Outer automorphism group

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).

The Schreier conjecture asserts 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.

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 D_n(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)	$|{\mbox{Out}}(G)|$
Z	infinite cyclic	Z2	2
Sn	n not equal to 6	trivial	1
S6		Z2 (see below)	2
An	n not equal to 6	Z2	2
A6		Z2 × Z2(see below)	4
Zn	n > 2	Zn×	φ(n) = $n\prod _{p|n}\left(1-{\frac {1}{p}}\right)$ elements
Zpn	p prime, n > 1	GLn(p )	(pn - 1)(pn - p )(pn - p2) … (pn - pn-1) elements
Mn	n = 11, 23, 24	trivial	1
Mn	n = 12, 22	Z2	2
PSL2(p)	p > 3 prime	Z2	2
PSL2(2n)	n > 1	Zn	n
PSL3(4) = M21		Dih6	(6? 12?)
Con	n = 1, 2, 3	trivial	1

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.

Outer automorphisms of S₆

To see that S₆ has an outer automorphism, recall that homomorphisms from a group G to a symmetric group S_n 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 S_n.

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).

No outer automorphisms for other symmetric groups

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 S_n for n > 6 must preserve the class of transpositions one can also proceed as follows. If one forms the products $\sigma =\tau _{1}\tau _{2}$ of two different transpositions $\tau _{1},\tau _{2}$ 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 $\sigma =\sigma _{1}\sigma _{2}$ of involutions $\sigma _{1},\sigma _{2}$ 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 $f(\tau _{1}),f(\tau _{2})$ under f of two different transpositions, an order divisible by 7, 4 or 6 therefore cannot occur.

The groups Out(S₆) and Aut(S₆)

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.

Outer automorphism groups of complex Lie groups

Let G now be a connected reductive group over an algebraically closed field. Then any two Borel subgroups are conjugate by an inner automorphism, so to study outer automorphisms it suffices to consider automorphisms that fix a given Borel subgroup. Associated to the Borel subgroup is a set of simple roots, and the outer automorphism may permute them, while preserving the structure of the associated Dynkin diagram. In this way one may identify the automorphism group of the Dynkin diagram of G with a subgroup of Out(G).

Puns

The term "Outer automorphism" lends itself to puns: the term outermorphism is sometimes used for "outer automorphism", and a particular geometry on which ${\mbox{Out}}(F_{n})$ acts is called outer space.

External links

• ATLAS of Finite Group Representations-V3

(contains a lot of information on various classes of finite groups (in particular sporadic simple groups), including the order of Out(G) for each group listed.

See also

• Mapping class group
• Out(F_n)