http://cameroncounts.wordpress.com/2011/04/21/the-symmetric-group-12/ always busy counting, doubting every figured guess . . . Fri, 24 May 2013 07:17:33 +0000 hourly 1 http://wordpress.com/ http://cameroncounts.wordpress.com/2011/04/21/the-symmetric-group-12/#comment-6404 Fri, 17 Aug 2012 13:57:06 +0000 http://cameroncounts.wordpress.com/?p=1266#comment-6404 Thank you. This with
http://www.maths.qmul.ac.uk/~raw/talks_files/ONStalk.ps
resolve my question.

]]> http://cameroncounts.wordpress.com/2011/04/21/the-symmetric-group-12/#comment-6400 Fri, 17 Aug 2012 12:24:26 +0000 http://cameroncounts.wordpress.com/?p=1266#comment-6400 The stabiliser of a point is a complement for T. In fact, since T acts regularly, the permutation domain can be identified with T, so that T acts by translation; then the stabiliser of the identity is a complement, acting by automorphisms of T (i.e. as a linear group).

]]> http://cameroncounts.wordpress.com/2011/04/21/the-symmetric-group-12/#comment-6397 Fri, 17 Aug 2012 10:13:26 +0000 http://cameroncounts.wordpress.com/?p=1266#comment-6397 It is a question. “If T is abelian then H is affine” . I seems to understand that T=Z_p^d, so H is an extension of Z_p^d by
Aut(Z_p^d). Why the extension split?
Thank you. Lev.

]]> http://cameroncounts.wordpress.com/2011/04/21/the-symmetric-group-12/#comment-2397 Tue, 26 Apr 2011 11:21:22 +0000 http://cameroncounts.wordpress.com/?p=1266#comment-2397 Yes, sorry – now fixed! This is not a good time of year for careful proofreading…

]]> http://cameroncounts.wordpress.com/2011/04/21/the-symmetric-group-12/#comment-2395 Tue, 26 Apr 2011 11:06:35 +0000 http://cameroncounts.wordpress.com/?p=1266#comment-2395 I think there is a typographical error in the fourth line of theO’Nan–Scott Theorem: should n=km not be n=k^m?

]]> http://cameroncounts.wordpress.com/2011/04/21/the-symmetric-group-12/#comment-2387 Thu, 21 Apr 2011 11:34:31 +0000 http://cameroncounts.wordpress.com/?p=1266#comment-2387 [...] Previous | Next [...]

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