The Outer Automorphism of S6

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 S_6.

Inner and Outer Automorphisms

If G is any group, then we can construct a new group \text{Aut}(G) consisting of all bijective group homomorphism \phi : G \to G. This set forms a group under function composition. There is a natural map from G to \text{Aut}(G). A group element g \in G is taken to the function \gamma_g : G \to G given by \gamma_g(h) = ghg^{-1}. Thus \gamma_g is conjugation by g.

The kernel of g \mapsto \gamma_g is called the center of G and is denoted by Z(G). That is Z(G) is the subgroup consisting of group elements that commute with every other group element. The image of this map is denoted by \text{Inn}(G) \subseteq \text{Aut}(G). An automorphism which is in \text{Inn}(G) 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 g \in G and an automorphism \phi \in \text{Aut}(G), then \phi \circ \gamma_g \circ \phi^{-1} = \gamma_{\phi(g)}. Thus we can form the quotient \text{Aut}(G)/\text{Inn}(G) which we will call the outer automorphisms of G and denote by \text{Out}(G).

Thus the inner automorphisms of G are those which come from G in the form of conjugation. The outer automorphisms of G are equivalence classes of all the automorphisms of G. Two automorphisms \phi, \phi' are equivalent in \text{Out}(G) if there is an inner automorphism \gamma_g such that \gamma_g \circ \phi = \phi'.

The Symmetric Group

For a natural number n \in \mathbb{N}, S_n will denote the group of permutations of the set \{1,2,\ldots,n\}. Any permutation \sigma \in S_n can be written as product of disjoint cycles which we will call its cycle type. For instance, if n=5 and \sigma(1)=3, \sigma(2)=5, \sigma(3)=4, \sigma(4)=1 and \sigma(5)=2, then \sigma can be written in cycle notation as (1,3,4)(2,5). Cycle notation is very useful for describing conjugation in S_n. If we have two permutations \sigma, \tau \in S_n and we know the cycle notation for \sigma, then the cycle notation for \tau \sigma \tau^{-1} can be derived by applying \tau to every entry in the cycle notation of \sigma. For instance, with \sigma = (1,3,4)(2,5), we have

\tau(1,3,4)(2,5)\tau^{-1} = (\tau(1),\tau(3),\tau(4))(\tau(2),\tau(5)).

To see why this result about conjugation is true, note the following. If \sigma(i)=j for some i,j \in \{1,2,\ldots,n\}, then


Thus two permutations \sigma, \sigma' are conjugate to one another if and only if \sigma and \sigma' have the same cycle type (ie they have the same number of cycles of any given size). This implies that for n \ge 3, the centre of S_n must be trivial as for any non-identity \sigma \in S_n, there is at least one other element of the same cycle type. Hence the map \sigma \mapsto \gamma_\sigma from S_n to \text{Aut}(S_n) is injective for n \ge 3.

Counting Conjugacy Classes

Before constructing an outer automorphism for S_6, we will see why such a map is special. In particular we will learn that \text{Aut}(S_n) consists entirely of inner automorphism for n \neq 6. The argument is based off studying the size of different conjugacy classes in S_n. This is important because if \phi is an automorphism of S_n and \sigma,\sigma' \in S_n are two permutations, then \sigma is conjugate to \sigma' if and only if \phi(\sigma) is conjugate to \phi(\sigma').

In S_n 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 S_n. In particular the number of permutations with n_k cycles of size k for k = 1,2,\ldots, n is

\displaystyle \frac{\displaystyle n!}{\displaystyle \prod\limits_{k=1}^n n_k! k^{n_k}}.

The transpositions (j,k) for 1 \le j < k\le n generate S_n and form a conjugacy class which we call T. For any automorphism \phi \in \text{Aut}(S_n) , the set \phi(T) must be a conjugacy class that is also of size | T | = n(n-1)/2. Since automorphism preserve order, the elements of \phi(T) must all be of order two. Thus the cycle type of the conjugacy class \phi(T) 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 \phi(T) is contained in the subgroup A_n \subseteq S_n. However this would imply that \phi(S_n) \subseteq \phi(A_n) which contradicts the fact that \phi : S_n \rightarrow S_n is a bijection. Thus the following lemma implies that \phi(T)=T when n \neq 6.

Lemma: Fix n \neq 6 and an odd number k such that k > 1 and 2k \le n. Let m_k denote the number of permutations \sigma \in S_n which have k cycles of size two and n-2k cycles of size one, then m_k \neq n(n-1)/2.

Proof: This claim can be manually checked for n < 6. For n > 6, assume in order to get a contradiction that m_k = \frac{n(n-1)}{2}. Note that m_k is equal to \frac{n!}{k! 2^k(n-2k)! 1^{n-2k}}=\frac{n!}{k! (n-2k)! 2^k}. Thus we have \frac{n(n-1)}{2} = \frac{n!}{k!(n-2k)!2^k} and hence

2^{k-1} = \frac{(n-2)(n-3)\ldots(n-2k+1)}{k!}.

Now if k \ge 5, then (n-4)(n-5)\ldots(n-2k+1) is a product of at least k consecutive integers and hence divisible by k! (see here). Thus

2^{k-1} = \frac{(n-2)(n-3)\ldots(n-2k+1)}{k!} = (n-2)(n-3)\frac{(n-4)(n-5)\ldots(n-2k+1)}{k!}

and hence \frac{(n-2)(n-3)\ldots(n-2k+1)}{k!} is divisible by an odd number other than one (either n-2 or n-3). Thus \frac{(n-2)(n-3)\ldots(n-2k+1)}{k!} cannot be a power of 2, contradicting the above equation.

When k = 3, the above equation becomes:

4= \frac{(n-2)(n-3)(n-4)(n-5)}{3!} = (n-2)\frac{(n-3)(n-4)(n-5)}{3!} = \frac{(n-2)(n-3)(n-4)}{3!}(n-5).

Thus if n is odd, n-2 is an odd number other than 1 that divides 4, a contradiction. If n is even, then since n > 6, n-5 is an odd number other than 1 that divides 4, which is again a contradiction. Thus we cannot have any case when m_k = \frac{n(n-1)}{2}.

S_n does not have an outer automorphism if n \neq 6

With the above lemma we can now prove that all the automorphism of S_n for n \neq 6 are inner. First note that the only automorphism of S_2 = \{ \text{id}, (12)\} is the identity which is an inner automorphism. When n \ge 3, the map \sigma \mapsto \gamma_\sigma from S_n to \text{Aut}(S_n) is injective. Thus there are exactly n! inner automorphisms of S_n. We will show that for n \neq 6 there are exactly n! automorphisms. Hence there can be no outer automorphisms.

Let \phi be an automorphism of S_n. The symmetric group S_n is also generated by the n-1 transpositions \sigma_i = (i,i+1) (for i =1, \ldots, n-1). By the previous lemma we know that if n \neq 6, then each \phi(\sigma_i) must be a transposition (j_i,k_i) for some j_i, k_i \in \{1,\ldots, n\} such that j_i \neq k_i. As the collection \sigma_1,\ldots,\sigma_{n-1} generates S_n, the automorphism \phi is determined by the transpositions \phi(\sigma_i)= (j_i,k_i). We will now show that there are exactly n! choices for these transpositions.

Since \sigma_1\sigma_{2} = (1,2,3), we know that \phi(\sigma_1)\phi(\sigma_2) = (j_1,k_1)(j_2,k_2) must be a three cycle. Thus one of j_2,k_2 must be equal to one of j_1,k_1 and the other must be distinct. By relabeling if necessary, we will require that k_1=j_2 and j_1 \neq k_2. Thus there are n choices for j_1, n-1 choices for k_1=j_2 and n-2 choices for k_2. This gives us n(n-1)(n-2) choices for the first two transpositions. Continuing in this manner, we note that \sigma_1\sigma_3 = (1,2)(3,4) and \sigma_2\sigma_3 = (2,3,4), we get that one of j_3,k_3 is equal to one of k_2 and the other one is distinct from j_1,k_1=j_2,k_2, thus giving us n-3 choices for the third transposition. In general we will have that \sigma_{i-1}\sigma_i is a three cycle and \sigma_l\sigma_i is a pair of disjoint two cycles if l < i-1. Thus one can show inductively that there are n-i choices for the transposition \phi(\sigma_i). Hence there are n(n-1)(n-2)\ldots 1 = n! choices for all the transpositions \phi(\sigma_i) and hence n! choices for \phi. Thus, if n \neq 6, all automorphisms of S_n are inner.

The outer automorphism of S_6.

The key part of showing that S_n didn’t have any outer automorphisms when n \neq 6 was showing that any automorphism must map transpositions to transpositions. Thus to find the outer automorphism of S_6 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 (a,b)(c,d)(e,f). There are \frac{6!}{3!2^3} = 15 such triple transpositions which is exactly \frac{6\cdot 5}{2} – the number of single transpositions.

The outer automorphism, \psi, will be determined by the values it takes on \sigma_1,\sigma_2,\sigma_3,\sigma_4 and \sigma_5 (where again \sigma_i = (i,i+1)). Each of these transpositions will get to a different triple transposition (a,b)(c,d)(e,f). To extend \psi to the rest of S_6 it suffices to check the Coxeter relations \psi(\sigma_k)^2 = (\psi(\sigma_i)\psi(\sigma_{i+1}))^3 = (\psi(\sigma_i)\psi(\sigma_j))^2=\text{id} for k = 1,2,3,4,5, i =1,2,3,4 and j \ge i+2.

The relation \psi(\sigma_k)^2=\text{id} will hold for any triple transposition \psi(\sigma_k). The other relations are a bit trickier. If we have two triple transpositions \tau = (a,b)(c,d)(e,f) and \tau' = (a',b')(c',d')(e',f'), then \tau\tau' will be a product of two three cycles if none of the transpositions of \tau is a transposition of \tau'. If \tau and \tau' share exactly one transposition, then \tau\tau' is a product of two two cycles. If \tau and \tau' share two or more transpositions they must actually be equal and hence \tau\tau' is the identity.

With this in mind, one can verify that following assignment extends to a valid group homomorphism from S_6 to S_6:

\psi(\sigma_1) = (1,2)(3,4)(5,6),
\psi(\sigma_2) = (1,3)(2,5)(4,6),
\psi(\sigma_3) = (1,5)(2,6)(3,4),
\psi(\sigma_4) = (1,3)(2,4)(5,6),
\psi(\sigma_5) = (1,6)(2,5)(3,4).

We now just need to justify that the group homomorphism \psi is an automorphism. Since S_6 is a finite group, \psi is a bijection if and only if \psi is an injection. Recall that only normal subgroups of S_6 are the trivial subgroup, A_6 and S_6. Thus to prove injectivity of \psi, it suffices to show that \psi(\tau) \neq \text{id} for some \tau \in A_6. Indeed \psi(\sigma_1\sigma_3) = (1,6)(2,5) \neq \text{id} 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!


The proof I give that for n \neq 6, the group S_n has no outer automorphisms is based off the one given in this paper On the Completeness of Symmetric Group.

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