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From: Al2009 on 3 Nov 2009 06:05
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.

Any thoughts?

Thanks.
From: Arturo Magidin on 3 Nov 2009 16:22
On Nov 3, 3:05 pm, Al2009 <algebra_whate...(a)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.
--
Arturo Magidin
From: Achava Nakhash, the Loving Snake on 3 Nov 2009 17:13
On Nov 3, 1:22 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Nov 3, 3:05 pm, Al2009 <algebra_whate...(a)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.
> --
> Arturo Magidin

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.


Regards,
Achava
From: Arturo Magidin on 3 Nov 2009 18:13
On Nov 3, 4:13 pm, "Achava Nakhash, the Loving Snake"
<ach...(a)hotmail.com> wrote:
> On Nov 3, 1:22 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
>
>
>
> > On Nov 3, 3:05 pm, Al2009 <algebra_whate...(a)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.

Aut(S_n) = S_n for n=/=6, and Aut(S_6) is a (non-direct) semidirect
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

From: Arturo Magidin on 3 Nov 2009 23:17
On Nov 3, 5:13 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Nov 3, 4:13 pm, "Achava Nakhash, the Loving Snake"
> > > > 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.

[...]

>
> Aut(S_n) = S_n for n=/=6, and Aut(S_6) is a (non-direct) semidirect
> product of S_6 by a cyclic group of order 2.

Note that S_2 has two elements, hence Aut(S_2) = {1}, so there is
another exception missing there. The OP misreported what Wikipedia
says: it says Aut(S_2) is trivial, but that S_2 is isomorphic to C_2.

--
Arturo Magidin
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