On Nov 3, 4:13 pm, "Achava Nakhash, the Loving Snake"
<ach
...@hotmail.com> wrote:
> On Nov 3, 1:22 pm, Arturo Magidin <magi
...@member.ams.org> wrote:
> > On Nov 3, 3:05 pm, Al2009 <algebra_whate...@yahoo.ca> wrote:
> > > Hi,
> > > I am trying to understand some automorphism groups of symmetric groups.
> > >http://en.wikipedia.org/wiki/Symmetric_group#Automorphism_group
> > > It says that
> > > Aut(S_2) = C_2,
> > > Aut(S_6) = S_6 \semidirect C_2
> > > Aut(S_n) = S_n, for n>7.
> > > I know that
> > > G/Z(G) = Inn(G), Out(G) = Aut(G)/Inn(G).
> > > But I can't figure out why Aut(S_6) = S_6 \semidirect C_2
> > > Aut(S_n) = S_n, for n>7.
> > Should be n>6.
> > This is nontrivial. That Aut(S_n)=S_n for n>7 follows by looking at
> > the conjugacy classes: any automorphism must send conjugacy classes to
> > conjugacy classes. A simple count shows that any automorphism of S_n
> > with n>6 must fix the conjugacy class of the transpositions, and then
> > you can leverage that to a proof. It will also show that there is a
> > possible non-inner automorphism for S_6. Constructing it is not
> > obvious, however.
> I have wondered about this inactively since grad school. Do you have
> a reference for this result? I also noticed that the OP left out S_3,
> S_4, and S_5. I suspect they are not difficult cases as it is quite
> easy to get one's hands on all the elements, but for completeness, it
> would be nice to know.
product of S_6 by a cyclic group of order 2.
Rotman (Introduction to the Theory of groups, 4th Edition), Lemma 7.4
shows that an automorphism of S_n is inner if and only if it preserves
transpositions. From there, you can just count how many elements of
order 2 there are in each conjugacy class in S_n to get that in all
cases except for n=6 and n=2, there is no conjugacy class of elements
of order 2 with the same number of elements as the class of
transpositions, which yields the result. He then explicitly
constructs an outer automorphism for S_6, but for that I like
"Combinatorial structure on the automorphism group of S_6", by T.Y.
Lam and David Leep, Expositiones Mathematicae 11 (1993), no. 4, pp.
289-308.
A sketch of the proof of the lemma: let phi be an automorphism that
preserves transpositions. It maps (1,2) to some (i,j); let g_2 be
conjugation by (1,i)(2,j); then g_2^{-1}phi fixes (1,2). Proceed by
induction to assume you have g_r^{-1}...g_2^{-1}phi fixing (1,2),...,
(1,r). This map still preserves transpositions, and must sned (1,r+1)
to some (t,v). But (1,2) cannot be disjoint from (t,v), because then
(1,2)(1,r+1) would map to (1,2)(t,v) of order 2, instead of something
of order 3. So (1,r+1) maps to either (1,k) or to (2,k) for some k.
Then k>r; now let g_{r+1} be conjugation by (k,r+1), so g_{r+1]^
{-1}...g_2^{-1}phi fixes (1,2),...,(1,r+1). Inductively, you can
compose phi with enough conjugations so that you get a map that fixes
every transposition, and hence is the identity. Thus, phi is a
conjugation.
--
Arturo Magidin
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