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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 21st 2013

moved the statement that a profinite group is a group internal to profinite sets from the “Examples”-section to “Definition”-section

• CommentRowNumber2.
• CommentAuthorZhen Lin
• CommentTimeNov 21st 2013

I vaguely recall that it’s not entirely trivial that pro-(finite algebras) are the same as internal algebras in profinite sets – there’s some restriction on the signature and axiomatisation, I think.

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeNov 25th 2013

Stone Spaces chapter VI, section 2.9 has a general result in this direction.

• CommentRowNumber4.
• CommentAuthorjesse
• CommentTimeMay 4th 2017

I added the remark that finite index subgroups of a profinite group are not generally open, and gave the usual example involving non-principal ultrafilters. I also briefly discussed when a group is isomorphic to its own profinite completion.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJan 8th 2022

Dumb question:

Is it right that the (integral, say) group cohomology of the profinite integers is the colimit of the group cohomologies of the cyclic groups

$H^k\big( \widehat \mathbb{Z} ;\, \mathbb{Z}\big) \;\simeq\; H^k\big( \underset{\longleftarrow}{\lim} C_\bullet ;\, \mathbb{Z}\big) \;\simeq\; \underset{\longrightarrow}{\lim} H^k\big( C_\bullet ;\, \mathbb{Z}\big)$

?

Hence that, in particular,

$H^2\big( \widehat \mathbb{Z} ;\, \mathbb{Z}\big) \;\simeq\; \underset{\longrightarrow}{\lim} H^2\big( C_\bullet ;\, \mathbb{Z}\big) \;\simeq\; \underset{\longrightarrow}{\lim} C_\bullet \;\simeq\; \mathbb{Q}/\mathbb{Z}$

?