From: jamesrheckman@gateway.netnospam (Jim Heckman) Subject: Re: Automorphism group of S_n Date: 10 Feb 2001 10:56:30 GMT Newsgroups: sci.math >From: alan_gross@my-deja.com >Date: 2/10/01 1:56 AM Pacific Standard Time >Message-id: <963396$oii$1@nnrp1.deja.com> > >What is the automorphism group of S_n, >the symmetric group on n letters ? For all n except n=2 and n=6, Aut(S_n) is S_n itself. I.e., S_n has no outer automorphisms. Aut(S_2) is trivial, and Aut(S_6) is of order 2*|S_6| -- there's a funky outer involution that swaps the (6) cycles with the (3)(2) elements, and thus the (3)(3)'s with the (3)'s and the (2)(2)(2)'s with the (2)'s. Of course, this also swaps all of the appropriate subgroups of S_6 containing these types of elements, perhaps most notably the transitive reps of S_5 with the non-transitive ones (which are permutation isomorphic to S_5 itself). -- Jim Heckman ============================================================================== From: Robin Chapman Subject: Re: Automorphism group of S_n Date: Sat, 10 Feb 2001 13:16:37 GMT Newsgroups: sci.math In article <20010210055630.14450.00000428@ng-fq1.news.gateway.net>, jamesrheckman@gateway.netnospam (Jim Heckman) wrote: [quote of previous message deleted --djr] Just to observe that this is quite easy to prove. The key is to show that for n =/= 6 each automorphism of S_n preserves the conjugacy class of transpositions (i j). -- Robin Chapman http://www.maths.ex.ac.uk/~rjc/rjc.html "His mind has been corrupted by colours, sounds and shapes." The League of Gentlemen Sent via Deja.com http://www.deja.com/