1.6 The Symmetric and Alternating Groups

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In this section we present some of the simpler properties of the symmetric

and alternating groups.

Recall that, by definition, Sn is the group of permutations of the set {1, 2,

. . . , n}. Let i1, i2, . . . , ik be distinct elements of {1, 2, . . . , n} and define

_ := (i1 i2 · · · ik) 2 Sn to be the permutation satisfying _(i1) = i2, _(i2) =

i3, . . . , _(ik) = i1, _(i) = i for all i 62 {i1, i2, · · · , ik}. We call _ a cycle in

Sn. Two cycles in Sn are said to be disjoint if the sets of elements that they

permute nontrivially are disjoint. Thus the cycles

(2 4 7) and (1 3 6 5) 2 Sn

are disjoint. One has the following:

Proposition 1.6.1 If _ 2 Sn, then _ can be expressed as the product of

disjoint cycles. This product is unique up to the order of the factors in the

product.

A transposition in Sn is simply a cycle of the form (a b), a 6= b. That

any permutation in Sn is a product of transpositions is easy; just note the

factorization for cycles:

(i1 i2 · · · ik) = (i1 ik)(i1 ik−1) · · · (i1 i2).

Let V be a vector space over the field of (say) rational numbers, and let

(v1, v2, . . . , vn) _ V be an ordered basis. Let G act on the set {1, 2, . . . , n}

and define _ : G ! GL(V ) by

_ 7! (vi 7! v_(i)), i = 1, 2, . . . , n.

One easily checks that the kernel of this homomorphism is precisely the

same as the kernel of the induced map G ! Sn. In particular, if G = Sn

the homomorphism _ : Sn ! GL(V ) is injective. Note that the image

_(i j) of the transposition (i, j) is simply the identity matrix with rows i

and j switched. As a result, it follows that det(_(i j)) = −1. Since det :

GL(V ) ! Q× is a group homomorphism, it follows that ker(det _ _) is a

normal subgroup of Sn, called the alternating group of degree n and denoted

An. It is customary to write “sgn” in place of det _ _, called the “sign”

homomorphism of Sn. Thus, An = ker(sgn).6 Note that _ 2 An if and only

6The astute reader will notice that the above passage is actually tautological, as the

cited property of determinants above depends on the well-definedness of “sgn.”

1.6. THE SYMMETRIC AND ALTERNATING GROUPS 23

if it is possible to write _ as a product of an even number of transpostions. (A

more elementary, and indeed more honest treatment, due to E. Spitznagel,

can be found in Larry Grove’s book, Algebra, Academic Press, New York,

1983, page 17.)

Let (i1 i2 · · · ik) be a k-cycle in Sn, and let _ 2 Sn. One has

Lemma 1.6.2 _(i1 i2 · · · ik)_−1 = (_(i1) _(i2) · · · _(ik)).

From the above lemma it is immediate that the conjugacy class of an

element of Sn is uniquely determined by its cycle type. In other words, the

elements (2 5)(3 10)(1 8 7 9) and (3 7)(5 1)(2 10 4 8) are conjugate in S10,

but (1 3 4)(2 5 7) and (2 6 4 10)(3 9 8) are not. It is often convenient to

use the notation [1e12e2 · · · nen] to represent the conjugacy class in Sn with a

typical element being the product of e1 1-cycles, e2 2-cycles, . . . , en n-cycles.

Note that

P

ei · i = n. Thus, in particular, the conjugacy class containing

the element (4 2)(1 7)(3 6 10 5) 2 S10 would be parametrized by the symbol

[12224]. Note that if _ is in the class parametrized by the symbol [1e1 · · ·]

then |Fix(_)| = e1.

Example. From the above discussion, it follows that

• The conjugacy classes of S5 are parametrized by the symbols [15],

[132], [123], [14], [122], [23], [5].

• The conjugacy classes of S6 are parametrized by the symbols [16],

[142], [133], [124], [1222], [123], [15], [23], [24], [32], [6].

Before leaving this section, we shall investigate the alternating groups

in somewhat greater detail. Just as the symmetric group Sn is generated

by transpositions, the alternating group An is generated by 3-cycles. (This

is easy to prove; simply show how to write a product (ab)(cd) as either a

3-cycle or as a product of two 3-cycles.) The following is important.

Theorem 1.6.3 If n _ 5, then An is simple.

For n = 5, the above is quite easy to prove. For n _ 6, see Exercise 19

below.

Recall that if G is a group having a subgroup H _ G of index n, then

there is a homomorphism G ! Sn. However, if G is simple, the image of the

above map is actually contained in An, i.e., G ! An. Indeed, there is the

composition G ! Sn ! {±1}; if the image of G ! Sn is not contained in An,

then G will have a normal subgroup of index 2, viz., ker(G ! Sn ! {±1}).

The above can be put to use in the following examples.

24 CHAPTER 1. GROUP THEORY

Example 1. Let G be a group of order 112 = 24 · 7. Then G cannot

be simple. Indeed, if G were simple, then G must have 7 2-Sylow

subgroups creating a homomorphism G ! A7. But |A7|2 = 8, so G

can’t “fit,” i.e., G can’t be simple.

Example 2. Suppose that G is a group of order 180 = 22 · 32 · 5. Again,

we show that G can’t be simple. If G were simple, it’s not too hard

to show that G must have 6 5-Sylow subgroups. But then there is

a homomorphism G ! A6. Since G is assumed to be simple, the

homomorphism is injective, so the image of G in A6 has index 360

180 = 2.

But A6 is a simple group, so it can’t have a subgroup of index 2.

As mentioned above, the conjugacy classes of Sn are uniquely determined

by cycle type. However, the same can’t be said about the conjugacy classes

in An. Indeed, look already at A3 = {e, (123), (132)}, an abelian group.

Thus the two 3-cycles are clearly not conjugate in A3, even though they

are conjugate in S3. In other words the two classes in A3 “fuse” in S3. The

abstract setting is the following. Let G be a group and let N/G. Let n 2 N,

and let C be the G-conjugacy class of n in N:

C = {gng−1| g 2 G}.

Clearly C is a union of N-conjugacy classes; it is interesting to determine

how many N-conjugacy classes C splits into. Here’s the answer:

Proposition 1.6.4 With the above notation in force, assume that C =

C1 [C2 [· · ·[Ck is the decomposition of C into disjoint N-conjugacy classes.

If n 2 C, then k = [G : CG(n)N].

The above explains why the set of 5-cycles in A5 splits into two A5-

conjugacy classes (doesn’t it? See Exercise 7, below.) This can all be cast

in a more general framework, as follows. Let G act on a set X. Assume that

X admits a decomposition as a disjoint union X = [X_ (_ 2 A) where for

each g 2 G and each _ 2 A, gX_ = X_ for some _ 2 A. The collection of

subsets X_ _ X is called a system of imprimitivity for the action. Notice

that there are always the trivial systems of imprimitivity, viz., X = X, and

X = [x2X{x}. Any other system of imprimitivity is called non-trivial. If

the action of G on X admits a non-trivial system of imprimitivity, we say

that G acts imprimitively on X. Otherwise we say that G acts primitively

on X.

1.6. THE SYMMETRIC AND ALTERNATING GROUPS 25

Consider the case investigated above, namely that of a group G and a

normal subgroup N. If n 2 N, then the classes CG(n) = CN(n) precisely

when the conjugation action of G on the set CG(n) is a primitive one. Essentially

the same proof as that of Proposition 1.6.4 will yield the result of

Exercise 15, below.

Exercises 1.6

1. Give the parametrization of the conjugacy classes of S7.

2. Let G be a group of order 120. Show that G can’t be simple.

3. Find the conjugacy classes in A5, A6.

4. Prove that A4 is the semidirect product of Z2 × Z2 by Z3.

5. Show that Sn = h(12), (23), . . . , (n − 1 n)i.

6. Let p be prime and let G _ Sp. Assume that G contains a transposition

and a p-cycle. Prove that G = Sp.

7. Let x 2 Sn be either an n-cycle or an n−1-cycle. Prove that CSn(x) =

hxi.

8. Show that Sn contains a dihedral group of order 2n for each positive

integer n.

9. Let n be a power of 2. Show that Sn cannot contain a generalized

quaternion group Q2n.

10. Let G act on the set X, and let k be a non-negative integer. We

say that G acts k-transitively on X if given any pair of sequences

(x1, x2, . . . , xk) and (x0

1, x0

2, . . . , x0

k) with xi 6= xj , x0

i 6= x0

j for all

i 6= j then there exists g 2 G such that g(xi) = x0

i, i = 1, 2, . . . , k.

Note that transitivity is just 1-transitivity, and double transitivity is

2-transitivity. Show that Sn acts n-transitively on {1, 2, . . . , n}, and

that An acts (n − 2) − transitively (but not (n − 1)-transitively) on

{1, 2, . . . , n}.

11. Let G act primitively on X. Show that G acts transitively on X.

12. Let G act doubly transitively on X. Show that G acts primitively on

X.

26 CHAPTER 1. GROUP THEORY

13. Let G act transitively on the set X and assume that Y _ X has the

property that for all g 2 G, either gY = Y or gY \ Y = ;. Show that

the distinct subsets of the form gY form a system of imprimitivity in

X.

14. Let G be a group acting transitively on the set X, let x 2 X, and

let Gx be the stabilizer in G of x. Show that G acts primitively on

X if and only if Gx is a maximal subgroup of G (i.e., is not properly

contained in any proper subgroup of G). (Hint: If {X_} is a system

of imprimitivity of G, and if x 2 X_, show that the subgroup M =

StabG(X_) = {g 2 G| gX_ = X_} is a proper subgroup of G properly

containing Gx. Conversely, assume that M is a proper subgroup of

G properly containing Gx. Let Y be the orbit containing {x} in X

of the subgroup M, and show that for all g 2 G, either gY = Y or

gY \ Y = ;. Now use Exercise 13, above.)

15. Let G act on the set X, and assume that N / G. Show that the Norbits

of N on X form a system of imprimitivity. In particular, if the

action is primitive, and if N is not in the kernel of this action, conclude

that N acts transitively on X.

16. Prove the following simplicity criterion. Let G act primitively on the

finite set X and assume that for x 2 X, the stabilizer Gx is simple.

Then either

(a) G is simple, or

(b) G has a normal transitive subgroup N with |N| = |X|. (Such a

subgroup is called a regular normal subgroup .)

17. Let G be the group of automorphisms of the “cubical graph,” depicted

below:

•

•

•

• •

•

•

............. • .....................................

..................................................

..................................................

..................................................

..................................................

..................................................

..................................................

1.6. THE SYMMETRIC AND ALTERNATING GROUPS 27

Show that there are two distinct decompositions of the vertices of the

above graph into systems of imprimitivity: one is as four sets of two

vertices each and the other is as two sets of four vertices each. In the

second decomposition, if V denotes the vertices and if V = V1 [ V2 is

the decomposition of V into two sets of imprimitivity of four vertices

each, show that the setwise stabilizer of V1 is isomorphic with S4.

18. Let G act on X, and assume that N is a regular normal subgroup of

G. Thus, if x 2 X, then Gx acts on X − {x} and, by conjugation, on

N# := N − {1}. Prove that these actions are equivalent.

19. Using Exercises 16 and 18, prove that the alternating groups of degree

_ 6 are simple.

20. Let G act on the set {1, 2, . . . , n}, let F be a field and let V be the

F-vector space with ordered basis (v1, v2, . . . , vn). As we have already

seen, G acts on V via the homomorphism _ : G ! GL(V ). Set

V G = {v 2 V | _(g)v = v}.

(a) Show that dim V G = the number of orbits of G on {1, 2, . . . , n}.

(b) Let V1 _ V be a G-invariant subspace of V ; thus G acts as a group

of linear transformations on the quotient space V/V1. Show that

if the field F has characteristic 0 or is prime to the order of |_(G)|,

then

(V/V1)G _= V G/V G

1 .

(c) Assume that G1,G2 _ Sn, acting on V as usual. If g1g2 = g2g1

for all g1 2 G1, g2 2 G2 show that G2 acts on V G1 and that

(V G1)G2 = V G1 \V G2 . Use this result to obtain another solution

of Exercise 11 of Section 1.2.

28 CHAPTER 1. GROUP THEORY

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