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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(^;)H k(limC ;)limH k(C ;) 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(^;)limH 2(C ;)limC / 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}