Automorphisms of the symmetric and alternating groups

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In group theory, a branch of mathematics, the automorphisms and outer automorphisms of the symmetric groups and alternating groups are both standard examples of these automorphisms, and objects of study in their own right, particularly the exceptional outer automorphism of S6, the symmetric group on 6 elements.

Contents

[edit] Summary

n Aut(Sn) Out(Sn)
n\neq 2,6 Sn 1
n = 2 1 1
n = 6 S_6 \rtimes C_2[1] C2
n Aut(An) Out(An)
n\geq 4, n\neq 6 Sn C2
n = 1,2 1 1
n = 3 C2 C2
n = 6 S_6 \rtimes C_2 V=C_2 \times C_2

Generic case:

  • n\neq 2,6: Aut(Sn) = Sn, and thus Out(Sn) = 1.
Formally, Sn is complete and the natural map S_n \to \mbox{Aut}(S_n) is an isomorphism.
  • n\neq 2,6: Out(An) = Sn / An = C2, and the outer automorphism is conjugation by an odd permutation.
  • n\neq 2,3,6: Aut(An) = Aut(Sn) = Sn
Formally, the natural maps S_n \to \mbox{Aut}(S_n) \to \mbox{Aut}(A_n) are isomorphisms.

Exceptional cases:

  • n = 1,2: trivial:
Aut(S1) = Out(S1) = Aut(A1) = Out(A1) = 1
Aut(S2) = Out(S2) = Aut(A2) = Out(A2) = 1
  • n = 3: Aut(A3) = Out(A3) = S3 / A3 = C2
  • n = 6: Out(S6) = C2, and \mbox{Aut}(S_6)=S_6 \rtimes C_2 is a semidirect product.
  • n = 6: \mbox{Out}(A_6)=C_2 \times C_2, and \mbox{Aut}(A_6)=\mbox{Aut}(S_6)=S_6 \rtimes C_2.

[edit] Generic case

[edit] The exceptional outer automorphism of S6

Among symmetric groups, only S6 has an outer automorphism, which one can call exceptional (in analogy with exceptional Lie algebras) or exotic. In fact, Out(S6) = C2.

This was discovered by Hölder in 1895.

This also yields another outer automorphism of A6, and this is the "only" exceptional outer automorphism of a finite simple group[2]: for the infinite families of simple groups, there are formulas for the number of outer automorphisms, and the simple group of order 360, thought of as A6, would be expected to have 2 outer automorphisms, not 4. However, when A6 is viewed as \operatorname{PSL}(2,9) the outer automorphism group has the expected order. (For sporadic groups (not falling in an infinite family), the notion of exceptional outer automorphism is ill-defined, as there is no general formula.)

[edit] Construction

There are numerous constructions, listed in Janusz and Rotman[3].

Note that as an outer automorphism, it's a class of automorphisms, well-determined only up to an inner automorphism, hence there is not a natural one to write down.

One method is:

  • Construct an exotic map (embedding) S_5 \to S_6
  • S6 acts by conjugation on the 6 conjugates of this subgroup;
yielding a map S_6 \to S_X, where X is the set of conjugates. Identifying X with the numbers 1,\dots,6 (which depends on a choice of numbering of the conjugates, i.e., up to an element of S6 (an inner automorphism)) yields an outer automorphism S_6 \to S_6
  • This map is an outer automorphism, since a transposition doesn't map to a transposition, but inner automorphisms preserve cycle structure

Throughout the following, one can work with the multiplication action on cosets or the conjugation action on conjugates.

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.

[edit] Exotic map S_5 \to S_6

There is a subgroup (indeed, 6 conjugate subgroups) of S6 which are abstractly isomorphic to S5, and transitive as subgroups of S6.

[edit] Sylow 5-subgroups

Janusz and Rotman construct it thus:

  • S5 acts transitively by conjugation on its 6 Sylow 5-subgroups, yielding an embedding S_5 \to S_6 as a transitive subgroup of order 120. (The obvious map S_n \to S_{n+1} fixes a point and thus isn't transitive.)

This follows from inspection of 5-cycles: each 5-cycle generates a group of order 5 (thus a Sylow subgroup), there are 5! / 5 = 120 / 5 = 24 5-cycles, yielding 6 subgroups (as each subgroup also includes the identity), and Sn acts transitively by conjugation on cycles of a given class, hence transitively by conjugation on these subgroups.

One can also use the Sylow theorems, which imply transitivity.

[edit] Frobenius group

Another way: 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).

The Frobenius group of affine transformations of \mathbf{F}_5 (maps x \mapsto ax + b where a\neq 0) has order 20=(5-1)\cdot 5 and acts on the field with 5 elements, hence is a subgroup of S5. (Indeed, it is the normalizer of a Sylow 5-group mentioned above, thought of as the order 5 group of translations of \mathbf{F}_5.)

S5 acts transitively on the coset space, which is a set of 120 / 20 = 6 elements (or by conjugation, which yields the action above).

[edit] Other constructions

Witt found a copy of Aut(S6) in the Mathieu group M12 (a subgroup T isomorphic to S6 and an element σ that normalizes T and acts by 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.

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).

[edit] Structure of outer automorphism

On cycles, it exchanges permutations of type (12) with (12)(34)(56) (class 21 with class 23), and of type (123) with (145)(263) (class 31 with class 32).

On A6, it interchanges the 3-cycles (like (123)) with elements of class 32 (like (123)(456)).

[edit] 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.)
  2. Second, show that every automorphism (other than the above for S6) stabilizes transpositions.

This 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:

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.

[edit] No other outer automorphisms of S6

S6 has exactly one (class) of outer automorphisms: Out(S6) = C2.

To see this, observe that there are only two conjugacy classes of S6 of order 15: the transpositions and those of class 23. Thus Aut(S6) acts on these two conjugacy classes (and the outer automorphism above interchanges these conjugacy classes), and an index 2 subgroup stabilizes the transpositions. But an automorphism that stabilizes the transpositions is inner, so the inner automorphisms are an index 2 subgroup of Aut(S6), so Out(S6) = C2.

More pithily: an automorphism that stabilizes transpositions is inner, and there are only two conjugacy classes of order 15 (transpositions and triple transpositions), hence the outer automorphism group is at most order 2.

[edit] Small n

[edit] Symmetric

For n = 2, S_2=C_2=\mathbf{Z}/2 and the automorphism group is trivial (obviously, but more formally because \mbox{Aut}(\mathbf{Z}/2)=\mbox{GL}(1,\mathbf{Z}/2)=\mathbf{Z}/2^*=1). The inner automorphism group is thus also trivial (also because S2 is abelian).

[edit] Alternating

For n = 1 and 2, A1 = A2 = 1 is trivial, so the automorphism group is also trivial. For n = 3, A_3=C_3=\mathbf{Z}/3 is abelian (and cyclic): the automorphism group is \mbox{GL}(1,\mathbf{Z}/3)=\mathbf{Z}/3^*=C_2, and the inner automorphism group is trivial (because it is abelian).

[edit] References

  1. ^ http://links.jstor.org/sici?sici=0002-9890(198206%2F07)89%3A6%3C407%3AOAO%3E2.0.CO%3B2-L
  2. ^ ATLAS p. xvi
  3. ^ http://links.jstor.org/sici?sici=0002-9890(198206%2F07)89%3A6%3C407%3AOAO%3E2.0.CO%3B2-L