This blog post is about why the number 6 is my favourite number – because the symmetric group on six elements is the only finite permutation group that contains an outer automorphism. In the post we will first look at why there are no other symmetric groups with outer automorphisms and then we will define an outer automorphism of .

## Inner and Outer Automorphisms

If is any group, then we can construct a new group consisting of all bijective group homomorphism . This set forms a group under function composition. There is a natural map from to . A group element is taken to the function given by . Thus is conjugation by .

The kernel of is called the center of and is denoted by . That is is the subgroup consisting of group elements that commute with every other group element. The image of this map is denoted by . An automorphism which is in is called an inner automorphism. One can check that the subgroup of inner automorphisms is a normal subgroup. Indeed if we have a group element and an automorphism , then . Thus we can form the quotient which we will call the outer automorphisms of and denote by .

Thus the inner automorphisms of are those which come from in the form of conjugation. The outer automorphisms of are equivalence classes of all the automorphisms of . Two automorphisms are equivalent in if there is an inner automorphism such that .

## The Symmetric Group

For a natural number , will denote the group of permutations of the set . Any permutation can be written as product of disjoint cycles which we will call its cycle type. For instance, if and , , , and , then can be written in cycle notation as . Cycle notation is very useful for describing conjugation in . If we have two permutations and we know the cycle notation for , then the cycle notation for can be derived by applying to every entry in the cycle notation of . For instance, with , we have

.

To see why this result about conjugation is true, note the following. If for some , then

.

Thus two permutations are conjugate to one another if and only if and have the same cycle type (ie they have the same number of cycles of any given size). This implies that for , the centre of must be trivial as for any non-identity , there is at least one other element of the same cycle type. Hence the map from to is injective for .

## Counting Conjugacy Classes

Before constructing an outer automorphism for , we will see why such a map is special. In particular we will learn that consists entirely of inner automorphism for . The argument is based off studying the size of different conjugacy classes in . This is important because if is an automorphism of and are two permutations, then is conjugate to if and only if is conjugate to .

In we know that two elements are conjugate if and only if they have the same cycle type. We can use this to count the size of different conjugacy classes of . In particular the number of permutations with cycles of size for is

.

The transpositions for generate and form a conjugacy class which we call . For any automorphism , the set must be a conjugacy class that is also of size . Since automorphism preserve order, the elements of must all be of order two. Thus the cycle type of the conjugacy class must contain only cycles of size two or size one. Also there must be an odd number of two cycles. If there was an even number of two cycles, then we would have that is contained in the subgroup . However this would imply that which contradicts the fact that is a bijection. Thus the following lemma implies that when .

**Lemma**: Fix and an odd number such that and . Let denote the number of permutations which have cycles of size two and cycles of size one, then .

**Proof**: This claim can be manually checked for . For , assume in order to get a contradiction that . Note that is equal to . Thus we have and hence

.

Now if , then is a product of at least consecutive integers and hence divisible by (see here). Thus

and hence is divisible by an odd number other than one (either or ). Thus cannot be a power of , contradicting the above equation.

When , the above equation becomes:

.

Thus if is odd, is an odd number other than that divides 4, a contradiction. If is even, then since , is an odd number other than that divides , which is again a contradiction. Thus we cannot have any case when .

## does not have an outer automorphism if

With the above lemma we can now prove that all the automorphism of for are inner. First note that the only automorphism of is the identity which is an inner automorphism. When , the map from to is injective. Thus there are exactly inner automorphisms of . We will show that for there are exactly automorphisms. Hence there can be no outer automorphisms.

Let be an automorphism of . The symmetric group is also generated by the transpositions (for ). By the previous lemma we know that if , then each must be a transposition for some such that . As the collection generates , the automorphism is determined by the transpositions . We will now show that there are exactly choices for these transpositions.

Since , we know that must be a three cycle. Thus one of must be equal to one of and the other must be distinct. By relabeling if necessary, we will require that and . Thus there are choices for , choices for and choices for . This gives us choices for the first two transpositions. Continuing in this manner, we note that and , we get that one of is equal to one of and the other one is distinct from , thus giving us choices for the third transposition. In general we will have that is a three cycle and is a pair of disjoint two cycles if . Thus one can show inductively that there are choices for the transposition . Hence there are choices for all the transpositions and hence choices for . Thus, if , all automorphisms of are inner.

## The outer automorphism of .

The key part of showing that didn’t have any outer automorphisms when was showing that any automorphism must map transpositions to transpositions. Thus to find the outer automorphism of we will want to find an automorphism that does not preserve cycle type. In particular we will find one which maps transpositions to the cycle type . There are such triple transpositions which is exactly – the number of single transpositions.

The outer automorphism, , will be determined by the values it takes on and (where again ). Each of these transpositions will get to a different triple transposition . To extend to the rest of it suffices to check the Coxeter relations for , and .

The relation will hold for any triple transposition . The other relations are a bit trickier. If we have two triple transpositions and , then will be a product of two three cycles if none of the transpositions of is a transposition of . If and share exactly one transposition, then is a product of two two cycles. If and share two or more transpositions they must actually be equal and hence is the identity.

With this in mind, one can verify that following assignment extends to a valid group homomorphism from to :

,

,

,

,

.

We now just need to justify that the group homomorphism is an automorphism. Since is a finite group, is a bijection if and only if is an injection. Recall that only normal subgroups of are the trivial subgroup, and . Thus to prove injectivity of , it suffices to show that for some . Indeed as required. Thus we have our outer automorphism!

This isn’t the most satisfying way of defining the outer automorphism as it requires using the generators of the symmetric group. There are many more intrinsic ways to define this outer automorphism which will hopefully be the topic of a later blog post!

## References

The proof I give that for , the group has no outer automorphisms is based off the one given in this paper On the Completeness of Symmetric Group.

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