Is Aut S_n isomorphic to S_n?
Derek Clegg
speculation. (I fully expect this to be both well known and trivial. In fact,
probably it's proved in a later chapter.) One of the exercises in Chapter 1
asks us to construct Aut S_3, the group of automorphisms of the symmetric group
on 3 elements. I tried generalizing this. I believe that I can prove that
|Aut S_n| is n!, and I have a vague hand-waving argument that Aut S_n is
isomorphic to S_n. I'd like to know whether my guess is correct. (Of course,
if Aut S_3 isn't isomorphic to S_3, which is the result I obtained for the
exercise, I'm completely wrong.) Any one care to help? (Incidentally, I wish
this were a homework problem, since then I'd find out the answer....)
Derek B Clegg (dcl...@next.com)
David Sibley
Aut S_6 is strictly bigger than S_6. But for n other than 6, every automorphism
of S_n is inner. See Huppert's Endliche Gruppen I, Kap. II, Satz 5.5.
David Sibley
sib...@math.psu.edu
Arthur Rubin
> Derek B Clegg (dcl...@next.com)
Aut S_1 = Aut S_2 = S_1, as the groups are cycles.
For n >=3, clearly S_n is a normal subgroup of Aut S_n, with the map:
-1
x -> (y -> x y x ).
(This is off the top of my head --- I could be wrong.)
For n != 6, you can show that the set of transpositions (a b) in S_n is
preserved by any automorphism, as it is the only conjugacy class of S_n
with elements of order 2 with cardinality n(n-1)/2. That should be
sufficient to show that any automorphism of S_n is inner, with some work.
For n = 6, there is an outer automorphism, as can be seen by embedding S_6
into a special linear group, but I can't remember which one.
--
Arthur L. Rubin: a_r...@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-...@mcimail.com 7070...@compuserve.com art...@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
Ich bin ein Virus. Mach' mit und kopiere mich in Deine .signature.
Charles Yeomans
The answer is yes, except when n = 6. All automorphisms of S_n are
inner, for n not 6. For S_6, [Aut:Inn] =2. There is a very nice
paper by David Leep and T.Y. Lam, "COmbinatorial Stucture of the
automorphism group on S_6" - actually I should say there will be a
nice paper, since it hasn't been published yet.
Charles Yeomans
cyeo...@ms.uky.edu
yeo...@austin.onu.edu
John C. Baez
>In article 38...@rosie.NeXT.COM, derek...@next.com (Derek Clegg) writes:
>>I'm reading Jacobson's Algebra, and one of his exercises led me to a wild
>>speculation. (I fully expect this to be both well known and trivial. In fact,
>>probably it's proved in a later chapter.) One of the exercises in Chapter 1
>>asks us to construct Aut S_3, the group of automorphisms of the symmetric group
>>on 3 elements. I tried generalizing this. I believe that I can prove that
>>|Aut S_n| is n!, and I have a vague hand-waving argument that Aut S_n is
>>isomorphic to S_n. I'd like to know whether my guess is correct.
>Aut S_6 is strictly bigger than S_6. But for n other than 6, every automorphism
>of S_n is inner. See Huppert's Endliche Gruppen I, Kap. II, Satz 5.5.
Far out! Maybe this is why 6 people is supposed to be the ideal number
for a conversation? What is Aut S_6, anyway?
Just in case Clegg or readers don't know what "inner" automorphisms are,
they are just the automorphisms of the form g -> hgh^{-1} for some h in
the group. Using inner automorphisms gives a homomorphism
S_n -> Aut(S_n)
and this is 1-1 except for n=2, because in every other case the
center of S_n is trivial. (By the way this means that S_2 is not equal
to Aut(S_2).) So proving "Clegg's conjecture" amounts to showing that
this homomorphism is onto, which apparently is true unless n is 6.
Is there a neat geometrical interpretation of the outer automorphisms of
S_6? Too bad it's not S_5, which is related to icosahedra...
Charles Yeomans
Yes. S6 can be identified with the group of nonsingular semilinear
fractional transformations with square determinant on the projective
line over F_9(=field with 9 elements). Using this, Aut(S6) can
be identified with the full group of nonsingular semilinear
fractional transformations of the same projective line.
Also, Aut(S6) contains three subgroups of index 2. They are
isomorphic to S6, PGL(2,9) and the non-simple Mathieu group
M_10.
Actually, I didn't quite answer your question, but I answered
some question.
Charles Yeomans
cyeo...@ms.uky.edu
yeo...@austin.onu.edu
Bradley W. Brock
> >Aut S_6 is strictly bigger than S_6. But for n other than 6, every
automorphism
> >of S_n is inner. See Huppert's Endliche Gruppen I, Kap. II, Satz 5.5.
>
> Far out! Maybe this is why 6 people is supposed to be the ideal number
> for a conversation? What is Aut S_6, anyway?
>
> S_6? Too bad it's not S_5, which is related to icosahedra...
>
transitive copies of S_5. The outer automorphism flips these with the usual
intransitive copies. Interestingly, it flips the following conjugacy classes:
6 and 321, each with 120 elements
33 and 3111, each with 40 elements
222 and 221111, each with 15 elements
where I have written the cycle structure of the conjugacy class.
The other conjugacy classes are fixed.
--
Bradley W. Brock, Department of Mathematics
Rose-Hulman Institute of Technology | I appropriate all my opinions,
br...@nextwork.rose-hulman.edu | so technically they're not mine.
Christopher J. Henrich
>
>>Aut S_6 is strictly bigger than S_6. But for n other than 6, every automorphism
>>of S_n is inner. See Huppert's Endliche Gruppen I, Kap. II, Satz 5.5.
>
>Far out! Maybe this is why 6 people is supposed to be the ideal number
>for a conversation? What is Aut S_6, anyway?
an outer automorphism:
Consider 6 things, e.g. the numerals 123456. There are 15
non-ordered pairs of them. (Call these "duads.") Also there are 15
ways to divide the original set into three duads, e.g.
{(12}{34}{56}}. Call these "synthemes." Five synthemes can be
chosen so as to contain each duad exactly once; and there turn out
to be exactly six ways to make this choice. You can label these
six sets ABCDEF. Then a permutation of 123456 induces a
permutation of ABCDEF. A two-cycle such as (12) induces a
product of three disjoint two-cycles such as (AB)(CD)(EF), so the
map from one permutation to the other cannot be an innter
automorphism.
If you like these ideas, you will greatly enjoy the paper I got them
from; it's by H. S. M. Coxeter , titled something like "A
configuration in PG(3, 5) with 95040 self-transformations), and is
reprinted in _Twelve Geometric Essays_.
>
>Is there a neat geometrical interpretation of the outer automorphisms of
>S_6? Too bad it's not S_5, which is related to icosahedra...
>
See the above-mentioned paper. (Of course, it depends on what you
regard as "geometrical".)
Regards,
Chris Henrich
Allen Knutson
>>Aut S_6 is strictly bigger than S_6. But for n other than 6, every
>>automorphism of S_n is inner. See Huppert's Endliche Gruppen I, Kap. II,
>>Satz 5.5.
jb...@riesz.mit.edu (John C. Baez) writes:
>Far out! Maybe this is why 6 people is supposed to be the ideal number
>for a conversation? What is Aut S_6, anyway?
A pleasant group theoretical way to produce this outer automorphism was shown
to me by Danny Goldstein. Let S_5 act on its Sylow 5-groups by conjugation, of
which one can check readily that there are 6. (There are 4! 5-cycles, and each
5-gp has 4 of them.) This gives a map of S_5 into S_6, which is not the obvious
way to fit S_n into S_{n+1}, i.e. by holding one of the elements fixed: we know
this because the action must be transitive (part of Sylow's theorem).
Now we know this weird copy of S_5 in S_6. It has 6 conjugates, and S_6 acting
on THOSE by conjugation gives a weird map from S_6 to S_6 (which one can see
must be an automorphism, since the only images of S_6 are S_6, Z_2, and {e}).
>Is there a neat geometrical interpretation of the outer automorphisms of
>S_6? Too bad it's not S_5, which is related to icosahedra...
Careful: it's A_5 which is the group of rotations of an icosahedron (or
its dual, the dodecahedron). The group of rotations and reflections is
A_5 x Z_2, not S_5. Allen K.
John C. Baez
mathematical physics rather than pure mathematics... without a slight
sense of duty to reality, I could easily abandon myself to the bliss of
pondering such things as the deep inner meaning of the outer
automorphisms of S_6... which for some reason I regard as irresponsible.
Anyway, I was trying to squeeze out some simple reason for the existence
of outer automorphisms of S_6. And I really mean SIMPLE -- for god's
sake, permuting 6 objects is a pretty fundamental sort of thing!
Maybe people can have fun helping me out here. After all, I never understood
Sylow's theorem very well, and there are a bunch of you who seem to know
this stuff inside and out.
So, first I wished the result was about S_5:
>>which is related to icosahedra...
>
>Careful: it's A_5 which is the group of rotations of an icosahedron (or
>its dual, the dodecahedron). The group of rotations and reflections is
>A_5 x Z_2, not S_5. Allen K.
Yes, I was worriying abou this, which is why I carefully hedged using
the word "related." I guess the point is that
1 -> A_5 -> A_5 x Z_2 -> Z_2 -> 1
can be fit inside
1 -> SO(3) -> O(3) -> Z/2 -> 1
(i.e. one has a commutative diagram with inclusions going down), while
1 -> A_5 -> S_5 -> Z_2 -> 1
can't, since there's no nontrivial homomorphism from S_5 to A_5.
Someone, however, (who shall remain nameless just in case he made a
mistake and might be embarassed) wrote to me saying that:
-----
Seeing S_5 acting on 6 points on the
icosahedron is really easy. The 6 points are the unordered pairs of
antipodal vertices.
The stabilizer of a pair of antipodal vertices is a subgroup of S_5 of
order 20. Thus, it must be the normalizer of a Sylow 5-subgroup. This
is the correct stabilizer for the action I described earlier. And the
stabilizer completely determines which transitive permutation action
you're looking at.
Not that it matters. Any representation on 6 objects will give a map of
S_5 into S_6. But in fact there's only one transitive representation of
S_5 on 6 letters anyway.
-----
I'm pretty confused now. Does the above assume that the symmetries of
the icosahedron including reflections is S_5?? I.e., is it inconsistent
with Allen K.'s remarks?
I'm over my head...
John C. Baez
folk like myself can enjoy... although it depends on a fact I don't
know:
From: to...@math.lsa.umich.edu
Message-Id: <920520141...@supernova.math.lsa.umich.edu>
To: jb...@math.mit.edu
Subject: geometrical interpretation of Aut(S_6)
Status: R
For some reason we can't post news from here.
One answer to your query is that the outer automorphism interchanges
the points and lines of a four dimensional symplectic space
over F_2, since Sp(4,2) = S_6.
Someone else mentioned that the outer automorphism interchanges the two
classes of S_5's, and that this is realized in the sporadic group M_12 of
degree 12. Similarly, M_12 has two classes of M_11 subgroups, one which
stabilizes a point and one which is transitive; these are interchanged
by Out(M_12), and this is realized in M_24.
Tom
-------
Now, sporadic groups are one of those things I've always thought would
be fun to spend years on if I were immortal, but as it is, I know next
to nothing about them. But symplectic vector spaces are what
mathematical physicists live with day and night... not over Z_2, of
course, but some of us make enough sign errors to be working in Z_2 for
all practical purposes!
John C. Baez
Percy P was a mathematician
whose "pureness" was never denied.
But he found one day, to his sorrow,
that his theorems had all been applied!
He had taken the standard precautions;
his papers were pointedly dry!
But his own esoteric notation
had been cracked by a physicist spy!
The colloquium buzzed with the gossip;
he could offer no valid excuse.
Percy P was a traitor of traitors,
for his work was of PRACTICAL USE!
Nobody dared to defend him.
Could it be that he'd plead the crime
That his work was just then needed
to effect quantization of time?
Ignored when he joined conversations;
one would think that he poisoned the air.
And he felt on his way to the office -
a new man might be in his chair.
A committee was in operation,
working twenty four hours a day,
Deleting his name from the journals,
and throwing his reprints away.
He knew where his future was leading,
no sense in prolonging the pain;
He left with a handful of papers,
and never was heard from again.
So take heed all you mathematicians
who pretend your endeavor is pure;
Tho' your luck may hold for a decade,
in the end you can never be sure.
John C. Baez
>Okay, now lets call a pair of axes a DUAD. Please consult your
>icosahedron and contemplate a duad. A duad forms the diagonal of a
^s
>rectangle whose four corners are vertices of the icosahedron.
Hal Lillywhite
>Passed on by my friend Ellen Spertus (and slightly improved):
>Percy P was a mathematician
> whose "pureness" was never denied.
>But he found one day, to his sorrow,
> that his theorems had all been applied!
>...
Just to give credit where it probably belongs, I think this was
printed in the _Princeton Tiger_ many years ago. I can't give an
exact date (or author) but I picked it up off the bulletin board in
the BYU math department in about 1963, they credited it to the
above publication.
Bradley W. Brock
> action is transitive, that is, any axis can be mapped to any other.
> Thus we have an interesting way of thinking of A_5 as a subgroup of S_6.
>
> Exercise 5: Does this inclusion of A_5 in S_6 extend to an inclusion of
> S_5 in S_6? [I don't know the answer to this one.]
Yes, there is a transitive representation of S_5 in S_6.
>
> Now, back to synthemes! We have been talking about 5 of them, the true
> crosses. But there are 5 of a different sort, the SKEW CROSSES, which
> look like
>
> A
>
> B C
> A
>
> C B
>
> and the rotated versions thereof. And there are also 5 which I'll call
> CRUDE CROSSES which look like
>
>
> A
>
> B C
> A
>
> B C
>
> and the rotated versions thereof.
Skew crosses and crude crosses are the same under A_5.
Bradley W. Brock
> in Coxeter in one form, but it might be nicer to see it in
> terms of crosses. I have not done this, so it would be great if
> someone posted a list of the 6 pentads, the first of which is:
>
> A B C C B
>
> B B C A C B B C A C
> A A A A A
>
> C C C B B A A B B C
A B C C B
B B C A B C C B A C
A A A A A
C C B C B A A B C B
and the other four are rotations of this pentad.
Bradley W. Brock
In article <1992May21.2...@cs.rose-hulman.edu>
>
> In article <1992May21....@galois.mit.edu> jb...@riesz.mit.edu (John
C.
> Baez) writes:
> > Now, back to synthemes! We have been talking about 5 of them, the true
CROSSES.
>
> Skew crosses and crude crosses are the same under A_5.
>
> > easy this summer and wishes to pay me back for all the hard work I put
> > into keeping sci.math and sci.physics interesting (it's a dirty job, but
> > somebody has to do it!), I would be delighted to get some Mathematica
> > code that would print out 3 beautiful pictures of an icosahedron with 1)
> > a true cross, 2) a skew cross, and 3) a crude cross inscribed in it,
> > perhaps with the golden rectangles shaded in gray. I can visualize the
> > true cross quite well, but have some trouble with the other two.
No Mathematica code here, but it is not difficult to see that in the case of
the skew/crude crosses the planes containing the three golden rectangles
intersect in a line at angles of Pi/3 radians to each other.
Bradley W. Brock
>
>
> (+-G, +-1, 0),
> (+-1, 0, +-G),
> (0, +-G, +-1)
>
> are the vertices of a regular icosahedron.
Corollary Exercise: Show that the 20 points
(+-1/G, +-G, 0),
(+-G, 0, +-1/G),
(0, +-1/G, +-G)
(+-1, +-1, +-1)
are the vertices of a regular dodecahedron.
RING, DAVID WAYNE
>Exercise 5: Does this inclusion of A_5 in S_6 extend to an inclusion of
>S_5 in S_6?
I think it does. Switching your first two True Crosses (an odd permutation)
is generated by switching axes 1 and 5, 2 and 3, 4 and 6, numbering the
axes 1-5 clockwise from the top of the page with axis 6 at the center of
the pentagon. The rest of the odd permutations can be reached from there
with the known even permutations.
Dave Ring
dwr...@zeus.tamu.edu
John C. Baez
>
>Baez) writes:
>> Note that the group A_5 acts as symmetries of the set of 6 axes. This
>> action is transitive, that is, any axis can be mapped to any other.
>> Thus we have an interesting way of thinking of A_5 as a subgroup of S_6.
>>
>> Exercise 5: Does this inclusion of A_5 in S_6 extend to an inclusion of
>> S_5 in S_6? [I don't know the answer to this one.]
>
>Yes, there is a transitive representation of S_5 in S_6.
Well, that's what we've heard recently on sci.math, but I'd really like
a proof that *this* inclusion of A_5 in S_6 extends to an inclusion of
S_5, and what I'm really looking for is a *geometrical* description of
the action of S_5 on the axes of an icosahedron.
>> Now, back to synthemes! We have been talking about 5 of them, the true
>> crosses. But there are 5 of a different sort, the SKEW CROSSES, which
>> look like
>>
>> A
>>
>> B C
>> A
>>
>> C B
>>
>> and the rotated versions thereof. And there are also 5 which I'll call
>> CRUDE CROSSES which look like
>>
>>
>> A
>>
>> B C
>> A
>>
>> B C
>>
>> and the rotated versions thereof.
>
>Skew crosses and crude crosses are the same under A_5.
Gulp - whoops - thanks! Let's just call 'em all skew crosses, since
that rhymes better. 5 true, and 10 skew.
John C. Baez
>No Mathematica code here, but it is not difficult to see that in the case of
>the skew/crude crosses the planes containing the three golden rectangles
>intersect in a line at angles of Pi/3 radians to each other.
Here we have someone truly in tune with the essence of the icosahedron!
Unfortunately I was composing my essay at the computer with no
icosahedron at hand, so some of the things I hadn't verified ahead of
time were screwed up. Thanks for all the corrections and clarifications!
John C. Baez
This post improves a cancelled earlier version. It's only worth reading
if you didn't read the earlier one, or want to print out a nicer version.
In his book Twelve Geometric Essays, Coxeter has an essay with an
ominous title, roughly, "12 points in PG(3, 5) with 95040
self-transformations." In the first two pages of this essay he shows
how to see that not all automorphisms of S_6 are inner, by explicitly
constructing an outer automorphism. In what follows I recount these
first two pages relate the ideas to the icosahedron. It is very elementary, so
finite group experts may be bored to tears, but anyone who has enjoyed
the Platonic solids will be able to follow it.
First, buy, borrow or steal, or at least draw, a regular icosahedron!
The regular icosahedron is a beautiful thing. It has 20 faces, each an
equilateral triangle. Since it is dual to the dodecahedron, it has 12
vertices. (If you don't believe me, count them.) And by Euler's
formula,
F - E + V = 2,
it has 30 edges. (Again, I encourage you to count them. In what
follows you will need to grok this fact, not just believe it.)
Homework exercise 1: Prove that the regular icosahedron exists.
What? Yes, prove there is a solid with 20 faces all congruent
equilateral triangles, five meeting at each vertex! For all you know,
it might not exist - the faces might not be able to be precisely
equilateral! We wouldn't want to be studying a nonexistent object,
after all.
Okay, now think of the regular icosahedron as being centered at the
origin. Then each vertex x has an antipodal vertex -x. Draw the
line segment from x to -x; let's call it an AXIS. There are 6 axes.
If we draw the "top" of the icosahedron, the vertices look like:
*
* *
*
* *
and each axis contains one of these 6 vertices as one of its endpoints.
We will use this picture to keep track of the axes.
Okay, now let's call a pair of axes a DUAD. Please consult your
icosahedron and contemplate a duad. A duad forms the diagonals of a
rectangle whose four corners are vertices of the icosahedron.
Exercise 2: Show that this is a golden rectangle, that is, the long
side is (1 + sqrt(5))/2 as long as the short side.
Now since a duad is determined by a choice of two axes, there are 6
choose 2 duads, that is, 15 duads. We can easily use duads to count how
many symmetries the icosahedron has - rotations, that is, but not
reflections. Any symmetry of the icosahedron maps a duad to another
duad. We can map a fixed duad to any other duad in four different ways
(since a rectangle, hence a duad, has 4-fold symmetry.) If you think
about it, knowing where a fixed duad goes completely determines the
rotation of the icosahedron. Thus there are 4 x 15 = 60 symmetries.
Let us call a set of 3 duads, no two having an axis in common, a
SYNTHEME. Since there are 6 axes, 2 per duad, a syntheme complete
accounts for all the axes. We can call the 3 duads in a syntheme A, B,
and C, and draw the syntheme thus:
A
B B
A
C C
(for example). Here we are just showing which of the top 6 vertices lie
in which duad. It is infinitely more beautiful, however, to stare at
your icosahedron and visualize the 3 interlocking duads in the above
syntheme. Each duad determines a golden rectangle, and in the syntheme
above the golden rectangles are all perpendicular! Let us call this
kind of syntheme a TRUE CROSS. There are 5 true crosses, namely the
above one and 4 more obtained by rotating it, e.g.,
B
C A
A
C B
and so on. Note that exercise 2 and the contemplation of the true cross
easily allow one to solve:
Exercise 3: Let G = (1 + sqrt(5))/2. Show that the 12 points
(+-G, +-1, 0),
(+-1, 0, +-G),
(0, +-G, +-1)
are the vertices of a regular icosahedron.
Of course exercise 3 implies exercise 1, but there is an infinitely more
simple and charming solution of exercise 1. If you understand how the
icosahedron is dual to the dodecahedron, you can easily use exercise 3
to do
Exercise 4: Show that the 20 points
(+-1/G, +-G, 0),
(+-G, 0, +-1/G),
(0, +-1/G, +-G)
(+-1, +-1, +-1)
are the vertices of a regular dodecahedron.
Now, let's use the fact that there are 5 true crosses to determine the
symmetry group of the icosahedron. Any symmetry of the icosahedron (again, not
including reflections) determines a permutation of the true crosses, and
knowing the permutation allows us to reconstruct the symmetry. It
follows that the symmetry group of the icosahedron is a subgroup of the
group of permutations of 5 objects, S_5.
Which permutations can we get? Well, by rotating the pentagon as we
were doing above, we clearly get all cyclic permutations. In fact, one
can check:
Exercise 5: The symmetries of the icosahedron correspond to the *even*
permutations of the 5 true crosses.
Thus the symmetry group of the icosahedron is the group of even
permutations of 5 objects, the alternating group A_5. This of course
has 5!/2 = 60 elements.
Note that the group A_5 acts as symmetries of the set of 6 axes. This
action is transitive, that is, any axis can be mapped to any other.
Thus we have an interesting way of thinking of A_5 as a subgroup of S_6.
Exercise 6: Does this inclusion of A_5 in S_6 extend to an inclusion of
S_5 in S_6? (See the end of the article for the solution.)
Now, back to synthemes! We have been talking about 5 of them, the true
crosses. But there are 5 of a different sort, which look like
A
B C
A
C B
and the rotated versions thereof. By applying other symmetries of the
icosahedron to the above one also gets
A
B C
A
B C
and the rotated versions thereof. There are thus 10 synthemes of this
sort in all; we'll call such a syntheme a SKEW CROSS. If there is anyone
who is taking it easy this summer and wishes to pay me back for all the
hard work I put into keeping sci.math and sci.physics interesting (it's
a dirty job, but somebody has to do it!), I would be delighted to get
some Mathematica code that would print out 2 beautiful pictures of an
icosahedron with 1) a true cross, and 2) a skew cross inscribed in it,
perhaps with the golden rectangles shaded in gray. I can visualize the
true cross quite well, but have some trouble with the skew cross.
Exercise 7: Every syntheme is a true cross or a skew cross.
There are thus 15 synthemes in all. Note that the symmetries of the
icosahedron act transitively on the 10 skew crosses, giving us an
interesting embedding of A_5 in S_10.
Exercise 8: Every duad lies in 3 synthemes - one true cross and two skew
crosses.
Thus we have a nice duality: 15 duads and 15 synthemes, 3 duads in each
syntheme, and 3 synthemes containing each duad. In Coxeter's essay he
draws a picture of this setup.
There are various ways to pick 5 synthemes such that each duad lies
in exactly one of the synthemes. For example, we may choose the 5 true
crosses: each duad lies in exactly one of the true crosses. Let us call
such a choice of 5 synthemes a PENTAD.
So, one pentad consists of the 5 true crosses:
A B C C B
B B C A C B B C A C
A A A A A
C C C B B A A B B C
Another pentad consists of 5 of the skew crosses, and looks like this:
A B C C B
B B C A B C C B A C
A A A A A
C C B C B A A B C B
and there are 4 more consisting of rotations of this pentad.
Exercise 9: There are only 6 pentads!
Now any permutation of the 6 axes induces a permutation of the 6
pentads. This defines a homomomorphism from S_6 to itself. As Chris Henrich
pointed out, this is the mysterious and sublime OUTER AUTOMORPHISM of
S_6. I might as well quote his post, which is infinitely more compact
than the above ramblings:
Consider 6 things, e.g. the numerals 123456. There are 15
non-ordered pairs of them. (Call these "duads.") Also there are 15
ways to divide the original set into three duads, e.g.
{(12}{34}{56}}. Call these "synthemes." Five synthemes can be
chosen so as to contain each duad exactly once; and there turn out
to be exactly six ways to make this choice. You can label these
six sets ABCDEF. Then a permutation of 123456 induces a
permutation of ABCDEF. A two-cycle such as (12) induces a
product of three disjoint two-cycles such as (AB)(CD)(EF), so the
map from one permutation to the other cannot be an inner
automorphism.
Exercise 10: Take the embedding of A_5 in S_6 described above and
compose it with the outer automorphism of S_6. What does the image
look like as a subgroup of S_6? [I haven't worked it out, but I bet
it's the group consisting of even permutations of (123456) which leave
the element 1 alone. Here 1,2,3,4,5,6 are the six pentads and 1 is the
pentad consisting of the true crosses.]
---------------------------------------------------------------------------
Thanks go to Derek Clegg for wondering whether S_n = Aut(S_n) for all n.
Thanks go to Allen Knutson for saying the answer was yes
unless n = 6, and to Dave Sibley for some important clues on the
relation to icosahedra. Thanks go to Christopher J. Henrich (see an
earlier for summarizing the relevant ideas from Coxeter's essay.
Coxeter mentions that he could have titled his essay "Some thoughts on
the number 6," so I decided to steal that title for this post. Thanks
go to Bradley W. Brock for clarifying the nature of the skew crosses,
finding the pentads consisting of skew crosses, and creating exercise 4!
Thanks go to Charles Yeomans and Tom R. for some facts that were too
sophisticated to find their way into this article. Thanks go to
Sylvester for inventing duads and synthemes.
Exercise 6 was solved by Dave Ring:
Does this inclusion of A_5 in S_6 extend to an inclusion of
S_5 in S_6?
Yes. Switching the first two true crosses, an odd permutation:
A B
B B C A
A A
C C C B
is generated by switching axes 1 and 5, 2 and 3, 4 and 6, numbering the
axes as follows:
1
5 2
6
4 3
Note that doing this doesn't affect the other three true crosses.
The rest of the odd permutations can be reached from this permutation
Bradley W. Brock
In article <21MAY199...@zeus.tamu.edu> dwr...@zeus.tamu.edu (RING, DAVID
In an obvious way this permutation corresponds to a skew cross. More
generally, the ten odd involutions in this S_5 correspond to the ten skew
crosses.
--
Bradley W. Brock, Department of Mathematics
Rose-Hulman Institute of Technology | "In humility consider others better than
br...@nextwork.rose-hulman.edu | yourselves."-Paul of Tarsus
Keith Ramsay
>Thanks go to Derek Clegg for wondering whether S_n = Aut(S_n) for all n.
>Thanks go to Allen Knutson for saying the answer was yes
To pick at a nit, n=2 is also an exception. What is true for n<>6 is
that all the automorphisms of S_n are inner. S_2 has a non-trivial
center, so it is not isomorphic to its group of (inner) automorphisms,
which is trivial.
--
Keith Ramsay
ram...@raven.math.ubc.ca