Automorphisms of The Symmetric and Alternating Groups - No Other Outer Automorphisms

No Other Outer Automorphisms

To see that none of the other symmetric groups have outer automorphisms, it is easiest to proceed in two steps:

1. First, show that any automorphism that preserves the conjugacy class of transpositions is an inner automorphism. (This also shows that the outer automorphism of S₆ is unique; see below.) Note that an automorphism must send each conjugacy class (characterized by the cyclic structure that its elements share) to a (possibly different) conjugacy class.
2. Second, show that every automorphism (other than the above for S₆) stabilizes the class of transpositions.

The latter can be shown in two ways:

• For every symmetric group other than S₆, there is no other conjugacy class of elements of order 2 with the same number of elements as the class of transpositions.
• Or as follows:

Each permutation of order two (called an involution) is a product of k>0 disjoint transpositions, so it has cyclic structure 2k1n-2k. What's special about the class of transpositions (k=1)?

If one forms the product of two different transpositions τ₁ and τ₂, then one always obtains either a 3-cycle or a permutation of type 221n−4, so the order of the produced element is either 2 or 3. On the other hand if one forms a product of two involutions σ₁, σ₂> that have type k>1, sometimes it happens that the product contains either

• two 2-cycles and a 3-cycle (for k=2 and n ≥ 7)
• a 7-cycle (for k=3 and n ≥ 7)
• two 4-cycles (for k=4 and n ≥ 8)

(for larger k, add to the permutations σ₁, σ₂ of the last example redundant 2-cycles that cancel each other). Now one arrives at a contradiction, because if the class of transpositions is sent via the automorphism f to a class of involutions that has k>1, then there exist two transpositions τ₁, τ₂ such that f(τ₁τ₂)=f(τ₁)f(τ₂) has order 6, 7 or 4, but we know that τ₁τ₂ has order 2 or 3.