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From: Achava Nakhash, the Loving Snake on 3 Nov 2009 23:35
On Nov 3, 3:13 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> 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-

Thanks for the info. I will study it when I get a chance, hopefully
soon. I was always a big fan of Professor Lam, but I am surprised
that wrote an article on group theory. Of course I was long gone from
Berkeley by 1993.


Regards,
Achava
From: Arturo Magidin on 4 Nov 2009 00:28
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.

A first step to understanding is being careful; the first line is
incorrect, and does not appear in Wikipedia: Aut(S_2)={1}, but S_2 is
isomorphic to C_2. Either you misread, or you misunderstood what you
read. You have shown a tendency to carelessness in several threads:
that is something you need to guard against if you want to understand
and succeed.

--
Arturo Magidin

From: Al2009 on 3 Nov 2009 14:41
Thanks for your help.

I borrowed the book Rotman's group theory 3rd edition from my school library.

I am currently reading the lemma you mentioned.


"... 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"

I am having hard time understanding the above.

What I understand is,

1. S_n can be generated by transpostions.
2. Inner automorphisms preserve cycle types of an element in S_n.
3. S_n can be partitioned to conjugacy classes. Therefor a conjugacy class can be represented by its cycle type.
3. There is a conjugacy class of S_n, which consist of disjoint transpositions only.

In Rotman's book,
"Let C_1 be the conjugacy class of S_n consisting of all transpostions,,,; every element of C_1 has order 2"
.........
"Let C_k be the conjugacy class consisting of all products of k disjoint transpostions",

So S_n for large n,
C_2 = {(i_1 i_2)(i_3 i_4), (i_5 i_6)(i_7 i_8),,,,,}
C_3 = {(j_1 j_2)(j_3 j_4)(j_5 j_6), (j_7 j_8)(j_9 j_10)(j_11 j_12) ,,,

Intuitively, for some k, C_k = C_1, but that is not the case for n=6.

Any simple counter example showing that why C_k \neq C_1 ?

Thanks.
From: Al2009 on 3 Nov 2009 14:53
> 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#Automorph
> ism_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.
>
> A first step to understanding is being careful; the
> first line is
> incorrect, and does not appear in Wikipedia:
> Aut(S_2)={1}, but S_2 is
> isomorphic to C_2. Either you misread, or you
> misunderstood what you
> read. You have shown a tendency to carelessness in
> several threads:
> that is something you need to guard against if you
> want to understand
> and succeed.
>
> --
> Arturo Magidin
>

Thanks for your advice :)
I checked wiki again, and you are correct. Hope I dont make this kind of mistakes again.
From: Arturo Magidin on 4 Nov 2009 01:05
On Nov 3, 11:41 pm, Al2009 <algebra_whate...(a)yahoo.ca> wrote:
> Thanks for your help.
>
> I borrowed the book Rotman's group theory 3rd edition from my school library.
>
> I am currently reading the lemma you mentioned.
>
> "... 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"
>
> I am having hard time understanding the above.
>
> What I understand is,
>
> 1. S_n can be generated by transpostions.
> 2. Inner automorphisms preserve cycle types of an element in S_n.
> 3. S_n can be partitioned to conjugacy classes. Therefor a conjugacy class can be represented by its cycle type.
> 3. There is a conjugacy class of S_n, which consist of disjoint transpositions only.

Wby do you have "disjoint" here? It is false. The conjugacy class of
S_n that contains (1,2) will also contain (1,3).

> In Rotman's book,
> "Let C_1 be the conjugacy class of S_n consisting of all transpostions,,,; every element of C_1 has order 2"
> ........
> "Let C_k be the conjugacy class consisting of all products of k disjoint transpostions",
>
> So S_n for large n,
> C_2 = {(i_1 i_2)(i_3 i_4), (i_5 i_6)(i_7 i_8),,,,,}

No; C_2 would contain (1,2)(3,4), and (1,3)(2,4), and (1,4)(2,3), and
lots of other things. You notation here is bad because it is likely to
confuse you.



> C_3 = {(j_1 j_2)(j_3 j_4)(j_5 j_6), (j_7 j_8)(j_9 j_10)(j_11 j_12) ,,,
>
> Intuitively, for some k, C_k = C_1,

Just for k=1.

> but that is not the case for n=6.

If you understand what you are writing, then you realize how utterly
silly what you have just written is. C_k consists of all elements of
C_k that are the product of EXACTLY k disjoint transpositions. C_k =
C_r if and only if k=r.

The point Rotman makes is that for n>2, the SIZE of C_1 is different
from the size of every *other* C_k, except in the case of n=6 (when
C_1 and C_3 happen to have the same size).

> Any simple counter example showing that why C_k \neq C_1 ?

You are being careless again. No wonder you are having trouble
understanding! THEY ARE ALL DIFFERENT! C_i contains *exactly* the
elements that can be written as a product of i disjoint
transpositions. They are ALL DIFFERENT from elements that are the
product of k disjoint transpositions for k=/=i. I mean: if sigma is a
product of k disjoint transpositions, then how many elements does it
"move"?

I also find your lack of initiative here troubling. Why did you not
try some small values of n? They would have quickly revealed to you
your carelessness and confusion.

--
Arturo Magidin
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