## 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.
• XP : (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
• PQ : (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.
• QY : (5,2). Y = {ZQ | ∀pP, ∃!qZ, pq}. 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 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,
• 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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