Outer automorphism group
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In mathematics, the outer automorphism group of a group G is the quotient of the automorphism group Aut(G) by its inner automorphism group Inn(G). The outer automorphism group is usually denoted Out(G). If Out(G) is trivial and G has a trivial center, then G is said to be complete.
Note that the elements of Out(G) are cosets of automorphisms of G, and not themselves automorphisms. This is an instance of the fact that quotients of groups are not in general subgroups. However, the elements of Aut(G) which are not inner automorphisms are usually called outer automorphisms; they are the elements of the non-trivial cosets in Out(G).
It was conjectured by Otto Schreier that Out(G) is always a solvable group when G is a finite simple group. This result is now known to be true as a corollary of the classification of finite simple groups, although no simpler proof is known.
This group is important in the topology of surfaces because there is a happy connection: the extended mapping class group of the surface is the Out of its fundamental group.
[edit] Out(G) for some finite groups
For the outer automorphism groups of all finite simple groups see the list of finite simple groups. Sporadic simple groups and alternating groups (other than the alternating group A6; see below) all have outer automorphism groups of order 1 or 2. The outer automorphism group of a finite simple group of Lie type is an extension of a group of "diagonal automorphisms" (cyclic except for Dn(q) when it has order 4), a group of "field automorphisms" (always cyclic), and a group of "graph automorphisms" (of order 1 or 2 except for D4(q) when it is the symmetric group on 3 points). These extensions are semidirect products except that for the Suzuki-Ree groups the graph automorphism squares to a generator of the field automorphisms.
| Group | Parameter | Out(G) |
|---|---|---|
| Sn | n not equal to 6 | trivial |
| S6 | Z2 (see below) | |
| An | n not equal to 6 | Z2 |
| A6 | Z2 × Z2(see below) | |
| Zn | n > 2 | Zn×, φ(n) =  elements |
| Zpn | p prime, n > 1 | GLn(p ), (pn - 1)(pn - p )(pn - p2) … (pn - pn-1) elements |
| Mn | n = 11, 23, 24 | trivial |
| Mn | n = 12, 22 | Z2 |
| PSL2(p) | p > 3 prime | Z2 |
| PSL2(2n) | n > 1 | Zn |
| PSL3(4) = M21 | Dih6 | |
| Con | n = 1, 2, 3 | trivial |
[edit] The outer automorphisms of the symmetric groups
The outer automorphism group of a finite simple group in some infinite family of finite simple groups can almost always be given by a uniform formula that works for all elements of the family. There is just one exception to this: the alternating group A6 has outer automorphism group of order 4, rather than 2 for the other simple alternating groups. Equivalently the symmetric group S6 is the only symmetric group with a non-trivial outer automorphism group.
To see that S6 has an outer automorphism, recall that homomorphisms from a group G to a symmetric group Sn are essentially the same as actions of G on a set of n elements, and the subgroup fixing a point is then a subgroup of index at most n in G. Conversely if we have a subgroup of index n in G, the action on the cosets gives a transitive action of G on n points, and therefore a homomorphism to Sn.
So to construct an outer automorphism of S6, we need to construct an "unusual" subgroup of index 6 in S6, in other words one that is not one of the six obvious S5 subgroups fixing a point (which just correspond to inner automorphisms of S6). We will do this by constructing an S5 subgroup acting transitively on the six points. A transitive action of S5 on six points comes from a subgroup of S5 of index 6, or equivalently of order 20. So we just need to find a group of order 20 that is a subgroup of S5. But this is easy: we just take the Frobenius group of order 20; this is the group of all permutations of the finite field of five elements of the form ax + b for a nonzero, and so is a subgroup of S5. So working backwards we see that S6 has a non-trivial outer automorphism.
Another way to see that S6 has an outer automorphism is to use the fact that A6 is isomorphic to PSL2(9), which has an outer automorphism group of order 4 (though there seems to be no really easy way to see this isomorphism).
To see that none of the other symmetric groups have outer automorphisms, it is easiest to proceed in two steps. First show that any automorphism that preserves the conjugacy class of transpositions is an inner automorphism. Then show that 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. To prove that an automorphism of the symmetric group Sn for n > 6 must preserve the class of transpositions one can also proceed as follows. If one forms the products σ = τ1τ2 of two different transpositions τ1,τ2 then one obtains either a 3-cycle or a permutation of type 1n−422. In particular the order of the produced elements is either two or three. On the other hand if one forms products σ = σ1σ2 of involutions σ1,σ2 each consisting of k ≥ 2 2-cycles it may happen (for n ≥ 7) that the product contains either
- a 7-cycle
- two 4-cycles
- a 2-cycle and a 3-cycle
Note here that any 7-cycle in S7 is the product of two involutions of class 11 23, any permutation of class 42 in S8 is a product of involutions of class 24, finally a permutation of class 2231 in S7 is a product of two involutions of class 13 22 (for larger k resp. larger n compose these permutations with redundant 2-cycles or fixed points acting on the complement of a 7-element subset or 8-element subset such that they cancel out in the product). Now one arrives at a contradiction because the automorphism f must preserve the order (which is either two or three) of elements given as the product of the images f(τ1),f(τ2) under f of two different transpositions, an order divisible by 7, 4 or 6 therefore cannot occur.
For S6 the class of cycle shape 23 happens to have the same number of elements (15) as the class of transpositions (of cycle shape 142), and in fact the non-trivial outer automorphisms exchange these two conjugacy classes. But a slight variation of this argument shows that S6 has an outer automorphism group of order at most 2 (and therefore exactly 2 as it has at least one non-trivial outer automorphism).
The full automorphism group of A6 appears naturally as a maximal subgroup of the Mathieu group M12 in 2 ways, as either a subgroup fixing a division of the 12 points into a pair of 6-element sets, or as a subgroup fixing a subset of 2 points.
