Outer Automorphism Group

Article Index for Outer

Website Links For Outer

Information About Outer Automorphism Group

	CATEGORIES ABOUT OUTER AUTOMORPHISM GROUP
	group theory
	group automorphisms

	APPAREL
	BABY
	BEAUTY
	BOOKS
	CAR TOYS
	CELL PHONES
	DVD'S
	ELECTRONICS
	GOURMET FOOD
	GROCERIES
	HEALTH & PERSONAL
	HOME & GARDEN
	JEWELRY
	MUSIC
	MUSIC INSTRUMENTS
	OFFICE PRODUCTS
	SOFTWARE
	SPORTING GOODS
	TOOLS & HARDWARE
	TOYS
	VIDEO GAMES
	SHOPPING HOME
	MORE SHOPPING...

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 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 ''A''6; 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 Product s except that for the Suzuki-Ree Groups the graph automorphism squares to a generator of the field automorphisms. Group Parameter Out(G) ''S''''n'' ''n'' not equal to 6 Trivial ''S''6   Z2 (see below) ''A''''n'' ''n'' not equal to 6 Z2 ''A''6   Z2 × Z2(see below) Z''n'' ''n'' > 2

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 ''A''₆ has outer automorphism group of order 4, rather than 2 for the other simple alternating groups. Equivalently the symmetric group ''S''₆ is the only symmetric group with a non-trivial outer automorphism group.

To see that ''S''₆ 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 ''S''₆, we need to construct
an "unusual" subgroup of index 6 in ''S''₆, in other words one that is not one of the six obvious ''S''₅ subgroups fixing a point (which just correspond to inner automorphisms of ''S''₆). We will do this
by constructing an ''S''₅ subgroup acting ''transitively'' on the six points. A transitive action of ''S''₅ on six points comes from a subgroup of ''S''₅ of index 6, or equivalently of order 20. So we just need to find a group of order 20 that is a subgroup of ''S''₅. 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 ''S''₅. So working backwards we see that ''S''₆ has a non-trivial outer automorphism.

Another way to see that ''S''₆ has an outer automorphism is to use the fact that ''A''₆ 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 ''S''₆, 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 $\backslash sigma=\; au\_1\; au\_2$ of two different transpositions $au\_1\; ,\; au\_2$ then one obtains either a 3-cycle or a permutation of type 1^''n''−42². In particular the order of the produced elements is either two or three. On the other hand if one forms products $\backslash sigma=\backslash sigma\_1\; \backslash sigma\_2$ of Involution s $\backslash sigma\_1,\; \backslash sigma\_2$ 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 ''S''₇ is the product of two involutions of class 1¹ 2³, any permutation of class 4² in ''S''₈ is a product of involutions of class 2⁴, finally a permutation of class 2²3¹ in ''S''₇ 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 $f(\; au\_1),\; f(\; au\_2)$ under ''f'' of two different transpositions, an order divisible by 7, 4 or 6 therefore cannot occur.

For ''S''₆ 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 ''S''₆ 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.