand of $Q$). A correct proof involves extending $f$ from $P$ to $Q$ one element at a time by transfinite induction, using the divisibility of $\mathbb{Q}$.

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  • $\begingroup$ Thanks a lot! But can I prove that Q is not a direct product by saying, if $\mathbb{Q}=\prod_{i \in I} G_i$ and $p/q$ and $r/s$ are generators of two different ciclic groups $G_i and G_j$ resp. then $(r.q)p/q=pr=(p.s)r/s \in G_i \cap G_j$ and thus the intersection is not trivial and thus is not a product? Thanks! $\endgroup$ – Werner Germán Busch Jan 11 '16 at 19:29
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    $\begingroup$ Sure, that works. $\endgroup$ – Eric Wofsey Jan 11 '16 at 20:51
  • $\begingroup$ Great thanks !!! $\endgroup$ – Werner Germán Busch Jan 11 '16 at 23:02

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