From: mareg@primrose.csv.warwick.ac.uk ()
Subject: Re: Sequence of groups starting from S_6
Date: Mon, 3 Dec 2001 09:54:51 +0000 (UTC)
Newsgroups: sci.math
Summary: Automorphism groups (and towers) of symmetric groups
In article <66f84b63c429b94605905b4264a0ab30.22128@mygate.mailgate.org>,
"Robin Chapman" writes:
>"Yoav Gross" wrote in message
>news:6n8fa6zn2p3e@legacy...
>
>> Define a sequence of groups by G_1=S_6 (symmetric group on 6 letters)
>> and for n>=2 G_n=Aut(G_(n-1)) .
>> Is this sequence becomes periodic eventually ?
>>
>
>Isn't G_3 = G_2?
Yes. G_3 = G_2 = Aut(A_6) = PGammaL(2,9).
For any finite nonabelian simple group S, it is not hard to show that
Aut(Aut(S)) =~ Aut(S).
There is also an old result that says that for any group G_1 with trivial
centre, the chain G_1 <= G_2 <= G_3 = ... (with above notation)
eventually becomes constant.
I seem to remember hearing (maybe it was on sci.math even!) that it had been
proved more recently that for any finite group whatsoever, the sequence
G_1, G_2, G_3 ... eventually becomes constant (or maybe periodic), but I
don't remember the details.
Derek Holt.
==============================================================================
From: "Jim Heckman"
Subject: Re: Sequence of groups starting from S_6
Date: Tue, 4 Dec 2001 10:16:29 GMT
Newsgroups: sci.math
Excellent! I was hoping Derek and/or Robin would take a swing at this!
On 3-Dec-2001, mareg@primrose.csv.warwick.ac.uk () wrote:
> In article <66f84b63c429b94605905b4264a0ab30.22128@mygate.mailgate.org>,
> "Robin Chapman" writes:
>
> >"Yoav Gross" wrote in message
> >news:6n8fa6zn2p3e@legacy...
> >
> >> Define a sequence of groups by G_1=S_6 (symmetric group on 6 letters)
> >> and for n>=2 G_n=Aut(G_(n-1)) .
> >> Is this sequence becomes periodic eventually ?
> >
> >Isn't G_3 = G_2?
>
> Yes. G_3 = G_2 = Aut(A_6) = PGammaL(2,9).
> For any finite nonabelian simple group S, it is not hard to show that
> Aut(Aut(S)) =~ Aut(S).
Cool! Since it's "not hard", could you show us? Pretty please? :-)
By the way, this result comes as a mild surprise to me, because somehow
I'd gotten the impression that Aut(Aut(S_6)) was *not* = Aut(S_6). Thinking
back on it now, I believe the impression came from considering the
other-than-S_6 index-2 subgroups of Aut(S_6), which for some reason I
thought were both =~ PGL(2,9). But clearly your result implies that they
*aren't* isomorphic, so is there a special name for the "other" one?
> There is also an old result that says that for any group G_1 with trivial
> centre, the chain G_1 <= G_2 <= G_3 = ... (with above notation)
> eventually becomes constant.
And this too? I can see why the G_i must be a chain (with each G_i
having trivial center), but not why they must become constant.
> I seem to remember hearing (maybe it was on sci.math even!) that it had been
> proved more recently that for any finite group whatsoever, the sequence
> G_1, G_2, G_3 ... eventually becomes constant (or maybe periodic), but I
> don't remember the details.
I think I remember this from sci.math.research, and I'm pretty sure it was
periodic.
--
Jim Heckman
==============================================================================
From: "Robin Chapman"
Subject: Re: Sequence of groups starting from S_6
Date: Tue, 4 Dec 2001 10:30:39 +0000 (UTC)
Newsgroups: sci.math
"Jim Heckman" wrote in message
news:u0p8egpqf7fufa@news.supernews.com...
> >
> > Yes. G_3 = G_2 = Aut(A_6) = PGammaL(2,9).
> > For any finite nonabelian simple group S, it is not hard to show that
> > Aut(Aut(S)) =~ Aut(S).
>
> Cool! Since it's "not hard", could you show us? Pretty please? :-)
>
> By the way, this result comes as a mild surprise to me, because somehow
> I'd gotten the impression that Aut(Aut(S_6)) was *not* = Aut(S_6). Thinking
> back on it now, I believe the impression came from considering the
> other-than-S_6 index-2 subgroups of Aut(S_6), which for some reason I
> thought were both =~ PGL(2,9). But clearly your result implies that they
> *aren't* isomorphic, so is there a special name for the "other" one?
This order 720 group is isomorphic to M_{10}, the stabilizer
of a point in the Mathieu group M_{11}.
Robin Chapman
--
Posted from olive.ulcc.wwwcache.ja.net [194.82.103.41]
via Mailgate.ORG Server - http://www.Mailgate.ORG
==============================================================================
From: mareg@primrose.csv.warwick.ac.uk ()
Subject: Re: Sequence of groups starting from S_6
Date: Tue, 4 Dec 2001 14:14:41 +0000 (UTC)
Newsgroups: sci.math
In article ,
"Jim Heckman" writes:
>Excellent! I was hoping Derek and/or Robin would take a swing at this!
>
>On 3-Dec-2001, mareg@primrose.csv.warwick.ac.uk () wrote:
>
>> In article <66f84b63c429b94605905b4264a0ab30.22128@mygate.mailgate.org>,
>> "Robin Chapman" writes:
>>
>> >"Yoav Gross" wrote in message
>> >news:6n8fa6zn2p3e@legacy...
>> >
>> >> Define a sequence of groups by G_1=S_6 (symmetric group on 6 letters)
>> >> and for n>=2 G_n=Aut(G_(n-1)) .
>> >> Is this sequence becomes periodic eventually ?
>> >
>> >Isn't G_3 = G_2?
>>
>> Yes. G_3 = G_2 = Aut(A_6) = PGammaL(2,9).
>> For any finite nonabelian simple group S, it is not hard to show that
>> Aut(Aut(S)) =~ Aut(S).
>
>Cool! Since it's "not hard", could you show us? Pretty please? :-)
Yes. In fact S does not need to be finite. We can embed S in A=Aut(S) and
regard A as acting on S by conjugation. So every nonidentity element
of A induces a nontrivial automorphism of S by conjugation, which means
that the centralizer C_A(S) of S in A is trivial.
Now S, being simple, is a minimal normal subgroup of A, and any other minimal
normal subgroup T would intersect S trivially, and so we would have [S,T] = 1,
contradicting C_A(S)=1. Hence S is the unique minimal normal subgroup of A.
Now let x be an automorphism of A. By uniqueness of S as minimal normal
subgroup, we must have x(S) = S. So x acts as an automorphism of S, hence
it acts on S in the same way as some element of A is acting by conjugation.
In other words, we can multiply x by an inner automorphism of A and assume
that x(s)=s for all s in S.
Let a in A. Then for any s in S, we have
a s a^-1 = x(a s a^-1) = x(a) s x(a)^-1,
so a^-1 x(a) centralizes S, and hence a^-1 = x(a). Thus x is the trivial
automorphism, and we are done.
>By the way, this result comes as a mild surprise to me, because somehow
>I'd gotten the impression that Aut(Aut(S_6)) was *not* = Aut(S_6). Thinking
>back on it now, I believe the impression came from considering the
>other-than-S_6 index-2 subgroups of Aut(S_6), which for some reason I
>thought were both =~ PGL(2,9). But clearly your result implies that they
>*aren't* isomorphic, so is there a special name for the "other" one?
Robin Chapman answered that one. The three subgroups of index 2 in
Aut(A_6) are S_6, PGL(2,9) and M_10.
>> There is also an old result that says that for any group G_1 with trivial
>> centre, the chain G_1 <= G_2 <= G_3 = ... (with above notation)
>> eventually becomes constant.
>
>And this too? I can see why the G_i must be a chain (with each G_i
>having trivial center), but not why they must become constant.
This is a result of Wielandt, 1939. I know that there is a proof in
Passman's book on Permutation groups, but unfortunately I do not have
a copy to hand.
>> I seem to remember hearing (maybe it was on sci.math even!) that it had been
>> proved more recently that for any finite group whatsoever, the sequence
>> G_1, G_2, G_3 ... eventually becomes constant (or maybe periodic), but I
>> don't remember the details.
>
>I think I remember this from sci.math.research, and I'm pretty sure it was
>periodic.
What I was remembering was a result of Joel Hamkins (1998), but in fact
he proves that the automorphism chain terminates only transfinitely. So
there is no guarantee (according to his result anyway) that the chain will
become periodic after finitely many steps.
Derek Holt.
==============================================================================
From: Robin Chapman
Subject: Re: Automorphism group of S_n
Date: Sun, 11 Feb 2001 10:26:17 GMT
Newsgroups: sci.math
In article <20010210221446.08836.00000412@ng-co1.news.gateway.net>,
jamesrheckman@gateway.netnospam (Jim Heckman) wrote:
>
> Yeah, I've seen that in a few places. Perhaps more to the OP's
> point is actually constructing Aut(S_6), which I've only seen done
> by laboriously exploring the detailed structure of S_6, or by
> making use of the (not obvious) isomorphism between A_6 and
> PSL(2,9) -- which requires knowing a lot about finite groups of
> Lie type. Is there a "simple" way to construct Aut(S_6), or at least
> to show that its order is 2*|S_6|?
The argument used to show that Aut(S_n) is usually S_n gives you that
Aut(S_6) has order |S_6| or 2|S_6|. So we have to find an automorphism
of S_6 not an inner automorphism. The classical approach is due to
Sylvester (or was it Cayley)?
S_6 acts on the numbers 1, 2, 3, 4, 5, 6. A duad is an unordered
pair of these numbers, eg 13 or 25. S_6 acts on these. A syntheme
is an unordered triple of disjoint duads e.g. 13/25/46. S_6 acts
on these too. A synthematic total is an unordered quintuple of
synthemes incorporating all 15 duads, e.g
13/25/46//14/26/35//12/34/56//15/24/36//16/23/45.
There are 15 duads, 15 synthemes and 6 synthematic totals.
Each permutation of S_6 gives rise to a permutation of the duads,
of the synethemes and of the synthematic totals. Labelling the
totals arbitrarily from 1 to 6 we see that each element of S_6 gives
rise to another via the permutation of the totals. Thus we get
a homomorphism from S_6 to S_6 which we see cannot be trivial.
But we can check
that a transposition (i j) permutes the totals without fixed points
so this automorphism is not linear.
The whole setup can be reversed. Each duad of totals corresponds to a
syntheme. Each syntheme of totals correponds to a duad and each
synthematic total of totals corresponds to a number. If we look at the
permutations of the set {numbers} u {totals} preverving the "structure"
we get a group of order 1440 having S_6 as a subgroup. This is
Aut(S_6).
--
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/
==============================================================================
From: mareg@mimosa.csv.warwick.ac.uk ()
Subject: Re: Automorphism group of S_n
Date: 11 Feb 2001 11:29:43 GMT
Newsgroups: sci.math
In article <20010210221446.08836.00000412@ng-co1.news.gateway.net>,
jamesrheckman@gateway.netnospam (Jim Heckman) writes:
>>From: Robin Chapman rjchapman@my-deja.com
>>Date: 2/10/01 5:16 AM Pacific Standard Time
>>Message-id: <963evi$vrg$1@nnrp1.deja.com>
>>
>>In article <20010210055630.14450.00000428@ng-fq1.news.gateway.net>,
>>jamesrheckman@gateway.netnospam (Jim Heckman) wrote:
>>
>>> >From: alan_gross@my-deja.com
>>> >Date: 2/10/01 1:56 AM Pacific Standard Time
>>> >Message-id: <963396$oii$1@nnrp1.deja.com>
>>> >
>>> >What is the automorphism group of S_n,
>>> >the symmetric group on n letters ?
>>>
>
>Yeah, I've seen that in a few places. Perhaps more to the OP's
>point is actually constructing Aut(S_6), which I've only seen done
>by laboriously exploring the detailed structure of S_6, or by
>making use of the (not obvious) isomorphism between A_6 and
>PSL(2,9) -- which requires knowing a lot about finite groups of
>Lie type. Is there a "simple" way to construct Aut(S_6), or at least
>to show that its order is 2*|S_6|?
I think that it is an exaggeration to say that knowledge of the
properties of PSL(2,9) requires knowing a lot about finite
groups of Lie type.
But if you want a quick proof that there exists an outer automorphism
of S_6, then the following works.
We make use of the fact that S_n has the presentation
i+1) >,
where x_i is the transposition (i,i+1).
When n=6, let y_i (1 <= i <= 5) be the elements
(1,2)(3,4)(5,6), (1,3)(2,5)(4,6), (1,2)(3,6)(4,5),
(1,6)(2,5)(3,4), (1,2)(3,5)(4,6).
and verify that the y_i satisfy the same relations as the defining relations
of S_6. This proves that the map that send x_i to y_i for 1 <= i <= 5
extends to an automorphism of S_6.
Derek Holt.