Outer automorphisms of S6
A famous fact of finite group theory, is that 6 is the only natural number n
such that the symmetric group Sn has outer automorphisms, making the group
Out(Sn) = Aut(Sn)/Inn(Sn) non-trivial.
In other words (though it is a different reason for n=2), such that there exists an effective action of
Sn on a set of n elements, other than the natural one.
So the exceptionally non-trivial one is Out(S6) with 2 elements.
A problem is to describe these outer automorphisms of S6 in an understandable way.
All I could find elsewhere on the topic seemed to me difficult to follow. But here is an explanation.
Rephrasing the problem : from a set X of 6 elements without any structure, construct another set Y
of 6 elements without any invariant bijection to X.
This construction is in 3 steps, forming a symmetric chain of 4 sets
X−P−Q−Y where each of P and Q is a set of 15 elements, and
each link represents a uniform binary relation (we may call incidence) between 2 neighbors.
Each incidence relation is marked by 2 numbers : the number of elements of one set related to each
element of the next, and conversely. The ratio of these numbers equals that of the elements of those sets.
- X−P : (2,5). P is the set of pairs in X, so each pair contains
2 elements, while each element is contained in 5 pairs. 2/5 = 6/15
- P−Q : (3,3). Q is the set of partitions of X into 3 pairs. Each pair is contained
in 3 such partitions. 3/3 = 15/15.
- Q−Y : (5,2). Y = {Z⊂Q | ∀p∈P, ∃!q∈Z, p∈q}.
5/2 = 15/6.
In words, each element of Y is a set of 5 partitions of X into pairs,
such that each pair is contained in exactly one of these partitions.
The projective line on the field F9
To understand that this Y indeed has 6 elements, let us choose a partition of
X into 2 sets A, B of 3 elements each.
Such a partition defines a bijection between
Y and the set of all 6 bijections between A and B as follows.
Between A = {a1,a2,a3} and B = {b1,b2,b3}, the bijection (a1−b1, a2−b2, a3−b3)
matches the following set of 5 partitions containing each pair exactly once:
{{a1,b1},{a2,a3},{b2,b3}}
{{a2,b2},{a1,a3},{b1,b3}}
{{a3,b3},{a1,a2},{b1,b2}}
{{a1,b2},{a2,b3},{a3,b1}}
{{a1,b3},{a2,b1},{a3,b2}}
The similar property with reversed relations is obvious : for each x∈X and
each q∈Q there exists exactly one pair containing x and contained in q.
The above directly gives some aspects of how permutations on X correspond to permutations on Y, as
a pair defines a transposition, while a partition in pairs defines the permutation having this partition
as its set of orbits. By the symmetry of roles between P and Q,
- The transposition of X switching b1 and b2, permutes Y by turning
any even bijection between A and B into an odd bijection, and conversely ;
- The permutation of X with 3 cyles {a1,b1}, {a2,b2}, {a3,b3} induces the transposition of Y
represented by switching the bijection (a1−b2, a2−b3, a3−b1) with the bijection (a1−b3, a2−b1, a3−b2), leaving
all others fixed.
I could not find a very clear picture of (thus did not actually verify) the isomorphism between
the alternating group A6 and PSL(2, 9) as I read existed, but assuming it exists, I found a few hints of
how it looks like. Here they are.
The set of all 10 partitions of X into 2 pairs of 3, forms a projective line
over the field with 9 elements F9 = {a+ib | a,b∈F3} where F3 is the field of 3 elements
{0,1,-1} where 1+1 = -1, and i2=-1. In the corresponding
2-dimensional vector space, the pair defined as quotient of the set of 8 non-zero values of
determinants (which forms a cyclical orbit by multiplication by nonzero
elements of F9) by the action of the subgroup with 4 elements {1,i,-1,-i}
(which are the squares of the 4 others), naturally matches the pair (X,Y).
5-Sylows
Usual accounts of Out(S6) speak about 5-Sylows, so we must say
something about it. For any choice of element x0 of X,
the set Y naturally corresponds to the set of possible 5-Sylows acting on
Z = X\{x0}of X.
Giving a 5-Sylow amounts to partitioning the set of edges of Z into two 5-cycles,
each of which is a way of visualizing Z as the set of vertices of a
regular pentagon while x0 is put at the center.
Then the 5 elements of Q for this choice, one for each pair with
x0, are given by
putting the rest of X into the pairs orthogonal to this one.
This method also works to find lists of pairings admitting each pair once, for
even numbers of elements other than 6 ; it just does not result in outer automorphisms then.
References
A page by John Baez
How automorphims act on permutations
another page
A pdf text
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