## Outer automorphisms of S_{6}

A famous fact of finite group theory, is that 6 is the only natural number *n*
such that the symmetric group S_{n} has outer automorphisms, making the group
Out(S_{n}) = Aut(S_{n})/Inn(S_{n}) non-trivial.
In other words (though it is a different reason for *n*=2), such that there exists an effective action of
S_{n} on a set of *n* elements, other than the natural one.
So the exceptionally non-trivial one is Out(S_{6}) with 2 elements.
A problem is to describe these outer automorphisms of S_{6} 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 F_{9}

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.

The the set of all 10 partitions of *X* into 2 pairs of 3, forms a projective line
over the field with 9 elements F_{9} = {a+ib | a,b∈F_{3}} where i^{2}=-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 F_{9}) 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(S_{6}) speak about 5-Sylows, so we must say
something about it. For any choice of element *x*_{0} of *X*,
the set *Y* naturally corresponds to the set of possible 5-Sylows acting on
*Z* = *X*\{*x*_{0}}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 *x*_{0} is put at the center.
Then the 5 elements of *Q* for this choice, one for each pair with
*x*_{0}, 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.

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