Overgroups of Intersections of Maximal Subgroups of the. Symmetric Group. Jeffrey Kuan

Transcription

1 Overgroups of Intersections of Maximal Subgroups of the Symmetric Group Jeffrey Kuan Abstract The O Nan-Scott theorem weakly classifies the maximal subgroups of the symmetric group S, providing some information about the lattice Λ of subgroups of S. We wish to obtain more information about Λ. We proceed by determining the intersection H of suitable pairs of maximal subgroups of S and the sublattice of overgroups of H in S. 1 Introduction A lattice is a partially ordered set such that any two elements x and y have a least upper bound x y and a greatest lower bound x y. For example, the set of subgroups of a group is a lattice when ordered by inclusion. Let Ω be a set with n elements and S = Sym(Ω) be the group of permutations of Ω. The maximal subgroups of S are known in a weak sense by the O Nan-Scott Theorem. For example, the stabilizers of proper nonempty subsets not of order n/2 are maximal, as are the stabilizers of regular partitions, where a partition is regular if each of its blocks is the same size. All other maximal subgroups are primitive, where a group G is primitive if the only partitions stablized by G are the trivial partitions 0 and. See Dixon and Mortimer, Permutation Groups, for a more comprehensive handling of the O Nan-Scott Theorem. To obtain more information about the lattice of subgroups of S, we will look at the overgroups of the intersection of two maximal subgroups of S. As a first step, we examine the case when the two maximal subgroups are subset stabilizers. We will prove a theorem classifying all overgroups in this case. 1

2 2 Preliminaries The set of partitions of Ω, which we will denote P, is a partially ordered set, where P Q if a Q b implies a P b, where P is the equivalence relation defined by P. Indeed P is a lattice: For P, Q P, we have P Q = {A B A P, B Q, and A B } and P Q is the partition such that P Q is generated by P and Q. P Q and P Q are also referred to as the join and meet of P and Q. Set 0 = {Ω} P and = {{ω} ω Ω} P. Then 0 is the least element of P and is the greatest element. Lemma 2.1. If G S, then (1) G is a group of automorphisms of the lattice of subsets of Ω by ga = {ga a A}, for g S and A Ω. (2) G is a group of automorphisms of the lattice of partitions of Ω by gp = {ga A P } for g S and P P. Proof. (1) For any g G and A, B Ω, we need to show that g(a B) = ga gb and g(a B) = ga gb. Since g is a permutation of the power set of Ω, then g is a bijection, so g(a B) = g(a B) = ga gb = ga gb and g(a B) = g(a B) = ga gb = ga gb. (2) Let g G and P, Q P. Since P Q = {A B A P, B Q, A B }, then g(p Q) = {g(a B) A P, B Q, A B } = {ga gb A P, B Q, A B } = {A B A gp, B gq, g 1 A g 1 B } = gp gq 2

3 Fix x P Q. Then y P Q iff there exist x 1, x 2,..., x k such that x P x 1 Q x 2 P... Q x k P y. But this happens iff gx gp gx 1 gq gx 2 gp... gq gx k gp gy. Thus x P Q y iff gx gp gq gy, so g(p Q) = gp gq. Given P P and G S, we can then define N G (P ) = {g G gp = P } and C G (P ) = {g G ga = A for all A P }. For A Ω, define N G (A) = {g G ga = A} and C G (A) = {g G ga = a for all a A}. Also write A for the complement Ω A to A in Ω. From these definitions, we get Lemma 2.2. If A Ω and P P, then (1) C S (A) = Sym(A). (2) N S (A) = N S (A) = C S (A) C S (A) = Sym(A) Sym(A). (3) C S (P ) = C S (X). X P (4) C S (P ) is the kernel of the action of N S (P ) on P, so C S (P ) N S (P ). Proof. (1) Define a map φ : C S (A) Sym(A) by φ(g)x = gx, for x A, and observe that φ is a homomorphism. If φ(g) = e, then gx = x for all x A. Since g C S (A), then we also have ga = a for all a A. So g = e, therefore φ is injective. If σ Sym(A), then define g C S (A) by gx = σx for x A and ga = a for a A. Then φ(g) = σ. Therefore φ is surjective. (2) Observe that for g S, ga = ga. Thus N S (A) = N S (A). Then define a homomorphism φ : N S (A) Sym(A) Sym(A) by φ(g) (a, x) = (ga, gx) for a A and x A. If φ(g) = e, then ga = a and gx = x for a A and x A, so g = e. Therefore φ is injective. If (σ 1, σ 2 ) Sym(A) Sym(A), define g by ga = σ 1 a and gx = σ 2 x for a A and x A. Then φ(g) = (σ 1, σ 2 ), so φ is surjective. (3) Here (as in (2)), we are using the convention G is the direct product of subgroups G 1, G 2,..., G k if the map (g 1, g 2,..., g k ) g 1 g 2... g k is an isomorphism. 3

4 Write P = {X 1, X 2,..., X m }. Define a homomorphism φ : C S (X) X P C S (P ) by φ(σ 1, σ 2,..., σ m ) = σ 1 σ 2... σ m. Suppose that φ(σ 1, σ 2,..., σ m ) = e. Then σ 1 σ 2... σ m x = x for all x Ω. Given any i, pick ω X i. Since σ j C S (X j ), then σ j ω = ω for j i. Therefore σ i ω = ω, so σ i = e. This holds for any i, so (σ 1, σ 2,..., σ m ) = e, so φ is injective. For σ C S (P ), define σ i C S (X i ) by σ i x = σx. Then σ 1 σ 2... σ m x = σx, so φ(σ 1, σ 2,..., σ m ) = σ, so φ is surjective. (4) Follows from the definitions. If H S, let P(H) = {P P H N S (P )} and P (H) = P(H) P, where P = P {0, }. For A Ω, write P A for the partition {A, A} of Ω. If Q P and B Q, write P B for {A P A B}. Note that P B partitions B. For a partition P, write P for {A P A = 1} and P for {x : x A P }. Let P be the partition (P P ) { P }. Note that P P. Lemma 2.3. (1) For G S and P a partition, N G (P ) N G (P ). (2) Either P = or P = { P }. (3) If P P, then P P. Proof. (1) Suppose g N G (P ). Then gp = P, and since g must map blocks of size one to blocks of size one, g P = P and g P = P. Then gp = g((p P ) P ) = (P P ) P = P, so g N G (P ). (2) Suppose A P. Then A = 1 with A (P P ) { P }. If A P P, this would imply A P, which would then imply A P, which is a contradiction. Thus A { P }. Therefore A = P, so P = { P }. (3) Suppose P = 0. Then (P P ) { P } = {{Ω}}. But then either P P = {Ω} or P = {Ω}, both of which are clearly impossible. If P =, then P P, which forces P =, which is again a contradiction. 4

5 Theorem 2.4. Let A be a proper nonempty subset of Ω. If A n/2, then N S (A) = C S (P A ) is a maximal subgroup of S. If P is a nontrivial regular partition (i.e. all the blocks of P are the same size), then N S (P ) is a maximal subgroup of S. Thus if A = n/2, then N S (P A ) = Sym(A) S 2, which contains C S (P A ) as a subgroup of index 2, is the only proper overgroup of C S (P A ). All other maximal subgroups of S are primitive. Proof. See [Dixon and Mortimer, Permutation Groups, p.137] Theorem 2.5. If G is a primitive subgroup of S which contains a tranposition, then G = S. Proof. See [Wielandt, Finite Permutation Groups, p.34] In particular, this tells us that no maximal subgroup which does not stabilize a subset or regular partition can contain a tranposition. Lemma 2.6. Let P P, H = C S (P ), and Q P(H). Then (1) For each X Q Q and A P with A X, we have H N S (X) and A X. (2) Q P. Proof. (1) Let x A X. For any a A {x}, the transposition (a x) is in H, which stabilizes Q. Since X Q with X > 1, then a must also be in X. Therefore A X. To show that H N S (X), suppose h H and x X. Then x A for some A P, so we must have hx ha = A X. This holds for all x X, so hx = X, as needed. (2) We need to show that if X Q and A P satisfy A X, then A X. If X Q Q, then part (1) shows this. Otherwise X = Q. In this case, suppose a A X. Then {a} Q, so h{a} = {ha} Q for all h H. Thus ha Q = X, and since the orbit of a under H is A, this tells us that A X. 5

6 3 The Intersection of Subset Stabilizers Lemma 3.1. Let A and B be distinct nonempty proper subsets of Ω and set P A,B = P A P B. Then (1) P A,B = {A B, A B, A B, A B}. (2) N S (A) N S (B) = C S (P A,B ). Proof. (1) This follows immediately from the respective definitions. (2) For notation s sake, let H = N S (A) N S (B). Suppose g H. Since g N S (A), by Lemma we have that g N S (A), so ga = A and ga = A. Similarly, gb = B and gb = B. Therefore g C S (P A,B ). Now suppose g C S (P A,B ). Then ga = g((a B) (A B)) = g(a B) g(a B) = (A B) (A B) = A So g N S (A). Similarly, g N S (B), so g H. Lemma 3.2. Let P, Q P and Q P. Then C S (P ) C S (Q). Proof. Since Q P, every block of Q is a union of blocks of P. Thus every element of C S (P ) is also an element of C S (Q). Lemma 3.3. Let P be a nontrivial partition of Ω and H = C S (P ). Suppose M is a subgroup of S which is not primitive. Let P= {R P(M) : R P } and Q= {Q P (M) : Q P and C S (Q) M}. Then P has a greatest member Q(M, P ) and the following are equivalent: (1) Q. (2) M is an overgroup of H. 6

7 (3) Q(M, P ) P and C S (Q(M, P )) M. Proof. By Lemma 2.1, P(M) is a sublattice of P, so P is closed under joins. Since P contains 0, it is nonempty, so we can construct a greatest element by taking the joins of all elements. Now assume that (1) holds. If Q Q, then by Lemma 3.2 H = C S (P ) C S (Q) M, so (2) holds. If (3) holds, then it follows immediately that Q(M, P ) Q, so (1) holds. Now suppose (2) holds. Let Q = Q(M, P ). First, I prove the following: Lemma 3.4. For X Q, N M (X) is primitive on X. Proof. If the action were not primitive, then there exists a nontrivial N M (X)- invariant partition Σ of X. Construct R by replacing the orbit X M of X in Q by θ = {Y m : Y Σ, m M}, so R = (Q X M ) θ. To show that R is also a partition of Ω, we need to show that A = Ω and distinct blocks of R intersect A R trivially. Writing Θ for Y, we see that if ω Θ, then ω Y θ R for Y θ some Y, and if ω / Θ then ω A (Q X M ) R for some A. We next show that if A and B are distinct elements of R then A B =. If A and B are both in (Q X M ), then this holds since Q is a partition. If A (Q X M ) and B θ, then every element of A is in Ω Θ and every element of B is in Θ, so A B =. If A and B are both in θ, then A = m 1 Y 1 and B = m 2 Y 2 for some m 1, m 2 M and Y 1, Y 2 Σ. If x m 1 Y 1 m 2 Y 2, then m 1 X = m 2 X, so m 1 1 m 2X = X. Then m 1 1 m 2 N M (X), and since m 1 1 x Y 1 m 1 1 m 2Y 2, and because Y 1, m 1 1 m 2Y 2 are elements of the partition Σ, this tells us that Y 1 = m 1 1 m 2Y 2, as needed. By Lemma 2.3, parts (1) and (3), Σ is also a nontrivial N M (X)-invariant partition of X, so M N S (R). Clearly Q < R, so to get a contradiction it remains to show R P. Suppose A P. Since Q P, A is a subset of some 7

8 X Q. If X / X M, then X R. If X = X m for some m, then we need to show that A is a subset of Y m for some Y Σ. If A = 1, then this is trivial, so suppose A > 1. Assume A Y m is nonempty. By Lemma either Σ = or { Σ}, so either Y m > 1 or Y m = Σ Q. In the latter case since Q P, this forces A P, which then contradcits A > 1. Therefore Y m > 1, so by Lemma A Y m. Since M N S (R) and R P, then R P with Q < R, which is a contradiction. The following lemma will also be helpful: Lemma 3.5. Suppose X 1, X 2 Q and σ M satisfies σx 1 = X 2. As N M (X i ) acts on X i, we have homomorphisms φ i : N M (X i ) Sym(X i ) defined by φ i (g) x = gx for g N M (X i ) and x X i, so φ 1 and φ 2 are quasiequivalent permutation representations. Then φ 1 (N M (X 1 )) = φ 2 (N M (X 2 )). Proof. Let α : Sym(X 1 ) Sym(X 2 ) be defined by α(g) x = σgσ 1 x. Then α and σ define a quasiequivalence between the two symmetric groups, so it suffices to show that α(φ 1 (N M (X 1 ))) = φ 2 (N M (X 2 )). But α(φ 1 (g)) x = σφ 1 (g)σ 1 x = σgσ 1 x Since g and σ are in M, then σgσ 1 M, so σgσ 1 x = φ 2 (σgσ 1 ) x, as needed. Returning to the proof of Lemma 3.3, I claim that N M (X) acts as Sym(X) on X. If P X m contains a block of size at least 2 for some m, then the action of H N M (X m ) on X m contains a transposition, so by Lemma 2.5 N M (X m ) acts as Sym(X m ) on X m. Then by Lemma 3.5 N M (X) acts as Sym(X) on 8

9 X. If P X m contains only blocks of size one for all m, then we can get a larger partition in P by replacing X M with P X m, which is a contradiction. m M I further claim that C S (X) M. If X = 1, then this is trivial. By the comments above, if X > 1, then there is A P such that A X and A > 1. Then 1 C H (A) C M (A) C M (X). Since N M (X) acts as Sym(X) on X, then C S (X) = C M (X) N M (X) M. Since this holds for all X Q, by Lemma C S (Q) M. Lemma 3.6. Let H and P be as above. Suppose M is a transitive imprimitive overgroup of H and set Q = Q(M, P ). Then M is transitive on Q, Q is regular, C S (Q) is the kernel of the action of M on Q, and M/C S (Q) is a transitive subgroup of Sym(Q). Proof. Suppose X 1, X 2 Q. Since M is transitive on Ω, some σ M must map some element of X 1 to some element of X 2, so σx 1 = X 2. Thus M is transitive on Q and X 1 = X 2, so Q is regular. By Lemma 3.3 and Lemma 2.2.4, C S (Q) = C M (Q) is the kernel of the action of M on Q. Thus the action of M on Q is equivalent to the action of M/C S (Q) on Q, so M/C S (Q) is a transitive subgroup of Sym(Q). Lemma 3.7. Let H and P be as above. If M is an overgroup of H and Y Q(M) with M N S (Y ) then M = C M (Y ) C S (Y ) and C CS (Y )(P Y ) C M (Y ). Proof. By Lemma 3.3 and Lemma 2.2.3, C S (Y ) = C S (Q) M, so C S (Y ) Y Q M. By [Aschbacher 1.14], M = M C S (Y )C S (Y ) = C S (Y )(M C S (Y )) = C S (Y )C M (Y ). By Lemma 2.2.2, N S (Y ) = C S (Y ) C S (Y ), and since M N S (Y ), then M C S (Y ) = C M (Y ) and M C S (Y ) = C S (Y ) are normal in M. Thus M = C M (Y ) C S (Y ) Now suppose σ C CS (Y )(P Y ). Then we need to show σ M and σy = y for all y Y. Since σ C S (Y ), the second part holds. To show that σ M, it 9

10 suffices to show that σ H = C S (P ). Since σ C S (Y ) and σ C S (P Y ), then σ C S (P ), as needed. Let O(Q) be the number of blocks of Q. Clearly, if Q P then O(Q) O(P ). Theorem 3.8. Let A and B be proper nonempty subsets of Ω. Assume H = N S (A) N S (B) G < S with G not primitive on Ω. Let P = P A,B, Q = Q(G, P ), and k = O(Q). Then one of the following holds: (1) k = 2 and G = C S (Q). (2) k = 2, Q is regular, and G = N S (Q). (3) k = 3, G is intransitive on Ω, and for some X Q, G = C S (X) C G (X), and C G (X) and its action on X satisfy (1) or (2). (4) k = 3, G is transitive on Ω, Q is a regular partition, C S (Q) is the kernel of the action of G on Q, and G is of index 1 or 2 in N S (Q). (5) k = 4, Q = P, and for some block X P, G = C S (X) C G (X), with C G (X) and its action on X described in one of (1)-(3). (6) k = 4, Q = P, G has two orbits Y and Y on Ω, and G/H is a subgroup of Sym(P ) with two orbits of length 2. (7) k = 4, Q = P is regular, G is transitive on Ω, H is the kernel of the action of G on P, and G/H is a transitive subgroup of Sym(P ). Proof. Suppose k = 2. By Lemma 3.3, C S (Q) G. If Q is not regular, then C S (Q) is maximal in S by Theorem 2.4, so (1) holds. If Q is regular, then by Theorem 2.4 G is either C S (Q) or N S (Q). So either (1) or (2) holds. Suppose k = 3. If G is intransitive, then some X Q is an orbit of G. Then G N S (X), so by Lemma 3.7 G = C S (X) C G (X). In addition, C CS (X)(Q X ) C G (X) < C S (X) and O(Q X ) = 2, so the action of C G (X) on X reduces to either (1) or (2). So (3) holds. If G is transitive, then by Lemma 3.6 G/C S (Q) is a transitive subgroup of Sym(Q), so G/C S (Q) has index 1 or 2 in 10

11 Sym(Q). Since G is transitive, Q is regular so N S (Q) acts as Sym(Q) on Q, so N S (Q)/C S (Q) = Sym(Q), thus (4) holds. Suppose k = 4. Then Q = P and by Lemma 3.3 G N S (Q) = N S (P ). If G has three or more orbits, then for some X P we have G = C S (X) C G (X), with the action of C G (X) on X satisfies one of (1)-(3). So (5) holds. Assume G has two orbits, Y and Y. If either Y or Y contains only one block of P, then (5) holds, so we can assume that Y and Y each contain two blocks of P. Since H = C S (P ) G N S (P ), by Lemma H is the kernel of the representation of G on P, so H G. Then G/H acts on P by gh A = ga for g G and A A, so G/H is a subgroup of Sym(P ). Since each orbit of G contains two blocks of P, each orbit of G/H also contains two blocks of P. Then G/H is ismorphic to a subgroup of C 2 C 2, so (6) holds. If G is transitive, then by Lemma 3.6 (7) holds. 4 The Map P(M) Let S be the set of subgroups of S, partially ordered by inclusion. Then S is a lattice, with H 1 H 2 = H 1 H 2 and H 1 H 2 = H 1, H 2. We can then define a map between lattices C S : P S by P C S (P ). Lemma 4.1. If X, Y Ω with X Y, then C S (X), C S (Y ) = C S (X Y ). Proof. Since X Y = X Y, we have X Y X. Thus C S (X) C S (X Y ) and likewise C S (Y ) C S (X Y ). So C S (X), C S (Y ) C S (X Y ). To show the other inclusion, it suffices to show that if x, y X Y, then the transposition (x y) C S (X), C S (Y ), since such transpositions generate C S (X Y ). If x and y are both in X or both in Y, then this is clear, so suppose x X and y Y. Letting z X Y, we find that (x y) = (x z)(y z)(x z) C S (X), C S (Y ), as 11

12 needed. Theorem 4.2. C S is an injective dual homomorphism of lattices. Proof. First, we need to show that C S (P 1 P 2 ) = C S (P 1 ) C S (P 2 ). Since P 1 P 2 P 1, by Lemma 3.2 C S (P 1 ) C S (P 1 P 2 ). Similarly C S (P 2 ) C S (P 1 P 2 ), so C S (P 1 ) C S (P 2 ) C S (P 1 P 2 ). For the other inclusion, by Lemma C S (P 1 P 2 ) = C S (A) : A P 1 P 2. We can write A = X 1 Y 1 X 2 Y 2... X k Y k, where X j P 1, Y j P 2, X i Y i and Y i X i+1. By Lemma 4.1, C S (A) = C S (X 1 ), C S (Y 1 ),..., C S (X k ), C S (Y k ) C S (P 1 ), C S (P 2 ), so C S (P 1 P 2 ) C S (P 1 ) C S (P 2 ). Next we want to show that C S (P 1 P 2 ) = C S (P 1 ) C S (P 2 ). Since P 1 P 1 P 2, by Lemma 3.2 C S (P 1 P 2 ) C S (P 1 ). Similarly C S (P 1 P 2 ) C S (P 2 ), so C S (P 1 P 2 ) C S (P 1 ) C S (P 2 ). If g C S (P 1 ) C S (P 2 ), then ga = A and gb = B for A P 1 and B P 2. Since every element of P 1 P 2 is of the form A B and g(a B) = ga gb = A B, then g C S (P 1 P 2 ) so C S (P 1 ) C S (P 2 ) C S (P 1 P 2 ). Suppose C S (P 1 ) = C S (P 2 ). Let A P 1 and B P 2 with x A B. Since the orbit of x under C S (P 2 ) is B and the orbit of x under C S (P 1 ) is A, we must have A = B. So P 1 = P 2, so C S is injective. For M S, define p(m) = {P : C S (P ) M}. Since p(m), p(m) is nonempty. Lemma 4.3. p(m) has a least element P(M). Proof. If P 1, P 2 p(m), then C S (P 1 ) M and C S (P 2 ) M, so by Lemma 4.2 C S (P 1 P 2 ) = C S (P 1 ) C S (P 2 ) M, so P 1 P 2 p(m). Then we can constrcut a least element P (M) by taking the meets of all elements of p(m). For M S, let q(m) = {Q : C S (Q) M N S (Q)}. 12

13 Lemma 4.4. P (M) is the least element of q(m). Proof. As M acts on p(m), M N S (P (M)). Further q(m) p(m), so P (M) is the least element of q(m). In particular, this tells us that M stabilizes P (M), so C S (P (M)) M N S (P (M)). It is worth noting that Lemma 4.4 provides a much shorter proof to Lemma 3.3: If (2) holds then C S (P ) M, so P p(m) hence P (M) P and by Lemma 4.4, P (M) q(m) so P (M) P(M). Thus P (M) P. Lettting Q = Q(M, P ), we get P (M) Q P. As P (M) Q, C S (Q) C S (P (M)) M. Since Q P, Q and since M S and C S (Q) M, we must have Q 0. Therefore (2) implies (3). Lemma 4.5. If M 1 M 2, then P (M 2 ) P (M 1 ). Proof. We know that p(m 1 ) p(m 2 ), so P (M 2 ) P (M 1 ). Theorem 4.6. The map P : S P defined by M P (M) is surjective and P (M 1 M 2 ) = P (M 1 ) P (M 2 ). Note that P C S is the identity map. Proof. Since M 1 M 2 M 1 and M 1 M 2 M 2, then by Lemma 4.5 P (M 1 ) P (M 1 M 2 ) and P (M 2 ) P (M 1 M 2 ). So P (M 1 ) P (M 2 ) P (M 1 M 2 ). For the other inclusion: We know that C S (P ) M 1 M 2 iff C S (P ) M 1 and C S (P ) M 2, so p(m 1 M 2 ) = p(m 1 ) p(m 2 ). Since C S (P (M 1 ) P (M 2 )) C S (P (M 1 )) M 1, and likewise with M 2, this tells us that P (M 1 ) P (M 2 ) p(m 1 ) p(m 2 ) = p(m 1 M 2 ). By the minimality of P (M 1 M 2 ), this shows the other inclusion. For R P, P (C S (R)) = R, so P is surjective. Define an equivalence relation on S by M 1 M 2 if P (M 1 ) = P (M 2 ). Define a partial order on S/ by [M 1 ] [M 2 ] if P (M 2 ) P (M 1 ). 13

14 Lemma 4.7. (1) S/ is a lattice. (2) For M 1, M 2 S, [M 1 M 2 ] = [M 1 ] [M 2 ]. Proof. First I prove (2). By Lemma 4.5 [M 1 M 2 ] is a lower bound for [M 1 ] and [M 2 ]. Suppose [L] is also a lower bound. Then P (M 1 ) P (L) and P (M 2 ) P (L), so by Theorem 4.6 P (M 1 M 2 ) = P (M 1 ) P (M 2 ) P (L), so [L] [M 1 M 2 ]. Thus [M 1 M 2 ] is the greatest lower bound. Since [M 1 ] and [M 2 ] have a upper bound, namely [S], we can construct [M 1 ] [M 2 ] by taking the meet of all upper bounds. Theorem 4.8. The map θ : S/ P defined by [M] P (M) is a dual isomorphism of lattices. Proof. If P (M 1 ) = P (M 2 ), then M 1 M 2, so [M 1 ] = [M 2 ]. Thus θ is injective. By Theorem 4.6, P is surjective, so θ is a bijection. Since the ordering on S/ is in essence defined by θ 1, then by construction θ is automatically an ismorphism. Lemma 4.9. P (gmg 1 ) = g P (M). Proof. This follows from Lemma 4.3 and the fact that p(gmg 1 ) = g p(m). Therefore we can define an action of S on S/ by g [M] = [gmg 1 ]. Lemma 4.9 tells us that the action of S on S/ is equivalent to the action of S on P. 5 Lattices of Overgroups of Subset Stabilizers For this section, let A and B be distinct proper nonempty subsets of Ω and let H = N S (A) N S (B). We will investigate O S (H), the lattice of overgroups of H. Recall the map P : S P of Theorem

15 Figure 1: The lattice Λ 1 Figure 2: The lattice Λ 2, Λ 3, and Λ 4 15

16 Theorem 5.1. The following are equivalent: (1) O(P (H)) = 3. (2) A B, A B, A B, or A B. (3) O S (H) is isomorphic (as a latttice) to either Λ 1, Λ 2, Λ 3, or Λ 4. Proof. First, we show that (1) and (2) are equivalent. By Lemma 3.1, P (H) = {A B, A B, A B, A B}. Since, for example, A B = iff A B, this tells us that (1) and (2) are equivalent. Next assume (1) holds. Write P (H) as {X 1, X 2, X 3 }. For notation s sake, also write Q 1, Q 2, and Q 3 for the partitions {X 1, X 2 X 3 }, {X 2, X 1 X 3 }, and {X 3, X 1 X 2 }. We apply Theorem 3.8 to G O S (H). Let k = O(G), and note that k = 1 iff G = S, k = 2 iff Q = Q i for some 1 i 3, and k = 3 iff Q = P (H). Let M i = N S (Q i ) for 1 i 3. Suppose P (H) is regular. If P (G) = Q i, then k = 2, so by Theorem 3.8 the fiber of P over Q i consists of M i, which is maximal by Theorem 2.4. The subgroups M 1, M 2, M 3 form the top row of Λ 1. Let F be the fiber of P over P (H). By Theorem 3.8, G F is transitive on P (H) iff case (4) of Theorem 3.8 holds, where G/H = S 3 or A 3. Thus G = M 4 = N S (P (H)) or G = K 4 is of index 2 in M 4. By Theorem 2.4, M 4 is maximal in S; M 4 is the group in the second row of Λ 1, and K 4 is below M 4 in the third row. Finally G F is intransitive on P (H) iff case (3) of Theorem 3.8 holds, where Q = P (H) or G = K i, 1 i 3, with K i /H = S 2, K i fixes X i, and K i is transitve on P (H) {X i }. Thus K i is the member of the third row of Λ i below M i. Thus O S (H) is isomorphic to Λ 1. Now suppose (up to choice of notation) that X 1 = X 2 and X 3 = X 1 + X 2. Then by Theorem 3.8, the fiber of P over P (H) consists of two elements: H and L 3 with L 3 /H = S 2, and L 3 fixing X 3 and transitive on P (H) {X 3 }. The group L 3 is the member of the fourth row of Λ 2. For i = 1, 2, the fiber over 16

17 Q i consists of the maximal subgroup M i. The second row of Λ 2 is M 1, M 3, M 2 in that order. Finally the fiber of Q 3 consists of M 3 and K 3 = C S (Q 3 ); K 3 is the member of the third row of Λ 2. Thus O S (H) is isomorphic to Λ 2. If X 1 = X 2 and X 3 X 1 + X 2. Then the fiber of P over P (H) contains two elements: H and L 3 with L 3 /H = S 2 and L 3 fixing X 3 and transitive on P (H) {X 3 }. The group L 3 is the member of the third row of Λ 3. For 1 i 3, the fiber over Q i consists of the maximal subgroup M i. The second row of Λ 3 is M 1, M 3, M 2 in that order. Then O S (H) is isomorphic to Λ 3. If X 1, X 2, X 3 are distinct with X 3 = X 1 + X 2, then the fiber of P over Q 3 consists of two elements: M 3 and K 3 with M 3 /K 3 = S2 and M 3 maximal in S. For i = 1, 2, the fiber over Q i consists of the maximal subgroup M i. Thus the second row of Λ 3 is M 1, M 3, M 2 and the third row of Λ 3 is K 3. The fiber over P (H) is just {H}, so O S (H) is isomorphic to Λ 3. If X 1, X 2, X 3 are distinct and X σ(3) X σ(1) + X σ(2) for each permutation σ, then each fiber of P consists of one element. The second row of Λ 4 is M 1, M 2, M 3, so O S (H) is isomoprhic to Λ 4. Therefore (1) implies (3). Now assume that (3) holds but (1) does not. Then O(P (H)) = 4. Since P is surjective and there are 15 partitions less than or equal to P (H), this tells us that O S (H) contains at least 15 elements. This contradicts (3); so therefore (3) implies (1). 17