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THE MAXIMAL SUBGROUPS of the SYMMETRIC GROUP Lecture 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) = fxg : g 2 Gg for any x 2 X. The action is called transitive if there is some x 2 X such that O(x) = X (i.e. there is only one orbit), in other words for any x; y 2 X there exists g 2 G with xg = y. The stabilizer of x 2 X is Gx = fg 2 G : xg = xg ≤ G. For example consider the action by right multiplication of G on X = fHt : t 2 T g g −1 Gg, the kernel is HG = g2G 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 2 X then jXj = jG : Gxj. 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 jGj = jXj. An example of a regular action is that of the Klein 4- group acting naturally on f1; 2; 3; 4g. 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 2 CG(N) is such that xg = x for some x 2 X then since every element of X has the form xn for some n 2 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 2 G 1 P acting on X. Then the number of orbits of the action is jGj g2G fg. For example if G acts semiregularly the number of orbits is jXj=jGj because f1 = jXj and fg = 0 for every 1 6= g 2 G. Therefore if the action is regular (semiregular and transitive) then jGj = jXj. 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 f1g < An < Sn and the normal subgroups ∼ of S4 are f1g < 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 jSn : Hj ≥ n. If H < An then jAn : Hj ≥ n. Proof. Let m = jSn : Hj. Sn acts transitively (hence non-trivially) by right multi- plication on X = fHx : x 2 Sng, which is a set of size m, this gives a homomor- phism ' : Sn ! Sm whose image is a transitive subgroup of Sm. Let K = ker('). Since H 6= An we have m ≥ 3 so jSn : Kj ≥ 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 = f1g. It follows that jSnj = n! divides jSmj = m! hence n ≤ m. A similar argument applies to show that if H < An then jAn : Hj ≥ 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 2 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 ' : Sn ! Sn, however '(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 '(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 '(x) is also a 3-cycle (x centralizes some An−4 so K = '(An−4) is centralized by the element '(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 '(x) must be a 3-cycle). Now show that '((123)), . '((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 '((12r)), '((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 ' : S5 ! S6 with transitive image ∼ H = '(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) = f1g 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) 2 Sn. The centralizer of σ in Sn is hσi. 3. The maximal subgroups of S5 Recall that the unique normal subgroups of Sn are f1g, 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 jG : Hj = n (counting principle - orbit-stabilizer lemma), the action is equivalent to the right multiplication action of G on fHx : T g x 2 Gg and setting HG = g2G H the normal core of H in G, the quotient G=HG is isomorphic to a subgroup of the symmetric group Sn, in particular jG : HGj 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 = fx : x 2 2 Sng (because (xN) = N in the quotient G=N).
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