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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeNov 21st 2013
   • PermaLink
   Author: Urs
   Format: MarkdownItexmoved the statement that a [[profinite group]] is a group internal to profinite sets from the "Examples"-section to "Definition"-section

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

2. • CommentRowNumber2.
   • CommentAuthorZhen Lin
   • CommentTimeNov 21st 2013
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   Author: Zhen Lin
   Format: MarkdownItexI 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.

   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.

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeNov 25th 2013
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   Author: DavidRoberts
   Format: MarkdownItex[[Stone Spaces]] chapter VI, section 2.9 has a general result in this direction.

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

4. • CommentRowNumber4.
   • CommentAuthorjesse
   • CommentTimeMay 4th 2017
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   Author: jesse
   Format: MarkdownItexI 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]].

   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.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJan 8th 2022
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   Author: Urs
   Format: MarkdownItexDumb 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, $$ H^2\big( \widehat \mathbb{Z} ;\, \mathbb{Z}\big) \;\simeq\; \underset{\longrightarrow}{\lim} H^2\big( C_\bullet ;\, \mathbb{Z}\big) \;\simeq\; \underset{\longrightarrow}{\lim} C_\bullet \;\simeq\; \mathbb{Q}/\mathbb{Z} $$ ?

   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,

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

   ?