ORGANIGRAMA ORGANIZACIONAL

Let ϕ be an action of a group G on a set Ω, that is, ϕ is a homomor- phism from G into the symmetric group SΩ. We call ϕ a permutation representation of G on Ω. If G acts on a set Ω transitively, then ϕ is called a transitive permutation representation of G on Ω.

Definition 2.4.1 Let ϕ1 : G1 → SΩ1 and ϕ2 : G2 → SΩ2 be two permutation representations or actions.

(1) ϕ1 and ϕ2 are called similar if there is an isomorphism σ:G1 →

G2 and a bijectionµ: Ω1 →Ω2 such that (αµ₎ϕ2(gσ₎

= (αϕ1(g)₎µ _{for ∀g ∈G}

1 and ∀α ∈Ω1, or equivalently, the diagram

Ω1 Ω2 Ω1 Ω2 - - µ µ ? ϕ1(g) ?ϕ2(g σ₎ is commutative.

(2) If two representations ϕ1 : G1 → SΩ1 and ϕ2 : G2 → SΩ2 are faithful and similar, then we have two permutation groups

ϕ1(G1)≤SΩ1 and ϕ2(G2)≤SΩ2 In this case, two groups ϕ1(G1) and ϕ2(G2) are isomorphic as permutation groups. We call them permutationally isomorphic.

(3) If G1 = G2 =G in (1) and one can take σ as the identity auto- morphism of G then ϕ1 : G →SΩ1 and ϕ2 : G→ SΩ2 are called equivalent orisomorphic. In this case, µ is called an equivalence orisomorphism.

A typical example of a transitive permutation representation is the group action on the right coset space of a subgroup, P(g) =

µ

Hx Hxg

¶ given in Example 2.1.3. In the following sense, it represents all transi- tive permutation representations of a group.

Theorem 2.4.1 Any transitive permutation representation ϕ : G → SΩ of a finite group G is equivalent to the permutation representation

P on H: P(g) = µ Hx Hxg ¶ for a subgroup H ≤G.

Proof: For a given representation ϕ : G → SΩ, take α ∈ Ω and let

H = Gα be the stabilizer. Since ϕ is assumed to be transitive, the

function µ : G/H =: {Hg | g ∈ G} → Ω defined by (Hg)µ _{= α}g _:=

αϕ(g) _{is bijective, as shown in the proof of the orbit-stabilizer theorem.} Then, the following diagram is commutative:

Transitive permutation representations 77 G/H Ω . G/H Ω - - ? ? µ µ P(g) ϕ(g) Hxg Hx αxg αx – ? –? ` Ã ` Ã

It implies that ϕand P are equivalent. ¤

Recall that the kernel Ker P of the permutation representation

P(g) = µ

Hx Hxg

¶

given in Example 2.1.3 is the core CoreG(H) =

T

g∈Gg−1Hg, which is the maximal normal subgroup of G contained

in H. It implies that for an abelian group G, the right translation of

G is the only faithful transitive representation ofG up to equivalence.

Theorem 2.4.2 Let P1 and P2 be two permutation representations of a group G on subgroups H and K, respectively:

P1(g) = µ Hx Hxg ¶ , P2(g) = µ Kx Kxg ¶ , ∀g ∈G. Then,

(1) P1 and P2 are equivalent if and only if H and K are conjugate in

G.

(2) When P1 and P2 are faithful, two groups P1(G) and P2(G) are permutationally isomorphic if and only if there is an automor- phism σ of G such that K =Hσ_.

Proof: Let Ω1 ={Hx|x∈G}and Ω2 ={Kx|x∈G}.

(1) (⇒) Assume that P1 and P2 are equivalent with an equivalence

µ : Ω1 → Ω2, and let Hµ = Ky. Then by the equivalency of P1 and

P2, the stabilizer of H in P1(G) and the stabilizer of Ky inP2(G) are equal, (verify it!) that is, H =y−1_Ky.

(⇐) Assume H =y−1_{Ky for some y∈G. Define}

Then µis a bijection. Since

(Hx)µP2(g)=Kyxg= (Hxg)µ= (Hx)P1(g)µ_,

we have µP2(g) = P1(g)µ, ∀g ∈G. Thus P1 and P2 are equivalent. (2) (⇐) Assume thatσis an automorphism ofGsuch thatK =Hσ_.

Let ν : Ω1 → Ω2 be a map defined by (Hx)ν = Hσxσ = Kxσ for all

x ∈ G. Then, it is easily shown that ν is onto. Moreover, since the indices of H and K in G are the same, ν is also one-to-one, which means that ν is a bijection. And, let ρ : P1(G) → P2(G) be a map defined by (P1(g))ρ=P2(gσ) for allg ∈G. Since P1 andP2 are faithful representations and σ is an automorphism of G, ρ is an isomorphism. For all g, x∈G,

((Hx)P1(g)₎ν _{= (Hxg)}ν _=K(xg)σ _=Kxσ_gσ _{= ((Hx)}ν₎P2(gσ)_{= ((Hx)}ν₎(P1(g))ρ_. So, two permutation groups P1(G) and P2(G) are permutationally isomorphic.

(⇒) Assume that there exist an isomorphism ρ : P1(G) → P2(G) and a bijection ν : Ω1 → Ω2 such that ((Hx)P1(g))ν = ((Hx)ν)(P1(g))

ρ

for all g, x ∈ G. Let Hν _{= Ky for some y ∈ G and (P}

1(g))ρ =P2(gβ) for all g ∈G and for some map β :G→G. Since two representations

P1 andP2 are faithful and ρis an isomorphism,β is an automorphism. If Hg =H then

Ky=Hν _{= (Hg)}ν _{= (H}P1(g)₎ν _{= (H}ν₎P2(gβ₎

=Kygβ_.

It is equivalent to y−1_{Ky =y}−1_Kygβ_{. So, g}β _∈y−1_{Ky. It means that}

Hβ _{⊆ y}−1_{Ky which is equivalent to yH}β_y−1 _{⊆K. Since the indices of}

H and K inG are the same,yHβ_y−1 _{=K. Therefore,K =H}σ_{, where}

σ is an automorphism ofGwhich is a composition of an automorphism

β of Gand conjugate under y−1_{. ¤}

Notice that there are differences between the equivalency of two permutation representations ϕ1 and ϕ2 of a group G and the permu- tational isomorphism of ϕ1(G) and ϕ2(G). In fact, if two permutation representations ϕ1 and ϕ2 of the same group G are equivalent, then two groups ϕ1(G) andϕ2(G) are permutationally isomorphic. But, the converse is not true.

Transitive permutation representations 79

For example, GL(3,2) acts on the projective points and on the pro- jective lines of the Fano plane naturally. (For the Fano plane, see Exercise 2.4.5 or Volume two, Chapter 8 for more information.)

These two actions are not equivalent, but as permutation groups, the images of these two actions are permutationally isomorphic. (Ex- plain why?) See Exercise 2.4.6.

In general, for n ≥ 3 and any Galois field GF(q), the actions of projective linear group PGL(n, q) on the projective points and on the projective hyperplanes of the (n−1)-dimensional projective geometry

P G(n −1, q) are not not equivalent, but as permutation groups, the images of these two actions are permutationally isomorphic.

Next we give some applications of permutation representations of groups.

Proposition 2.4.3 (Poincar´e’s argument) Let H ≤ G and |G :

H|=n. Then |G/CoreG(H)| is a divisor of (n!,|G|), where CoreG(H)

is the core of H in G.

Proof: From the transitive permutation representationP(g) =

µ Hx Hxg ¶ , we have G/Ker P =G/CoreG(H)∼=P(G).Sn,

implying the desired result. ¤

Corollary 2.4.4 (1) If a simple groupG has a subgroup of index n, then |G| is a divisor of n!.

(2) Let G be a finite group, and p the smallest prime divisor of |G|. If H ≤G and |G:H|=p, then HEG.

Proof: We prove only (2). By Proposition 2.4.3, |G/CoreG(H)| is a

divisor of (p!,|G|) = p. This forces that CoreG(H) =H, that is HEG.

¤

Proof: SinceGis simple and|G|= 60 = 22_{·3·5,Ghas no subgroups} of index 2,3 or 4, by Corollary 2.4.4(1).

First, we claim thatGhas a subgroup of index 5, that is, a subgroup of order 12. By Sylow theorems, the number of Sylow 2-subgroups of

G is n2 = 3,5 or 15. Since n2 =|G:NG(P)| for any P ∈Syl2(G) (see Theorem 1.1.7), n2 6= 3 as mentioned already. If n2 = 5, then we are done again by Theorem 1.1.7. Ifn2 = 15, we have two cases. As the first case, let any two Sylow 2-subgroups ofGhave a trivial intersection. In this case the number of 2-elements inGis 1 + 3×15 = 46. However, by Sylow theorems, the number of Sylow 5-subgroups isn5 = 6, and hence the number of nontrivial 5-elements is 4×6 = 24. So, G would have 46+24 = 70 2-elements and 5-elements, a contradiction. As the second, assume that there are two Sylow 2-subgroups whose intersection is a subgroup A of order 2. Consider the centralizer CG(A) of A in G. It

contains at least two Sylow 2-subgroups whose orders are 4, so its order

|CG(A)|is greater than 4 and is a multiple of 4 by Lagrange’s Theorem.

Hence, |CG(A)|= 12,as a divisor of |G|, and it is a subgroup of index

5.

Now, let H be a subgroup of G of order 12, i.e., of index 5. Then the permutation representation of G onH gives an isomorphism from

GintoS5. However,S5 has only one subgroup of order 60, which isA5,

(prove it!) so we get G∼=A5. ¤

Example 2.4.1 Any group G of order 144 is not simple.

Solution: Note that 144 = 24_·32_{. By Sylow theorems, n}

3(G) = 4 or 16. If n3 = 4, then|G:NG(P)| = 4, where P ∈Syl3(G). Consider the permutation representation ofGonNG(P). Then G/CoreG(NG(P)).

S4. Thus CoreG(NG(P)) is a nontrivial normal subgroup ofG, andGis

not simple. Ifn3 = 16, we distinguish two cases: (1) Let any two Sylow 3-subgroups have a trivial intersection. In this case, the number of 3- elements in G is 16×(9−1) + 1 = 129. It follows that the 2-elements inGis at most 144−129 + 1 = 16. SinceGhas a subgroup of order 16, which is the Sylow 2-subgroup, Gshould have only one such subgroup, and this subgroup should be normal. (2) Let there exist two Sylow 3-subgroups with nontrivial intersection A > 1. In this case we have

Transitive permutation representations 81

|A| = 3. Let H =NG(A). Then H contains at least those two Sylow

3-subgroups containing A, that is n3(H) > 1. By Sylow theorems,

n3(H) ≥ 4 and |H| = n3(H)· |NH(P)| ≥ 4·32 for P ∈ Syl3(H) by Theorem 1.1.7. Hence |G : H| ≤ 4. If H = G then A EG, and G

is not simple. If H < G, by Poincar´e’s argument, |G/CoreG(H)| | 4!,

and hence CoreG(H) is a nontrivial normal subgroup ofGand Gis not

simple. ¤

In Theorem 2.3.4, we show that any two distinct prime numbers

p and q, every group of order pqn _{is solvable for all n. The Burnside}

pa_qb_{-Theorem 2.3.5 says that every group of orderp}a_qb _{is solvable. The}

next theorem is a special case of Burnside pa_qb_-Theorem.

Theorem 2.4.6 Let |G| = p2_qn_{, where p < q are primes. Then G is}

solvable.

Proof: If there exists a Sylow q-subgroup of G which is normal in

G, then G is clearly solvable. Let all Sylow q-subgroups Q of G be non-normal. Then the third Sylow Theorem and the condition p < q

imply that nq(G) =|G|/|NG(Q)|=p2 and q |p2−1 = (p−1)(p+ 1).

Again byp < q, we havep= 2 andq = 3. Hence|G:Q|= 4. Consider the permutation representation of G on Q. Then G/CoreG(Q) . S4, and hence G/CoreG(Q) is solvable. Since CoreG(Q) is a q-group and

solvable, G should be solvable. ¤

Theorem 2.4.7 If G has a Sylow 2-subgroup P which is cyclic, then

Ghas a normal subgroup N such that P∩N = 1 andP N =G. In this case, we say that P has a normal complementN in G.

Proof: Let P = hai and o(a) = 2n_{. Consider the right regular rep-}

resentation R(a) of G. Then R(a) is a product of m 2n_{-cycles as a}

permutation of S|G|, where m= ₂1n|G| is an odd integer. HenceR(a) is

an odd permutation. Thus all even permutations in R(G) form a sub- group of index 2 of R(G), which is normal in R(G). Since G ∼=R(G),

G has a subgroup G1 of index 2. Since every Sylow 2-subgroup of G1 is contained in a Sylow 2-subgroup of G, it is also cyclic. Let P1 be

one of the Sylow 2-subgroup of G1. Now, as an induction hypothesis on |G|, one may assume that G1 has a normal subgroup N such that

P1 ∩N = 1 and P1N = G1. It follows that |N| is odd and N is char- acteristic in G1 (because (|N|,|G;N|) = 1, see Exercise 1.2.3). Hence

N is normal in G. Let P be a Sylow 2-subgroup of G containing P1. Then, obviously,P∩N = 1.By comparing the orders with 2|P1|=|P|,

we have P N =Gas desired. ¤

It will be shown later that Theorem 2.4.7 remains true when the even primep= 2 is replaced by the smallest prime divisor of |G|. (See Theorem 2.5.3.)

Exercises

2.4.1. Prove that any group of order 2·3·5·7 is solvable. (Compare this with Exercise 2.3.4.)

2.4.2. Prove that any group of order 24 is solvable.

2.4.3. Let n_p(G) be the number of Sylow p-subgroups of G. Assume that np(G) 6≡1(mod p2). Then there exist two Sylow p-subgroupsP1 and P2 of Gsuch that|P1 :P1∩P2|=p.

2.4.4. Prove that any group of order 432 is solvable.

2.4.5. (Finite projective planes) Given a non-empty finite set P whose elements are called points and a collection L of subsets of P whose elements are called lines, the pair P = (P,L) is called a projective planeif the following hold:

(1) For any two distinct linesL1, L2 ∈L,L1∩L2 consists of a unique point.

(2) For any two distinct pointsP1, P2 ∈P, there exists a unique line L∈Lcontaining these two points.

(3) There exist four points P₁, . . . , P₄ ∈ P such that any three of them are not contained in the same line.

The following theorem is basic.

Theorem Let n≥2 be an integer andP = (P,L) a projective plane. Then the following properties are equivalent:

(1) Each line contains exactly n+ 1 points. (2) Each point is contained in exactlyn+ 1lines.

Transitive permutation representations 83

(3) P contains n2_{+n+ 1points.} (4) Lcontains n2_{+n+ 1 lines.}

This number n is called theorder of the projective planeP.

Not for everynthere exists a projective plane of ordern. For example, no projective plane of order 10 exists; see C.W.H. Lam, L. Thiel and S. Swiercz (1989). The smallest projective plane has 7 points and 7 lines, called theFano plane, depicted below.

(1,0,0) (1,0,1) (1,1,0) (0,0,1) (0,1,0) (1,1,1) ¾ [0,1,0] [0,1,1] [1,1,0] [1,0,0] [0,0,1] [1,0,1] [1,1,1] w j j À ¼ 9 *

Figure 2.1: The Fano plane

Its point setP={1,2,3,4,5,6,7}and the line setL={{1,2,4},{2,3,5}, {3,4,6},{4,5,7},{5,6,1},{6,7,2},{7,1,3}}. Check (P,L) is a projec- tive plane of order 2.

2.4.6. (The Fano plane and its automorphism group) We first give a construction of projective planes of ordern=ps_{, a prime power.}

Construction Let F = GF(ps_{), the Galois field with p}s _elements.

Let V be a 3-dimensional vector space over F. Let P be the set of 1-dimensional subspaces and L the set of 2-dimensional subspaces of V. Then P = (P,L) is a projective plane of order n. It is called the Desarguesian planeof ordern.

The Fano plane is the Desarguesian plane of order 2. TakeF=GF(2) and a 3-dimensional vector spaceV overF. Since each 1-dimensional subspace contains only one non-zero vector, we may use that vector to denote the subspace. In the above picture, we let 1 = (1,0,0), 4 = (0,1,0), 5 = (0,0,1), 2 = (1,1,0), 6 = (1,0,1), 7 = (0,1,1) and 3 = (1,1,1). Then the three points in each line span a 2-dimensional subspace.

A map α : P → P is called an automorphism (or a collineation in geometry) of P if α permutes lines. All collineations of P form a group Aut(P).

For Desarguesian planesP, the linear transformations of V =V(3,F) induce collineations of the plane in a natural way. Since only the scalar transformations induce the trivial collineation, PGL(3, ps_{) can}

be viewed as a subgroup of Aut(P). This group has an action onPand an action onL. These two actions provide two transitive permutation representations of PGL(3, ps_).

For the Fano plane, PGL(3,2) ∼= GL(3,2). The images of the two actions ofG:= GL(3,2) on the point setP and the line setLofP are denoted byG_P andG_L, respectively. We have

(1) as permutation groups on seven symbolsG_P and G_L are permu- tationally isomorphic.

(2) These two permutation representations of Gare not equivalent. Sketch of Proof: LetP = (1,0,0)∈P and H the stabilizer of P in G. Then H =     1_v 0_{x y}0 w z u   ¯ ¯ ¯ ¯ ¯ x, y, z, u, v, w∈F, xu−yz= 1   .

We use [a, b, c] to denote the line consisting of the points (x, y, z) with ax+by+cz = 0. Let L= [1,0,0]∈L and HT _{the stabilizer of L in}

G. Then HT =     1_{0 x}v w_z 0 y u   ¯ ¯ ¯ ¯ ¯x, y, z, u, v, w ∈F, xu−yz = 1   .

Note that HT _{consists of those matrices in G that are transposes of}

the matrices in H.

Show thatHandHT are not conjugate inG. Then by Theorem 2.4.2, (2) holds. (Hint: any conjugate ofH fixes a point, but HT _{does not}

fix any point.)

To prove (1), consider the automorphismτ ofG defined by gτ = (gT)−1, ∀g∈G,

Transfer and Burnside Theorem 85

where g is viewed as a 3×3 matrix and gT _{denotes the transpose of}

g.

Then µ : Hg 7→ Hτ_gτ _{is a bijection from the coset space G/H to}

the coset space G/Hτ_{. Let P and P}

1 be the transitive permutation representations ofGon G/H and G/Hτ, respectively. Then the map P(x) 7→ P1(xτ) is an isomorphism from P(G) to P1(G), (1) holds. (Verify it!)

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