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 XPQY 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. 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:


The similar property with reversed relations is obvious : for each xX and each qQ 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, 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).


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.


A page by John Baez
How automorphims act on permutations
another page
A pdf text
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