Let $G$ be a group, let $\widehat{G}$ be its profinite completion, and let $G^{\text{ab}}$ be its abelianization.

Is is true that abelianization commutes with profinite completion, in the sense that $(\widehat{G})^{\text{ab}} = \widehat{G^{\text{ab}}}$?

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    $\begingroup$ I think it may depend on exactly what you mean by the abelianization of a profinite group. The commutator subgroup of a profinite group is not in general closed, so the abelianization in the category of profinite groups (i.e., the quotient by the closure of the commutator subgroup) may be different from the abelianization as an abstract group. If you mean the topological abelianization, then I think the answer to your question is "yes". $\endgroup$ – Jeremy Rickard Jul 17 '16 at 10:32
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    $\begingroup$ Thanks for the helpful response! Here is a somewhat related question: does abelianization commute with infinite products? I think the answer is yes, even if one does not take closures of the commutator subgroups, but I'm not sure if I'm overlooking something. $\endgroup$ – Ashvin Swaminathan Jul 18 '16 at 1:24
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    $\begingroup$ The product of abelianizations can be a proper quotient of the abelianization of the product, in general. Suppose for each $n\in\mathbb{N}$ we have a group $G_n$ with an element $x_n$ that is the product of $n$ commutators but no fewer. Then $(x_n)_{n\in\mathbb{N}}\in\prod_{n\in\mathbb{N}}G_n$ is in the product of the commutator subgroups of the $G_n$, but not in the commutator subgroup of the product. $\endgroup$ – Jeremy Rickard Jul 18 '16 at 10:45

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