THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP
MARTINO GARONZI
Lecture notes of the Minicourse “The maximal subgroups of the symmetric group” given at the “Escola de Algebra”´ in Unicamp (Campinas, S˜aoPaulo, Brazil) on December 3-7 2018. Version 1. I will improve the notes and release a more com- plete version in the future. I am grateful for any comment, criticism and correction.
Contents 1. Basic notions about group actions 1 2. The symmetric group 2 3. The maximal subgroups of S5 3 4. Imprimitivity blocks 4 5. Maximal imprimitive subgroups 5 6. Intransitive maximal subgroups 7 7. Primitive maximal subgroups 7 7.1. Characteristically simple groups 7 7.2. Primitive groups 8 7.3. Multiple transitivity 10 7.4. Jordan groups 11 7.5. Primitive actions: O’Nan-Scott 13 References 15
1. Basic notions about group actions All group actions we consider are on the right. G acts on X by the rule (x, g) 7→ xg if and only if the map γg : X → X, x 7→ xg induces a homomorphism G → Sym(X), g 7→ γg. The kernel of this homomorphism is the kernel of the action and the action is called faithful if it has trivial kernel. The orbits of the action are the sets O(x) = {xg : g ∈ G} for any x ∈ X. The action is called transitive if there is some x ∈ X such that O(x) = X (i.e. there is only one orbit), in other words for any x, y ∈ X there exists g ∈ G with xg = y. The stabilizer of x ∈ X is Gx = {g ∈ G : xg = x} ≤ G.
For example consider the action by right multiplication of G on X = {Ht : t ∈ T g g −1 G}, the kernel is HG = g∈G H where the notation is H = g Hg. HG is called the normal core of H in G.
Counting principle (Orbit-Stabilizer lemma): if G acts transitively on X and x ∈ X then |X| = |G : Gx|.
1 2 MARTINO GARONZI
The action is called semiregular if the point stabilizers are trivial, and it is called regular if it is semiregular and transitive. By the counting principle if G acts regularly then |G| = |X|. An example of a regular action is that of the Klein 4- group acting naturally on {1, 2, 3, 4}. More in general if G is a semidirect product N o H then the action of N of right multiplication on the right cosets of H is regular.
An important example is the following. If G acts faithfully on X and N is a normal subgroup of G acting transitively then the centralizer CG(N) acts semireg- ularly. Indeed if g ∈ CG(N) is such that xg = x for some x ∈ X then since every element of X has the form xn for some n ∈ N we have xn = xgn = xng therefore g fixes all the points in X, so g = 1 because the action of G is faithful.
Exercise (Burnside lemma). Let fg be the number of fixed points of g ∈ G 1 P acting on X. Then the number of orbits of the action is |G| g∈G fg. For example if G acts semiregularly the number of orbits is |X|/|G| because f1 = |X| and fg = 0 for every 1 6= g ∈ G. Therefore if the action is regular (semiregular and transitive) then |G| = |X|.
2. The symmetric group
Let Sn be the symmetric group on n letters. It is well-known that if n ≥ 2 and n 6= 4 then the alternating group An is a simple group that has index 2 as a subgroup of Sn. Moreover An is nonabelian if n ≥ 5. This easily implies that if n 6= 4 the normal subgroups of Sn are {1} < An < Sn and the normal subgroups ∼ of S4 are {1} < K < A4 < S4 where K = C2 × C2 is the Klein group. In particular An is the unique subgroup of Sn of index 2.
Proposition 1. If H < Sn and H 6= An then |Sn : H| ≥ n. If H < An then |An : H| ≥ n.
Proof. Let m = |Sn : H|. Sn acts transitively (hence non-trivially) by right multi- plication on X = {Hx : x ∈ Sn}, which is a set of size m, this gives a homomor- phism ϕ : Sn → Sm whose image is a transitive subgroup of Sm. Let K = ker(ϕ). Since H 6= An we have m ≥ 3 so |Sn : K| ≥ 3 (a transitive group on m points has always order at least m). The case n ≤ 4 can be done by hand, and we know that if n ≥ 5 the unique proper nontrivial normal subgroup of Sn has index 2, so K = {1}. It follows that |Sn| = n! divides |Sm| = m! hence n ≤ m. A similar argument applies to show that if H < An then |An : H| ≥ n.
We will now discuss the subgroups of Sn of index n.
Proposition 2. If n 6= 6 then the unique subgroups of Sn of index n are the point stabilizers. Moreover S6 has two conjugacy classes of subgroups of index 6.
Proof. We need to recall that if x ∈ An then the conjugacy class of x in An is not equal to its conjugacy class in Sn if and only if x is the products of disjoint cycles of pairwise distinct odd lengths. So for example if n ≥ 5 then all the 3-cycles are conjugated in An. ∼ Let H be a subgroup of Sn of index n. First observe that H = Sn−1. Indeed Sn acts faithfully on the set of the n right cosets of H by right multiplication and this gives ϕ : Sn → Sn, however ϕ(H) is contained in the stabilizer of H, which has THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP 3 order (n − 1)!. It follows that H is a group of order (n − 1)! isomorphic to ϕ(H) which is contained in a point stabilizer, that is isomorphic to Sn−1. It follows that ∼ H = Sn−1. Similarly any subgroup of An of index n is isomorphic to An−1. ∼ Now consider an isomorphism f : An−1 = H < Sn. The idea is to show that if x is a 3-cycle in An−1 then ϕ(x) is also a 3-cycle (x centralizes some An−4 so K = ϕ(An−4) is centralized by the element ϕ(x) of order 3 and has an orbit of size n − t ≥ n − 4 - it cannot act nontrivially on less than n − 4 elements - now proceed to show that ϕ(x) must be a 3-cycle). Now show that ϕ((123)), . . . ϕ((1, 2, n − 1)) must be of the form (abx) where the x’s are n − 3 elements, pairwise distinct, and a, b are fixed, so they generate the stabilizer of the one element fixed by all of them (obs: if r, s 6= 1, 2 then ϕ((12r)), ϕ((12s)) are 3-cycles that generate some A4 so they are of the form (abx), (aby) where a, b are fixed and x, y depend on r, s). Generalize this to Sn. S5 acts by conjugation on the set of its 6 Sylow 5-subgroups and this action is faithful. This gives an injective homomorphism ϕ : S5 → S6 with transitive image ∼ H = ϕ(S5). It follows that H = S5 has index 6!/5! = 6 in S6 and it is not a point stabilizer because it is transitive. It is possible to show that apart from point stabilizers this is indeed the only other class of subgroups of index 6. ∼ ∼ Corollary 1. If n ≥ 3 and n 6= 6 then Aut(Sn) = Sn = Aut(An). Moreover A6 has index 4 in Aut(A6).
Proof. The idea is that Aut(Sn) acts on the family of subgroups of Sn of index n, which, being n 6= 6, are all the n point stabilizers. This action is faithful, hence we get Aut(Sn) → Sn injective. On the other hand being Z(Sn) = {1} the ∼ canonical “conjugation” homomorphism Sn → Aut(Sn) is injective hence Sn =≤ ∼ ∼ Aut(Sn) =≤ Sn implying Aut(Sn) = Sn. The case n = 6 is treated separately.
Aut(A6) u: O fLL uu LL uu LL uu LLL uu L M10 S6 PGL2(9) I d II O r8 II rr II rrr II rr I rrr A6 ∼ Out(A6) = Aut(A6)/A6 = C2 × C2.
Exercise: let σ = (1 . . . n) ∈ Sn. The centralizer of σ in Sn is hσi.
3. The maximal subgroups of S5
Recall that the unique normal subgroups of Sn are {1}, An and Sn except when n = 4, in which case we have an additional normal subgroup, the Klein 4 group.
Generalized Cayley theorem: if G acts transitively on a set Ω of size n with point stabilizer H then |G : H| = n (counting principle - orbit-stabilizer lemma), the action is equivalent to the right multiplication action of G on {Hx : T g x ∈ G} and setting HG = g∈G H the normal core of H in G, the quotient G/HG is isomorphic to a subgroup of the symmetric group Sn, in particular |G : HG| divides 4 MARTINO GARONZI n!. In particular the symmetric group Sn has no nontrivial proper subgroup, distinct from An, of index less than n, otherwise it would be isomorphic to a subgroup of a symmetric group of degree less than n.
Observation: An is the unique subgroup of Sn of index 2. To see this let N 2 be a subgroup of Sn of index 2. Then it is normal and it contains Q = {x : x ∈ 2 Sn} (because (xN) = N in the quotient G/N). However An is generated by Q, therefore An ≤ hQi ≤ N and since |An| = |N| we deduce An = N. To see that An is generated by Q observe that any element of An is a product of an even number of 2-cycles, and a product of two 2-cycles is either a 3-cycle or a product of two disjoint 2-cycles, so it belongs to Q in either case (consider the cases (12)(34) = (1324)2 2 and (123) = (132) ). Therefore any element of An is a product of elements of Q.
Let G = S5. We want to determine the maximal subgroups of G. Observe first that the only normal subgroups of G are {1}, A5 and G (as is the case for all symmetric groups of degree at least 5) and A5 is the only subgroup of G of index 2. We claim that the only maximal subgroups of G are A5, the point stabilizers, the intransitive subgroups of type S3 ×S2 and the normalizers of the Sylow 5-subgroups.
Let M be a maximal subgroup of G. Suppose first that 5 does not divide |M|. Then M acts intransitively hence it is one of S4, S3 × S2 (it cannot have more than two orbits otherwise it would be properly contained in an intransitive subgroup admitting precisely two orbits). We may now assume that 5 divides |M|. M contains a Sylow 5-subgroup P of G. Suppose 3 divides |M|. Then |G : M| divides 8 hence |G : M| ∈ {2, 4, 8}. If |G : M| = 4 then G is isomorphic to a subgroup of S4, a contradiction. If |G : M| = 8 then |M| = 15 and M is cyclic, but S5 has no elements of order 15. We deduce that |G : M| = 2 hence M = A5. Now assume 3 does not divide |M|. Then |M| = 5 · 2n with n ∈ {0, 1, 2, 3}, now by Sylow theorem P E M so M ≤ NG(P ) hence M = NG(P ).
The normalizer NG(P ) is a semidirect product C5 o C4 and it is isomorphic to ∼ AGL(1, 5), the affine group of dimension 1 over F5. Observe that Aut(F5) = ∼ U(F5) = C4. Affine groups are defined to be those primitive groups with abelian socle.
Exercise: find all the maximal subgroups of S4.
4. Imprimitivity blocks
Let G be a subgroup of Sn acting on Ω = {1, . . . , n}.A block of imprimitivity, or simply block, of the action of G Is a subset B of Ω with the property that Bg = B or Bg ∩ B = ∅ whenever g ∈ G. In particular any orbit of G is a block. Observe that Ω is a block, and {ω} is a block for every ω ∈ Ω: these are called the trivial blocks. Also, {ω} is an example of block that, in general, is not an orbit. If B is a block then Bg is a block for all g ∈ G and the “translates” of B (the blocks Bg) form a partition of Ω, moreover |B| = |Bg| for all g ∈ G hence |B| divides |Ω| and the partitions consists of |Ω|/|B| blocks. Calling a = |B| and b = |Ω|/|B| we obtain that if G acts transitively n = ab, therefore we immediately deduce that any transitive group of prime degree is primitive (see for example our discussion of THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP 5
S5 above). The set of translates {Bg : g ∈ G} is called a block system of G. G is said to be primitive if it does not admit any nontrivial block, and imprimitive otherwise. Observe that if n > 2 and G is primitive then it is transitive because the G-orbits are blocks and n > 2 (observe that if n = 2 then all the subsets of {1, 2} are blocks for G independent of G).
Easy example: consider σ = (123456) and G = hσi < S6, as a permutation group of degree 6. Looking at σ2 = (135)(246) and σ3 = (14)(25)(36) we realize that the nontrivial imprimitivity blocks of G are {1, 3, 5}, {2, 4, 6} (which form one block system) and {1, 4}, {2, 5}, {3, 6} (which form another block system). This is how the cycle (123456) acts on each non-trivial block system.
1 / 2 1 gN / 2 / 3 W.. NNN ppp . NNN ppp . ppNNN . . ppp NN Ð . wppp NN 3 .. / 4 4 / 5 / 6 .. . .. Ð 5 / 6 Proposition 3. Suppose n > 2 and G acts transitively. G is primitive if and only if the point stabilizer M = Gα (for any α ∈ Ω) is a maximal subgroup of G.
Proof. Observe that if B is a block containing α then GB = {g ∈ G : Bg = B} (the setwise stabilizer of B) is a subgroup of G containing Gα. Indeed if g ∈ Gα then α ∈ B∩Bg hence B = Bg since B is a block. This proves that if Gα is maximal then G is primitive. Conversely if M < K < G the set B = {Mk : k ∈ K} is a nontrivial block for G. Indeed if g ∈ G and Mk ∈ B ∩ Bg there is some t ∈ K with −1 Mk = Mtg so g ∈ t Mk ⊆ K and this implies Bg = B. An easy example of a primitive group is h(1 . . . p)i acting on {1, . . . , p} where p is a prime. The point stabilizer is {1}.
Exercise: G ≤ Sn is called 2-transitive on Ω = {1, . . . , n} if it is transitive and its point stabilizer has precisely two orbits on Ω. Show that any 2-transitive group is primitive. Show that the converse does not hold (consider the dihedral group of prime degree p).
5. Maximal imprimitive subgroups
If H and K are two groups and K ≤ Sn, then H o K denotes the wreath product between H and K, i.e., the semidirect product Hn o K, where K acts on Hn by permuting the coordinates. More specifically π ∈ K acts on Hn by π (x1, . . . , xn) = (x1π−1 , . . . , xnπ−1 ). This may look strange but it is necessary to have a well-defined action on the right, indeed defining ti = xiπ−1 we have tiτ −1 = xiτ −1π−1 = xi(πτ)−1 hence πτ τ πτ (x1, . . . , xn) = (x1π−1 , . . . , xnπ−1 ) = (x1(πτ)−1 , . . . , xn(πτ)−1 ) = (x1, . . . , xn) . Recall that exponentiating by π means conjugation (π−1gπ). The following result is due to Frobenius. 6 MARTINO GARONZI
Theorem 1. Let H be a subgroup of the finite group G, let x1, . . . , xn be a right transversal for H in G, and let ξ be any homomorphism with domain H. Then the map f : G → ξ(H) o Sn given by −1 −1 x 7→ (ξ(x1xx1π ), . . . , ξ(xnxxnπ ))π, where π ∈ Sn is the unique permutation that satisfies xix ∈ Hxiπ for all i = 1, . . . , n, is a well-defined homomorphism with kernel equal to the normal core (ker ξ)G.
Proof. Since xi ∈ Hxi the permutation corresponding to the identity is 1 hence −1 −1 f(1) = 1. Now let x, y ∈ G and assume xixxiπ ∈ H, xiyxiτ ∈ H for all i = −1 1, . . . , n, then applying the second to iπ we find xiπyxiπτ ∈ H for all i = 1, . . . , n, −1 −1 −1 −1 so xixyxiπτ = (xixxiπ )(xiπ yxiπτ ) ∈ H. It follows that the permutation corre- sponding to xy is πτ and −1 −1 −1 f(xy) = (ξ(xixyxiπτ ))πτ = (ξ(xixxiπ )ξ(xiπyxiπτ ))πτ −1 −1 −1 = f(x) · π (ξ(xiπyxiπτ ))πτ = f(x)(ξ(xiyxiτ ))τ = f(x)f(y). f(x) = 1 if and only if the permutation π corresponding to x is the identity and −1 xixxi ∈ ker(ξ) for all i = 1, . . . , n, in other words x belongs to the normal core of ker(ξ) in G, because x ∈ HG and ker(ξ) E H.
Now assume G ≤ Sn acts imprimitively on Ω = {1, . . . , n}. This means that there is a nontrivial imprimitivity block B for G, let a = |B|. Let H = {g ∈ G : g(B) = B}, the setwise stabilizer of B. Observe that G acts transitively on the set of blocks {Bg : g ∈ G} with H as point stabilizer, so |G : H| equals the number of translates of B, call it b. Since the translates of B partition Ω we have ab = n. ∼ Of course we have a homomorphism ξ : H → Sym(B) = Sa induced by the action of H on B. By Theorem 1 we deduce a homomorphism f : G → ξ(H) o Sb ≤ Sa o Sb with kernel the normal core of ker(ξ) in G. Observe that if h ∈ ker(ξ) then h fixes B pointwise, and if h ∈ ker(ξ)g then ghg−1 ∈ ker(ξ) so h fixes Bg pointwise. This implies that (ker(ξ))G = {1} hence f is injective. Also, we may restrict the codomain of f to ξ(H) o K where K is the largest subgroup of Sb for which the restriction makes sense. ξ(H) and K could be called “components” of G.
This means that we can always embed any imprimitive group G into the wreath product of the so-called “primitive components” of G. Specifically, we start with a block B whose setwise stabilizer acts primitively on it (that is, a “minimal block”), meaning that ξ(H) acts primitively on B, then we apply the above construction giving G ≤ ξ(H) o K with K ≤ Sb transitive of degree b, and we repeat the process with K. This gives an embedding in the so-called iterated wreath product
G ≤ P1 o P2 o P3 o ... o Pk where the notation is A o B o C = A o (B o C).
For example the dihedral group of order 8 (inside S4), D = h(1234), (24)i acts imprimitively on {1, 2, 3, 4} having B = {1, 3} as a block. The above argument shows that D embeds into C2 o C2 and this actually shows that D and C2 o C2 are isomorphic (they have the same order). THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP 7
The full wreath product Sa oSb embeds into Sn (where n = ab) as an imprimitive subgroup. Actually it is a maximal imprimitive subgroup, meaning that it is not properly contained in any imprimitive subgroup of Sn, in other words if Sa o Sb is contained in an imprimitive subgroup Sc o Sd then a = c and b = d.
We will see that if a ≥ 5 then Sa oSb is also a primitive group (abstractly) but the degree of primitivity is much larger: ab (the point stabilizer being the normalizer b of (Sa−1) . This falls into a broader concept which is the following: the primitive groups of degree n, other than An and Sn, are “small” (their order is less than 4n, as proved by Wielandt 1969, Praeger and Saxl 1980, taken to 2n by Mar´otiin 2002).
Exercise: let n = pk be a prime power. The iterated wreath product P = Cp o Cp o ... o Cp of k copies of Cp is isomorphic to the Sylow p-subgroups of Sn.
Exercise: let n = mpk with m not divisible by p. Let Ω = {1, . . . , n} and let P be a Sylow p-subgroup of Sn. Show that • The action of P on Ω is transitive if and only if m = 1. • The action of P on Ω is primitive if and only if m = k = 1.
6. Intransitive maximal subgroups The subgroups defined by the action on two orbits A, B whose union is {1, . . . , n} are called “maximal intransitive subgroups” and have the shape H = Sa ×Sb where a = |A|, b = |B|, a+b = n. We study the maximality of G = Sa ×Sb inside Sn. If G is not maximal then it is properly contained in K < Sn which therefore is transitive on Ω (to take t ≤ a to r > a use something taking the first orbit to the second). Suppose K is imprimitive. Then clearly A = {1, . . . , a} and B = {a + 1, . . . , n} are contained in maximal imprimitivity blocks, however since they are orbits they are maximal imprimitivity blocks. This is only possible if a = b = n/2, hence Sa × Sb is maximal whenever a 6= b and a + b = n. If a = b = n/2 then Sa × Sb = Sa × Sa is not maximal: it is properly contained in the wreath product Sa o S2, which (as we will see) is a maximal subgroup of S2a (imprimitive). If K is primitive then it contains a 2-cycle and Jordan theorem (see below) implies that K = Sn.
7. Primitive maximal subgroups 7.1. Characteristically simple groups. A group G is called characteristically simple if its only characteristic subgroups are {1} and G. For example if N is a minimal normal subgroup of a finite group G then N is characteristically simple (because characteristic in normal implies normal). Proposition 4. If S is a nonabelian simple group the normal subgroups of Sn are its subproducts Si1 × · · · × Sim where m ≤ n and {i1, . . . , im} ⊆ {1, . . . , n}. n Proof. Let N be a normal subgroup of S with a nontrivial element g = (s1, . . . , sn) and suppose s1 6= 1. Then conjugating with (x, 1, 1,..., 1) we find that N contains x −1 −1 all the elements gx = (s1 , s2, . . . , sn), so that N 3 gxg = ([x , s1], 1, 1,..., 1). −1 Let t = [x , s1]. Of course t 6= 1 at least for some x (being S simple) so the conjugates of t generate S hence N contains S ×{1}×...×{1}. The same argument 8 MARTINO GARONZI
n shows that N contains the i-th factor of S whenever si 6= 1, and proves the claim. Proposition 5. Let G be a finite group. G is characteristically simple if and only if there exist a simple group S and a natural number n such that G =∼ Sn.
Proof. See [5, Theorem 8.10]. Exercise: let M be a maximal subgroup of a finite solvable group. Prove that M has prime power index (Hint: by induction on |G|; let N be a minimal normal subgroup of G, if M ≥ N then work in G/N and use induction, otherwise observe that N is characteristically simple).
Exercise: if |G| ≥ 3 then Aut(G) 6= {1}.
Exercise: Let G be a finite group. Then G is elementary abelian if and only if the natural action of Aut(G) on G − {1} is transitive.
Exercise: Let T be a nonabelian simple group. Prove that Aut(T n) =∼ Aut(T ) o Sn. Use the embedding theorem with H = NAut(T n)(R) → Aut(R) where R = T × {1} × · · · × {1}.
7.2. Primitive groups. A finite group G is called primitive of degree n if it admits a maximal subgroup M of index n such that MG = {1}. We shall see such group as a permutation group by means of its right multiplication action on Ω = {Mx : x ∈ G}. The corresponding homomorphism γ : G → Sym(Ω) is injective because its kernel is ∼ MG = {1}, and the permutation group γ(G) = G on Ω is primitive in the sense that it has no nontrivial blocks (as we have seen). We list some properties of such a group. First observe that if A, B are minimal normal subgroups of a group X and A ∩ B = {1} then ab = ba for all a ∈ A, b ∈ B. Indeed aba−1b−1 = a(ba−1b−1) = (aba−1)b−1 belongs to both A and B. Also, it is easy to see that if N is any normal subgroup of a group A then its centralizer in A is also a normal subgroup of A. Recall also that if H ≤ Sn acts regularly then |H| = n. Now let G be a primitive group of degree n.
(1) If H ≤ G is transitive then CG(H) is semiregular. Indeed if g ∈ CG(H) fixes α ∈ Ω then whenever h ∈ H we have (αh)g = (αg)h = αh therefore g fixes αh for all h ∈ H; being H transitive g fixes every ω ∈ Ω hence g = 1. (2) If N is a nontrivial normal subgroup of G then it is transitive. Indeed if ω ∈ Ω then ωN is a block: if g ∈ G and n ∈ N with ωng ∈ Nω then there is m ∈ N with ωng = ωm therefore if ` ∈ N then ω` = ωngm−1` = ωn(gm−1`g−1)g ∈ ωNg, this implies ωNg ⊆ ωN hence they are equal (having the same size). (3) G has at most two minimal normal subgroups. Indeed if N is a minimal normal subgroup then it is transitive, L = CG(N) is normal in G (hence transitive if non-trivial) and semiregular so it is regular if non-trivial, and it contains every minimal normal subgroup of G distinct from N because the intersection between two minimal normal subgroups is trivial. However N is contained in CG(L) hence N, CG(L) are both regular so N = CG(L) THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP 9
(they both have size n). If N = L then N is the unique minimal normal subgroup of G, otherwise the two minimal normal subgroups are N and L. (4) If G has two minimal normal subgroups then N is nonabelian. Indeed if N is abelian then N ≤ CG(N) and |N| = |CG(N)| = n hence N = CG(N), in particular there cannot be other minimal normal subgroups. Exercise: Let S be a group, prove that ∆ = {(s, s): s ∈ S} is a maximal subgroup of S × S if and only if S is a simple group. Moreover in this case S × S is a primitive group and |S| is its unique primitivity degree (meaning that every core-free maximal subgroup of S × S has index |S|).
If a minimal normal subgroup N of G is abelian then it is the unique minimal normal subgroup, let us study the structure of G in this case. The core-free maximal subgroup M intersects N trivially because M ∩ N is normal in M (being N normal in G) and it is normal in N (being N abelian) hence it is normal in MN, however MN = G because M is maximal and does not contain N (being MG = {1}). By minimality of N since M ∩ N E G we deduce M ∩ N = {1} hence G is a semidirect m product N o H. Since N is characteristically simple N = Cp for some prime p and some positive integer m, hence we may see N (additively) as a vector space of dimension m over Fp. Since there are no other minimal normal subgroups H acts faithfully on N by conjugation and minimality of N translates into irreducibility of the linear action of H, in other words H an irreducible subgroup of GL(m, p). Conversely it is easy to see that if V is a finite vector space over the field Fp (p is a prime!) and H is an irreducible subgroup of GL(V ) then V o H with the natural action of H is a primitive group whose only primitivity degree is |V | (which in particular is a prime power). Such type of group is called an affine group.
As a consequence we obtain the following. Suppose the group G has nontrivial center. Then G is primitive if and only if G is cyclic of prime order p, which therefore is its only primitivity degree. To see this let N be a minimal normal subgroup of G contained in the center of G (so that N has prime order p), then by the above discussion G is a semidirect product N o M which therefore is a direct product being N central, G = N × M. It follows that M E G, in other words ∼ M = MG, however MG = {1} by assumption so M = {1} and G = N = Cp.
On the opposite side of the spectrum N is nonabelian, in which case N = T n for some non-abelian simple group T . Suppose CG(N) = {1}, then the conjugation action of G on N embeds G in Aut(N), so that N ≤ G ≤ Aut(N). If n = 1 then N = T is a nonabelian simple group. In this case G is called almost- simple. Equivalently, an almost-simple group is a group G admitting a nonabelian simple normal subgroup T with the property that CG(T ) = {1}. Equivalently, G lies between S and Aut(S). An obvious example of almost-simple group is the symmetric group itself, Sn, when n ≥ 5, indeed An is a simple non-abelian normal subgroup of Sn and CSn (An) = {1}.
Proposition 6. Let n ≥ 5 and m a positive integer. G = Sn o Sm (the wreath product) is a primitive group of degree nm. m Proof. First we prove that G = Sn oSm = (Sn) oSm is a primitive group. For this m define N = An , it is a non-abelian normal subgroup of G (being a ≥ 5), and it is a minimal normal subgroup because since An is a non-abelian simple group (being 10 MARTINO GARONZI a ≥ 5) the normal subgroups of N are its subproducts and they are not normal in G because of the fact that Sm acts transitively on the set of the b factors, which are the minimal normal subgroups of N. Moreover CG(N) = {1}. To see this m first observe that CG(N) ≤ Sn because any element with a nontrivial component in Sb does not centralize all the elements that have 1 in all positions except one, m so CG(N) ≤ CSn (An) but CSn (An) = {1} (because of the normal structure of the symmetric group) hence CG(N) = {1}. This implies that N is the unique minimal normal subgroup of G: any other minimal normal subgroup would have to centralize N. We are left to construct a maximal subgroup M with trivial normal core and of b m index a . Let M := NG(K) where K = Sn−1, being Sn−1 any point stabilizer in
Sn. We have M ∩ N = NN (K) = K being NSn (Sn−1) = Sn−1 being Sn−1 maximal in Sn. If M is a maximal subgroup of G then since it does not contain N its normal core is trivial, being N the unique minimal normal subgroup of G, also MN = G for the same reason hence
m m m |G : M| = |MN : M| = |N : M ∩ N| = |Sn : Sn−1| = n . We are then left to show that M is a maximal subgroup of G. If H is a maximal subgroup of G containing M then H ∩N contains K, however it must be equal to K because since NH = G and N acts (by conjugation) trivially on its direct factors, H acts transitively on the factors of N hence it induces isomorphisms between the projections πi(H ∩ N) (where πi : N → {1} × · · · × {1} × An × {1} × · · · × {1}, the i-th factor), which therefore are all isomorphic and contain Sn−1 (for more details see below, O’Nan-Scott theorem). Since Sn−1 is maximal in Sn we deduce that H ∩ N = K. Since H ∩ N is normal in H, H is then contained in the normalizer NG(K) which equals M, so H = M.
For example S5 oS2, which is imprimitive of degree 5·2 = 10, is a primitive group of degree 52 = 25. Observe that nm is comparatively much smaller than nm. This is because if a group is primitive, its primitivity degrees are “large” (in a sense to be defined).
7.3. Multiple transitivity. Obvious examples of primitive groups are Sn and An (of degree n), for this observe that the point stabilizers are maximal in both cases: they have index n and we have seen that Sn has no proper subgroup of index less than n other than An, similarly An has no proper subgroups of index less than n.
Let G ≤ Sn act naturally on Ω = {1, . . . , n}. For 1 ≤ k ≤ n define Ok(Ω) to be the set of k-tuples (α1, . . . , αk) of pairwise distinct elements of Ω. We have a natural action of G on Ok(Ω). We say that G is k-transitive if it acts transitively on Ok(Ω). We say that G is sharply k-transitive if it acts regularly on Ok(Ω). Since |Ok(Ω)| = n(n − 1) ··· (n − k + 1) this is the order of any sharply k-transitive group of degree n, and it is a divisor of the order of any k-transitive group. Obviously if G is k-transitive then it is m-transitive for all 1 ≤ m ≤ k.
Proposition 7. Let k ≥ 2. If G acts faithfully and k-transitively on a set of size n ≥ 3 then the action is primitive. THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP 11
Proof. Since G is 2-transitive we may assume k = 2. Let B ⊆ X be such that |B| > 1 and B 6= X, and take x, y ∈ B distinct and z 6∈ B. Since G acts 2- transitively there exists g ∈ G with xg = x and yg = z, so B is not an imprimitivity block for G.
Proposition 8. G ≤ Sn is (n − 2)-transitive if and only if G = An or G = Sn.
Proof. Clearly Sn is n-transitive and An is (n − 2)-transitive (if you don’t believe: exercise). Also |On−2(Ω)| = n(n − 1) ··· (n − (n − 2) + 1) = n!/2 hence An is actually sharply (n − 2)-transitive. Since An is the only subgroup of Sn of index 2, it follows that An is the only sharply (n−2)-transitive group of degree n. Moreover if G ≤ Sn is (n − 2)-transitive then |On−2(Ω)| = n!/2 divides |G|, so for the same reason G = An or G = Sn.
Example: the normalizer in S5 of a Sylow 5-subgroup of S5 is sharply 2-transitive of degree 5.
Lemma 1. Let 1 ≤ k ≤ n. The transitive group G ≤ Sn is k-transitive if and only if the stabilizer of α ∈ Ω in G is (k − 1)-transitive on Ω − {α}. ∼ Exercise: if G is a solvable 4-transitive permutation group then G = S4. Hint: show that any minimal normal subgroup N is regular and study the conjugation action of a point stabilizer on N (N is a vector space over a field with p elements, p prime, and the conjugation action of the point stabilizer on N − {1} is 3-transitive and linear). 7.4. Jordan groups. Let G act faithfully on a set Ω. A subset ∆ of Ω is called a Jordan set if the pointwise stabilizer of Ω − ∆ acts transitively on ∆. Let G(∆) be such pointwise stabilizer. Since any Jordan set is contained in a G-orbit we may assume G is transitive on Ω. Obviously Ω itself is a Jordan set and every one element subset of Ω is also a Jordan set. Ω and the one-element subsets of Ω are called “improper” Jordan sets, all the others are called “proper”. If G acts k-transitively then ∆ ⊆ Ω is a Jordan set whenever |Ω − ∆| < k.
If two Jordan sets ∆1, ∆2 intersect then their union is a Jordan set because the pointwise stabilizer of Ω − (∆1 ∪ ∆2) contains hG(∆1),G(∆2)i. Therefore if ∆1 is a maximal Jordan set then one of the following occurs: ∆1 ∩ ∆2 = ∅, ∆2 ⊆ ∆1 or ∆1 ∪ ∆2 = Ω. If B ⊆ Ω is an imprimitivity block of G and ∆ is a Jordan set with B ∩ ∆ 6= ∅ then one of B and ∆ contains the other. Indeed if α ∈ B ∩ ∆ and there exist β ∈ B − ∆ and γ ∈ ∆ − B then a permutation mapping α to δ cannot fix β (by definition of block), and this contradicts the fact that ∆ is a Jordan set. Theorem 2 (Jordan’s Theorem). A finite primitive group having a proper Jordan set is 2-transitive. Proof. Suppose G is primitive on Ω. If ∆ is a proper subset of Ω with |∆| > 1 let B = {β ∈ Ω: β ∈ ∆g ⇔ α ∈ ∆g ∀g ∈ G}. Then B is an imprimitivity block for G: if γ ∈ B ∩ Bg then γ, γg−1 ∈ B, so if h ∈ G then γ ∈ ∆h if and only if γg−1 ∈ ∆h, if and only if α ∈ ∆h, if and only if γg−1 ∈ ∆h, if and only if γ ∈ ∆hg, if and only if α ∈ ∆hg, if and only if δ ∈ ∆hg, 12 MARTINO GARONZI if and only if δg−1 ∈ ∆h. Since B 6= Ω and G is primitive, B = {α}. Now call n = |Ω|, and let ∆ be a maximal Jordan set of G, with |∆| = k. By the above observation for any α ∈ Ω the translates of ∆ not containing α cover Ω − {α} (if not some β would be left uncovered hence α ∼ β), so since no two of them have union Ω (their union cannot contain α) their are pairwise disjoint hence k divides n−1. Since G is transitive any point α ∈ Ω is outside the same number (n−1)/k of translates of ∆. The total number of translates of ∆ is therefore n(n − 1)/k(n − k) (multiplying n(n − 1)/k would count each point n − k times). Since k divides n − 1 both k and n − k are coprime to n (writing kt = n − 1 we have n − kt = 1 and (n − k)t − n(t − 1) = 1). This implies that k(n − k) divides n − 1, and this can only happen if k = 1 or n − k = 1. If k = 1 then ∆ is improper, so n − k = 1 which means that the stabilizer of one point acts transitively on the other points, in other words the action of G on Ω is 2-transitive. The more general version of Jordan theorem is the following (see Isaacs, Finite Group Theory).
Theorem 3 (Jordan). Suppose G ≤ Sn acts primitively on Ω = {1, . . . , n} and ∆ is a Jordan set for G. If |∆| ≥ 2 and G(∆) acts primitively on ∆ then the action of G on Ω is (|Ω − ∆| + 1)-transitive. We will not prove this theorem. Instead we will study some of its consequences.
If G ≤ Sn acts primitively on Ω = {1, . . . , n} and contains a transposition (α, β) then G = Sn. Indeed {α, β} is a Jordan set of G and the action on G on {α, β} is primitive because 2 is a prime number. By Jordan theorem the action of G is (n − 1)-transitive hence |On−1(Ω)| = n! divides |G| so G = Sn.
Similarly if G ≤ Sn acts primitively on Ω = {1, . . . , n} and contains a 3- cycle (α, β, γ) then G = An or G = Sn. Indeed {α, β, γ} is a Jordan set of G and the action on G on {α, β, γ} is primitive because 3 is a prime number. By Jordan theorem the action of G is (n − 2)-transitive hence |On−2(Ω)| = n!/2 divides |G| so G ≥ An.
More in general we have:
Theorem 4. Let G ≤ Sn act primitively on Ω = {1, . . . , n} and assume G contains a p-cycle γ, where p is a prime such that p ≤ n − 3. Then G = An or G = Sn. Proof. The cases p = 2 and p = 3 have already been considered above, so now assume p ≥ 5. Let ∆ be the set of points moved by γ, so that |∆| = p. ∆ is a Jordan set and since p is a prime G(∆) acts primitively on ∆. By Jordan theorem we deduce that G is (n − p + 1)-transitive, so it is (n − p)-transitive. Let H ≤ G be the setwise stabilizer of Ω − ∆, then H/G(∆) is isomorphic to Sym(Ω − ∆) (this is because any permutation of Ω − ∆ is induced by an element of G, being G (n − p)-transitive and |Ω − ∆| = n − p).
Let P = hγi ≤ G. Then P ≤ G(∆) and since G(∆) acts faithfully on ∆ we deduce that |G(∆)| divides p!. Therefore P is a Sylow p-subgroup of G(∆) and by the Frattini argument, since G(∆)EH, we obtain H = G(∆)N where N = NH (P ). Therefore N/N(∆) =∼ H/G(∆) =∼ Sym(Ω − ∆). Since |Ω − ∆| ≥ 3 the symmetric THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP 13 group of Ω − ∆ contains a 3-cycle, which therefore belongs to the derived subgroup of Sym(Ω − ∆) (the alternating group). Let then x ∈ N 0 = [N,N] be such that the permutation induced by x on Ω − ∆ is a 3-cycle.
Let C = CN (P ). Since the automorphism group of P is abelian (being P cyclic of order p), N/C is abelian, so x ∈ N 0 ≤ C hence x commutes with γ. So the permutation induced by x on ∆ commutes with a p-cycle acting on the p points of ∆. Since the centralizer of a p-cycle in Sp is generated by such p-cycle, we deduce that p divides the order of x and xp acts trivially on ∆. Since x induces a 3-cycle on Ω − ∆ and 3 does not divide p we deduce that xp induces a 3-cycle on Ω − ∆. Therefore xp fixes the elements of ∆ and acts as a 3-cycle on Ω−∆, so x is a 3-cycle belonging to G. We know that this implies that G ≥ An.
This easily implies that the maximal imprimitive subgroups of Sn are maximal subgroups of Sn. Indeed a proper subgroup properly containing them would be primitive and would contain a 2-cycle. Similarly the same is true for An (the proof of this is a bit more tricky).
7.5. Primitive actions: O’Nan-Scott. See also [6, Chapter 7].
We have seen that primitive groups with abelian socle are precisely the affine groups. Now let G be a primitive group with nonabelian socle N = soc(G). We know that N is a product of at most two minimal normal subgroups, but in this discussion we will assume N itself is a minimal normal subgroup of G. We know that N = Sm for some nonabelian simple group S and some positive integer m.
Let G be a primitive monolithic group with socle N = soc(G) = T1 × · · · × Tm = m T , where T is a nonabelian simple group. We apply Theorem 1 to H = NG(T1) and ξ : H → Aut(T1), the conjugation action. Observe that ker(ξ) = CG(T1) and since the conjugation action of G on N permutes the factors T1,...,Tm, the normal core of CG(T1) in G is CG(N) = {1}. Define X := NG(T1)/CG(T1), which ∼ is an almost-simple group with socle T := T1CG(T1)/CG(T1) = T1. Obviously X is m isomorphic to the image of ξ. The minimal normal subgroups of T = T1 ×...×Tm are precisely its factors T1,...,Tm. Since automorphisms send minimal normal subgroups to minimal normal subgroups, it follows that G acts on the m factors of N. Let ρ : G → Sm be the homomorphism induced by the conjugation action of G on the set {T1,...,Tm}. The group K := ρ(G) is a transitive permutation group of degree m. By Theorem 1 G embeds in the wreath product X o K.
Let X be an almost-simple group with socle S and let K be a faithful transitive group of degree m. The wreath product G = X o K is itself a primitive monolithic group. To see this let N = soc(G). We argue that N is the unique minimal normal subgroup of G. Indeed another minimal normal subgroup would lie in CG(N) so it is enough to show that CG(N) = {1}. Let (x1, . . . , xn)π lie in CG(N). By considering the conjugates of (s, 1,..., 1) where 1 6= s ∈ S we see that π = 1 and if i ∈ {1, . . . , n} then xi ∈ CX (s) for all s ∈ S hence xi ∈ CX (S) = {1}.
This means that any primitive group G with nonabelian socle is a subgroup of X o K where X is an almost-simple group with socle S, K is a transitive group 14 MARTINO GARONZI of degree m and G projects surjectively onto K. Of course these are not all the conditions G must satisfy to be primitive, but they are necessary conditions.
Let G be a primitive group with nonabelian socle N = Sm. What we want to do now is to study the primitive actions of G. Let M be a maximal subgroup of G with trivial normal core, so that MN = G. Then N ∩ M is normal in M. We claim that N ∩ M is a maximal proper M-invariant subgroup of N. Indeed if N ∩ M < L < N is M-invariant then M < LM < G, contradicting the maximality of M. Indeed: • If M = LM then L ≤ M ∩ N a contradiction. • If LM = G then L E G contradicting the fact that N is a minimal normal subgroup of G. We want to show that the maximal subgroups M of G which supplement the socle (i.e. MN = G) are of the following three types: • Complement. M ∩ N = {1} so that G is a semidirect product N o M. m • Product type. A conjugate of NG(H ) where H is the intersection be- tween S and a maximal subgroup of X. • Diagonal type. M = NG(∆1 × ... × ∆l) where each ∆i is a diagonal and m/l is a prime number.
NB: the subgroups of diagonal type NG(∆1 × ... × ∆l) with m/l not a prime are not maximal. Actually each of them is contained in the normalizer of a refined product of diagonals. NB: the primality of m/l is not sufficient to establish that the diagonal type subgroup is maximal. But if it is not maximal then it is contained in a maximal subgroup which contains the socle. Take a maximal subgroup M of G supplementing the socle N, i.e. such that m MN = G. Call π1, ..., πm the projections πi : S = S1×...×Sm → Si. Observe that since N is a (the) minimal normal subgroup of G and the normalizer in G of M ∩N is a subgroup of G containing M, either M complements N or NG(M ∩ N) = M. ∼ Proposition 9. πi(M ∩ N) = πj(M ∩ N) for every i 6= j.
Proof. Notice that M is transitive on the factors Si of N, because G is transitive, N acts trivially on each factor and G = NM. So there exists an element h of M such that the conjugation by h sends Si to Sj, so it determines an automorphism of M ∩ N which induces an isomorphism ∼ ∼ πi(M ∩ N) = M ∩ N/ ker(πi|M∩N ) / M ∩ N/ ker(πj|M∩N ) = πj(M ∩ N).
This proves the statement.
There are three possibilities for π1(M ∩ N).
(1) π1(M ∩ N) = {1}. This implies that πi(M ∩ N) = 1 for every i, so M ∩ N = 1. In other words M complements N, so G = N o M and the primitivity degree is |N| = |S|m.
(2) π1(M ∩ N) = S. Then each πi|M∩N is surjective. Consider
Dij = πi(ker(πj|M∩N ))
for every i, j. Observe that Dij is a normal subgroup of Si. Indeed if x ∈ ker(πj|M∩N ) and s ∈ Si then there exists t ∈ M ∩ N with πi(t) = s THE MAXIMAL SUBGROUPS OF THE SYMMETRIC GROUP 15
−1 −1 −1 −1 hence s πi(x)s = πi(t xt) and πj(t xt) = πj(t) πj(x)πj(t) = 1 being −1 πj(x) = 1. This proves that s πi(x)s ∈ Dij so that Dij E Si. Since Si is simple we have two cases: Dij = Si or Dij = {1}. Observe that Dii = {1}. Fix i and assume that Dij = {1} for all j ∈ {1, . . . , m}. This means that ker(πj|M∩N ) = {1} for all j, so the projections ϕj = −1 πj|M∩N : M ∩ N → Sj are all isomorphisms. Let αij := ϕj ◦ ϕi , it is an isomorphism Si → Sj, and M ∩ N is contained in {x ∈ N : αij(πi(x)) = πj(x): ∀i, j}. By maximality of M ∩ N as proper M-invariant subgroup of N this proves that equality holds. In other words M ∩ N is a “diagonal α2 αm subgroup” {(s, s , . . . , s ): s ∈ S} where αi is an isomorphism S → S for all i. Up to conjugation in Aut(N) it follows that M ∩ N is isomorphic to {(s, s, . . . , s): s ∈ S}. In the general case M ∩N will be a direct product of diagonals as above. Specifically there exists a number l dividing m and l diagonals ∆1, ..., ∆l of length m/l such that M ∩ N = ∆1 × ... × ∆l (by maximality of M ∩ N as proper M-invariant subgroup of N it is enough to show ≤). In this case the maximality of M implies that m/l is a prime. To understand this it is sufficient to give an example: if m = 4, the normalizer of the diagonal ϕ ϕ ϕ ϕ ϕ−1ϕ (x, x 2 , x 3 , x 4 ) is contained in the normalizer of (x, y, x 3 , y 2 4 ).
∼ (3) 1 < π1(M ∩ N) < S. Let Vi := πi(M ∩ N) ≤ Si. We have Vi = Vj for every i, j, and every such isomorphism can be realized as the conjugation by an element of M. In this case M ∩ N ≤ V1 × · · · × Vm and it is possible ∼ to show that in fact equality holds, and V = Vi is the intersection between X and a maximal subgroup of S. [More details in the next version of the notes.] Using this we may list the primitivity degrees of a given primitive group. This is easy now because if G is a monolithic primitive group with socle N and point stabilizer M then being MN = G we have |G : M| = |N : N ∩ M| (being |G| = |MN| = |M||N|/|M ∩ N|). For example let S be a nonabelian simple group. The n n/p degrees of primitivity of G = S o Cn are |S : M| for M maximal in S and |S| where p is a prime dividing n (see the diagonal type case). Observe that in principle |N| could be a primitivity degree, you need to check the specific case. For example 2 2 2 the degrees of primitivity of A5 o C2 are 5 = 25, 6 = 36, 10 = 100 and |A5| = 60 2 (|A5| is not a primitivity degree because the complements of the socle are not maximal: exercise).
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