maximal subgroups of infinite symmetric groups

MAXIMAL SUBGROUPS OF INFINITESYMMETRIC GROUPS

MARCUS BRAZIL, JACINTA COVINGTON, TIM PENTTILA,CHERYL E. PRAEGER, and ALAN R. WOODS

[Received 2 July 1991—Revised 2 November 1992]

1. Introduction

By a maximal subgroup, we mean a maximal proper subgroup of the symmetricgroup S = Sym(Q) of all permutations of a set Q. Maximal subgroups of finitesymmetric groups have been studied intensively; however our work was largelymotivated by a recent paper [7] of Dugald Macpherson and Peter Neumann inwhich they considered maximal subgroups when Q is infinite. (Some of the resultsthere were anticipated by Ball [2] and Richman [9] in early papers.) Taking [7] asour starting point, we have extracted what we believe to be the fundamentalconcepts. This has led to several new constructions, and in some cases,classification, of maximal subgroups which lie above the stabilizers of subsets,filters and partitions of an infinite set Q.

The structure of subgroups, in particular maximal subgroups, of finitesymmetric groups was investigated by M. O'Nan and L. L. Scott, see [10], andlater by M. Aschbacher and Scott [1]. The approach used by them was extendedand, in some sense completed, by Liebeck, Praeger and Saxl [6], to show thatmaximal subgroups of finite symmetric groups are well understood in the sensethat there are several types of maximal subgroup, and their structure iscompletely determined modulo the classification of finite simple groups. Moreprecisely, given a maximal subgroup of a finite symmetric group, either it is the(setwise) stabilizer of a subset, partition, product decomposition, or affinestructure; or it is a group in diagonal action; or it is almost simple, that is, liesbetween a non-abelian simple group and its automorphism group. The originalaim was to find new classes of maximal subgroups of infinite symmetric groupswhich were in some sense analogues of classes of maximal subgroups of finitesymmetric groups. Ball [2] showed that the setwise stabilizer 5{ r ) of a finite subsetF is maximal in 5. But Neumann and Macpherson proved that any maximalsubgroup which is not the stabilizer of a finite set is highly transitive, and hencecannot stabilize a partition or preserve a geometry. Thus at first sight it seemsthat almost none of the classes of maximal subgroups of finite symmetric groupshave analogues in the infinite case. However, one can still ask for maximalsubgroups containing the stabilizer of a set, partition, etc. Typically these mustcontain all permutations which in some sense almost stabilize the structure. Forinstance, Richman [9] showed that the almost stabilizer of a partition of Q intofinitely many parts of equal cardinality is a maximal subgroup, and we investigatefurther maximal subgroups which are, or lie above, the almost stabilizers ofsubsets, filters and partitions.

The key to understanding these is the study of groups which contain thepointwise stabilizer 5(A) of some set A <= Q. (These include the stabilizers of filters

This research was partially supported by ARC grants A68931532 and A68830987.1991 Mathematics Subject Classification: 20B35, 20E28.

Proc. London Math. Soc. (3) 68 (1994) 77-111.

78 MARCUS BRAZIL ET AL.

and partitions.) In § 2, we fix our notation, give definitions and state some knownresults which will be used later. In § 3, we give definitions of filters andquasifilters, and then discuss the role of stabilizers of filters and almost stabilizersof partitions in the study of maximal subgroups. In § 4, we address the problem ofdetermining which filters have stabilizers which are maximal. This leads us tointroduce the concepts of the almost stabilizer of a filter and of closed andsuperclosed filters. A basic theorem arising from and typical of these sections is:

THEOREM 1.1. Suppose |Q| = K where K is an infinite cardinal. If G is a maximalsubgroup containing the pointwise stabilizer 5(A) of some set A c: Q with |AC| = K{and this includes the case where G is a maximal subgroup lying above thestabilizer 5{3F} of some non-trivial filter 3*), then either:

(i) G is the almost stabilizer of a partition of Q into finitely many parts; or

(ii) G = S{<S) the stabilizer of a non-trivial filter <$ which is unique, closed, and,if G is not the almost stabilizer of a set, superclosed.

Moreover in the case where G = S{<S} 2* S{3p), we have &<=.<§.

In § 5, we construct a maximal subgroup containing the stabilizer of a partitionwith infinitely many parts of the same cardinality. In § 6, we consider filters with alinearly ordered filter base and analyse the maximal subgroups containing thestabilizers of such filters. Finally, in § 7, we give an example of a superclosed filterwhose stabilizer is not maximal.

We wish to thank Peter Neumann and Dugald Macpherson for helpfuldiscussions on this area, and in particular to thank Dugald Macpherson for hiscontribution to Theorem 5.9.

2. Notation and preliminaries

Throughout this paper Q will denote an infinite set of cardinality K, and S orSym(Q) the symmetric group on Q. The degree deg(g) of a permutation g is thenumber of points moved by g. The groups of bounded permutations are thesubgroups Bx = {g e S: deg(g) < A} for infinite cardinals A =£ K. For convenience,we define Bx = 1 for finite A. The bounded subgroups are normal subgroups of S.When A = X(), Bx is the finitary symmetric group, usually denoted FS(Q).

If G^S and FcQ then G{r) = {geG: P = F} and G(r)={geG: as = a forall a e F} are respectively the setwise and pointwise stabilizers of F in G. Also,for such a set F we identify 5 ( r ) with Sym(P), where P = Q - F. A set F is fullfor a group G if G{r) induces the full symmetric group on F. We will also say thatG is full for r . If 2 c Q and |2 | = |Q - 2 | , then 2 is called a moiety of Q.

Let A and 2 be subsets of Q and let A be an infinite cardinal. Then A is said tobe X-contained in 2 , written A c A 2 , if |A — 2 | < A. For finite A, we define A c A 2by A c 2 . Two sets A and 2 are said to be k-almost equal, written A = A 2 , ifA c A 2 and 2 c A A. That is, A = A 2 means |2 A A| < A for infinite A, and A = 2for finite A. Note that =A is an equivalence relation.

A subset or increasing sequence J in a linear ordering (/, < ) is cofinal in / iffor every / e I there is / eJ such that / *£/. The cofinality fi of (/, < ) is the leastcardinal such that there is some / c / with \J\ = n which is cofinal in (/, < ) . Theletters a, /?, y, S will always denote ordinal numbers, and K, A, pi, v, rj will always

MAXIMAL SUBGROUPS 79

denote cardinals. We will use the common convention of identifying a cardinal Kwith the initial ordinal of cardinality K. The cofinality of a cardinal K is denotedby cf(/c). A cardinal K is regular if cf(/c) = K, and singular otherwise. For anycardinal K, cf(*r) is a regular cardinal. The two main properties of cofinality whichwe use are:

(a) cf(ic) is the smallest cardinal ju such that a set of size K can be written as aunion of ft sets of cardinality less than K;

(b) if K is a regular cardinal then any subset of K of cardinality K is cofinal. Wewill mainly apply this when K = cf(A) for some cardinal A.

One of the most useful tools in proving the maximality of a subgroup of aninfinite symmetric group is the following lemma of Dixon, Neumann and Thomas[5, Lemma 2.1]:

PROPOSITION 2.1 (Dixon, Neumann and Thomas). / / F , , F 2 c Q and |F, n F2| =min{|r,|, |r2|} then

Sym(r, U r2) = (Sym(r,), Sym(r2)>.

Thus if A U 2 =£* Q, then putting F, = Q — A and F2 = Q — 2 , we obtain S(AnZ) =(5( A ), S(X))- In particular, if A and 2 are disjoint sets of cardinality less than K and5(A), 5(2) =£ G then G = S.

We will also use the following superficially similar lemma [7, Lemma 2.3]:

PROPOSITION 2.2 (Macpherson and Neumann). Let F,, F2 be subsets of Q suchthat |F, fl F2| = K and F, U F2 = Q. If G is a subgroup of S which is full for F, andF2 then G = S.

We will be interested in stabilizers and almost stabilizers of sets, filters andpartitions. In general, we can define the stabilizer of any collection ^ of subsetsof Q by

Sm = {xeS: A ' e ^ ^ A e ^ f o r a l l A c Q } .

To define the almost stabilizer of a collection (€, we start by defining the depthA(^) of % to be A(^) = min{|A|: &€<€}. Then the almost stabilizer of acollection ^ of depth A is defined by

AStab(^) = {geS: (VA e ^)(320, 2, e ^ (A* =A20 A A*" =A2,)}.

It is easily checked that AStab(<<o) is a subgroup of S. We think of the almoststabilizer as the group of permutations which stabilize ^ modulo sets of smallcardinality, where 'small' means having cardinality less than the depth of thecollection.

If <# = {2}, then 5{<g} is just the setwise stabilizer S{x) of 2. Likewise, weabbreviate AStab({2}) to AStab(2). Note that

but that equality does not always hold. This is because, although for anyg e AStab(^) and A e <£ we have Ag =A2 for some 2 e <€, the 'error' A* A 2 maybe quite different for different sets A. Thus, although for any particular set A we

can find some h e Bx such that Ahg e %, we cannot necessarily find a single h suchthat Ah8 e <€ for every A e <€.

3. Filters, quasifilters and finite partitions

Our main results concern permutation groups that stabilize certain partitions ofQ and filters on Q. Let 9= {Qa: a e 1} be a partition of Q. The stabilizer of 0>is

S{9) = {geS: (Qa)8 e 9 for all a e / } .

We call 0* a finite partition if there are finitely many parts, each of cardinality K.If 0> = {Qx, Q2,•••, Qn) is a finite partition then the almost stabilizer of 9 is

AStab(3*>) = {geS: (

We have

with equality if and only if K is uncountable. Richman proved that this group is amaximal subgroup of S. (See also [7].)

PROPOSITION 3.1 (Richman [9]). Let 9 be a partition of a set Q of cardinality K

into finitely many parts, each of size K. Then AStab(^) is a maximal subgroup ofSym(Q).

We now turn our attention to filters. A collection &> of subsets of Q is called afilter on Q if it has the following properties:

(a) 0 $ & and Q e ^ ;

(b) if T e ^ a n d T<= T c Q then V e &\

(c) if T, r ' e S M h e n m r ' e ^ .

It will sometimes be more convenient to deal with the dual notion of an ideal.A collection 3> of subsets of Q is called an ideal on Q if it has the followingproperties:

(a) 06^;

(b) if remand T 'cTthenT'e^;(c) if r, reS> then r u T ' e ^ .

An ideal S> is called proper if it is a proper subset of &(Q), that is, if Q i 2,.Ideals and filters are dual in the sense that ^ is a filter if and only if^ = {A: Q - A e f } is a proper ideal. Note the asymmetry in the conventionthat we allow improper ideals but not improper filters.

A collection 38 of subsets of Q is a filter base for 9 if & = {2 c Q: A c 2 forsome A e 38}. Then we say that SF is the filter generated by 58. A collection 38 is afilter base if and only if the intersection of any two sets in 38 contains someelement of 38. Consequently, any collection 38 which is closed under intersectionsis a filter base. A filter generated by a single set A is called a principal filter; it hasthe form ^ = { I c Q : A c l } .

Recall that the depth A(^) of a filter 9 is given by A(^) = min{|A|: A e 9).Note that every filter of finite depth is principal. A uniform filter is a filter ofdepth K, that is, one in which every member has cardinality K. The filters we are

interested in in this section correspond to filters over the quotient of the Booleanalgebra of subsets of Q by the equivalence relation =A. For this reason, we referto filters in which A = K Q for every element A in this filter as trivial filters. Theseare just filters which contain no moieties. Obviously if $F is a non-trivial filter and^ is a filter with f c <S, then ^ is also non-trivial.

We will use the following elementary fact about filters:

LEMMA 3.2. Let A be a set in a filter 2F on Q. Then for any F c Q ,

r e 9 if and only if T n A e $\

Recall that the stabilizer of a filter 3* is defined by

5{SF> = {jce5: Tx e f o r e 9 for all T c Q}.

The stabilizer of a filter f is a subgroup of S and is a proper subgroup unlessSF = {A: |Ac| < A} for some A such that Ko =s A =s K. In the case where A = No, thisis the Frechet filter of all cofinite sets. These filters are all trivial, so the stabilizerof any non-trivial filter is a proper subgroup of 5. The stabilizer of an ideal $> isdefined in a similar manner. If 3> is the dual ideal to a filter $F then 5{gf} = S w .

Notice that a permutation g belongs to 5{gt:} if and only if both Ts e 2F andP " ' e f for all F e f . However, to show that a group G is a subgroup of S{3J:} itsuffices to show that T8 e SF for any g e G. Observe also that if we have a filterbase 58 then it suffices to consider only those sets r e 58.

Simon Thomas [3] has shown that it is consistent with ZFC that there aresubgroups of infinite symmetric groups which are not contained in any maximalsubgroup. However, Macpherson and Neumann have proved:

PROPOSITION 3.3 (Macpherson and Neumann [7]). Let G be a proper subgroupof S. If there is a moiety HofQ such that G{X} induces the full symmetric group on2 then there is a maximal subgroup H of S such that G =£ H.

One way to satisfy the hypothesis of the proposition is to require the strongercondition that 5 ( A ) ^ G for some A with |AC| = K. It can easily be seen that thiscondition is actually stronger—moreover, Covington and Mekler [4] have recentlyshown that this condition is stronger even when G is assumed to be maximal. Thecondition is satisfied, for example, if G is the stabilizer of a finite partition. (Tosee this, take A to be the union of all but one of the parts.) We have already seenin Proposition 3.1 that these subgroups are maximal.

Richman [9] was perhaps the first to see the significance of Stabilizers Of filtersin the study of maximal subgroups, but he was only concerned with stabilizers ofcollections of ultrafilters. Then Semmes [11] proved (cf. Theorem 3.5 below) thatif H is maximal subgroup such that 5(A) H for some A with |A| < K, then thereis a filter 2F such that H = S{3f). We will continually exploit this representation,which we show to be unique. However, we will also be interested in the casewhere |A| = K. Then it is not necessarily true that H is the stabilizer of a filter.However, as we show below, if it is not, then H is the almost-stabilizer of a finitepartition of Q. In order to prove this we introduce the notion of a quasifilter.

A collection 2. of subsets of Q is called a quasifilter on Q if it has the following

properties:

(a) 0^andQeS;

(b) if T e 2 and T <= T' <= Q then T' e 2;

(c) if T, r e ^ then either YC\T' e 2 ox r u F = " Q .

We say that 2 is uniform if every element of i2 has cardinality K.Recall that the stabilizer of a quasifilter is defined as:

We will also use quasiideals, which are the duals of quasifilters. A collection 2of subsets of Q is called a quasiideal on Q if it has the following properties:

(a) 0e2;

(b) if T e 2 and T' c T then T e 2;

(c) if r,T' e2 then either r U V e 2 or | r n T'\ < K.

A quasiideal 2 is called proper if it is a proper subset of 9*{Q.), that is, if Q £ .2.If 5(A) =s G < S for some set A with | Ac| = K, then by Proposition 3.3 there is a

maximal subgroup H with G =s // . We now characterize these subgroups.

THEOREM 3.4. Suppose G is a proper subgroup of S and there is some set A with|AC| = K such that 5(A) =s G. Then

where 2 is the quasifilter defined by

2 = {A: 5(A)^G}.

Moreover, if G is maximal then G = S{&).

Proof. We first show that 2 is a quasifilter. By assumption 2 is non-empty,0$ 2 since S ^ G, and it is clearly closed under supersets. Take 2, Te 2. If2 U I V Q then by Proposition 2.1, S(Ynr)^G whence S f l T e l Thus 2 is aquasifilter.

Since G preserves 2, we have G ^ 5 { 2 > . Thus it remains to show that 5{2} =£S.Take A e S with |Ac| = *:and choose F c A c such that |r | = |A| and A U I V Q .Then r $ 2, so there is no g e S{&) such that A8 = T. Thus G 5{2} < 5.

Finally, if G is maximal, we must have G = 5{2}.

In the special case where |A| < K, we obtain:

THEOREM 3.5 (Macpherson and Neumann [7]). / / G is a proper subgroup of Swith 5(A) G for some set A such that \ A| < K, then G^S{^)<S where &> is thenon-trivial filter generated by the filter base

m = {A: \A\<KandSiA)^G}.

In particular, if G is maximal then G = S^y

Proof Clearly, G preserves 38. So we need to show that 39 is a filter base.Since 38 is non-empty, it suffices to show that it is closed under intersections. Now|2UA|</c, so 2 U A ^ * Q , and by Proposition 2.1 we then have S(2nA)=s G;hence 2 n A e <&.

Theorem 3.4 applies if G ^S{3/:] for some non-trivial filter 9, since:

LEMMA 3.6 (Richman [9]). Let A be a set in a filter ZF on Q. Then 5(A) =£ S{3f}.

Proof. Let g e 5 ( A ) and let 2 e ^ . We need to show that 2 s e 9. But thisfollows from the fact that I s D ( 2

So combining Lemma 3.6 and Theorem 3.4, we obtain:

THEOREM 3.7. Let 9 be a non-trivial filter: If S{3P} *£ G <S then

G ^ S{

where Si is a quasifilter with &* c= Si.In particular, if G is a maximal subgroup then G =

From Lemma 3.6 and Theorem 3.5 we have:

COROLLARY 3.8. If 9 is a filter of depth A(^) < K and G is a proper subgroup ofS with S{3F}^G then there is a non-trivial filter ^ 3 9 such that G ^ S{<3} < S.

In particular, if G is maximal then G = S{<s).

THEOREM 3.9. Suppose G is a proper subgroup of S. If 5(A) ^ G for some set Asuch that \AC\ = K then either:

(a) G =£ 5<3F} < 5 for some non-trivial filter 9>; or

(b) G ^H where H is the almost stabilizer of a partition 9 of Q into finitelymany parts of size K.

Proof. If S ^ ) ^ G for some A with |A| < K then (a) holds by Theorem 3.5, sowe may assume that if 5 ( A ) ^ G then |A| = K. Therefore by Theorem 3.4, G iscontained in the stabilizer of a uniform quasifilter. It will actually be moreconvenient to expound the proof in terms of the dual quasiideal 2,. The collection$ of all finite unions of members of <£ is an ideal. If this is a proper ideal, thensince there is a moiety in <£ we have 5{ 2 } =s S{J,} < S, so letting 9 be the dual filterto 3> we have G ^ 5{SF} < 5.

Otherwise, there are finitely many sets Q,, Q2, ..., Qne 2, such that

We may assume that the Q, are disjoint and that the union of any two does notbelong to Si. Discarding any Q, with |Q,| < *", we may at the cost of replacing =by =K yielding Q, U Q2U ... U Q n

= l fQ, also assume that each Q, has cardinalityK. Note that we still have n ^ 2 since the quasifilter dual to Si is uniform.

We will show that 5{ 2 }, and hence G, is contained in the almost stabilizer ofthe partition & = {Q,, Q2> •••> &n}- (Strictly speaking 9> is a partition of a setwhich is only ic-almost equal to Q. However since it is trivial to check that thealmost stabilizer of ^ is identical to the almost stabilizer H of the partition of Qobtained from <& by enlarging Q, (say) by a set of cardinality less than K, this willestablish the theorem.)

Suppose there is some set 2 e Si which intersects both Q, and Qy in a set of sizeK for some / , ; such that l^i<j^n. Then, since Si is a quasiideal, 2 U Q, e Si

and 2 U Qy e Si. But since these two sets intersect in 2, their union Q, U Q, mustalso lie in Si, which contradicts our assumptions. Thus, any set in Si must beic-contained in some Q,. In particular, for any geS{Si} we have ( Q ^ c ^ Q y forsome y = l ,2 , . . . , / i . Then also (QJy~lclcQk for some k, and so i = k and(Qiy=KQj. Thus S{SI]^H.

We can now give more information about the form of maximal subgroupscontaining the stabilizer of a non-trivial filter. The closed almost principal filtergenerated by a set A of cardinality A is ^ = { 2 c Q : A c A 2 } . The dual is aclosed almost principal ideal.

COROLLARY 3.10. If S{3f) =s G <S for some non-trivial filter 2P then either:

(a) G^Sm<S for some filter <S 3 9; or

(b) G «s AStab(^) where & is a partition of Q into finitely many parts of size K.Moreover, this can occur only if & is contained in a uniform closed almostprincipal filter and S^^ < AStab(^).

Proof. In the case where the proof of Theorem 3.9 gives a filter (which we willnow call <§) satisfying G^S^, that filter ^ contains 9*. For if 5(A) =s G for someA with |A| < K, then ^ c ^ is guaranteed by Theorem 3.5. On the other hand, ifevery element of the quasifilter {A: 5 ( A ) ^G} given by Theorem 3.4 hascardinality K, then the filter ^ obtained (if any) contains this quasifilter, andtherefore by Lemma 3.6, ^ c CS.

Now suppose S^^H where H = AStab(^) for some partition ^ of Q intoparts Q,, Q2, ..., &„ with |Q,| = K for each /. Let 3 be the ideal dual to &, and Sibe the quasiideal dual to the quasifilter {A: 5(A)=^//}. By Lemma 3.6, thisquasifilter contains & so $ c Si. Clearly each Q, e Si, and arguing as in the proofof Theorem 3.9, we see that for every Ye Si there is some / such that £ c * Q,. Inparticular, this holds for all l e i . Let S be a moiety belonging to 3, andconsider any r € S. Then 2 U T e $, so is if-contained in some Q,. Hence 2 and Fare jc-contained in the same Q, which must be unique since 2 is a moiety. Thus 3>is contained in the closed almost principal ideal generated by Q,, so 8F iscontained in the dual closed almost principal filter {A: Qf e * A} which is uniform(and really a filter) since |Q^| = K.

Finally, to see S{p} < AStab(^), let 2 be a moiety in 3> and let h be apermutation interchanging Q, and some other Qy. Then h e AStab(^), butIh £KQ; solh^^ and therefore h $ S{^} = 5{SF}.

We remark that the above theorems (without appealing to Proposition 3.3)have as an immediate consequence a result of Ball. (See also [7].)

COROLLARY 3.11 (Ball [2]). If & is an ultrafilter {that is, a filter which ismaximal in the sense that it is not contained in any other filter) then S{3f) ismaximal.

Proof. Assume 5{Sf} *£ G <S. If ^ is a principal ultrafilter, then its stabilizer isthe stabilizer of a one element set and is therefore maximal. If ^ is anon-principal ultrafilter, then 3> is clearly not contained in any closed almost

principal filter. (To see this, use the fact that the complement of any set not in &,is in cF.) So by the previous result, G S{cs) for some filter ^ 3 SF. But then since2F is a maximal filter, ^ = SF. That is, S{&} =£ G S{c3) = 5{3C}. Thus S{g;) = G, so5<SFj is maximal.

The next results will give us uniqueness of non-trivial filters whose stabilizersare maximal.

LEMMA 3.12. Suppose 9* is a non-trivial filter and let <& be any filter properlycontaining SF. Then 5{^ — S<Sf} =£0.

In particular, if SF^^ and 5{3t} s* Sm, then &=<§.

Proof. Choose a set A € 9 such that |AC| = K. Take 2 e <§- 9. By Lemma 3.2,S D A e ^ - ^ , so we may assume 2 c A . Let g be a permutation which fixes 2pointwise and interchanges A - 2 with some subset of Ac. Then g e 5(2:) =£ S{<S}, byLemma 3.6, but ASDA = I ^ , so g $ S w .

THEOREM 3.13. Let G be a maximal subgroup such that G = S{3P} for somenon-trivial filter 2F. Then 9< is the unique non-trivial filter whose stabilizer is G.

Proof. By virtue of the maximality of G, the proof of Corollary 3.10 showsthat G = S{<S) for a filter ^ 3 ^ , where ^ is defined in terms of the group Gwithout reference to 9. Therefore G = 5{3C} = S^j, so by Lemma 3.12, we haveSF=C§. But this holds with the same ^ for any non-trivial filter 9 havingG = S(&y Therefore 9 is unique.

Finally, we now have a complete characterization of maximal subgroupsof 5 which contain the pointwise stabilizer of a set whose complement hascardinality K.

THEOREM 3.14. Suppose H is a maximal subgroup of S. If 5(A) ^H for some setA such that |AC| = K then either:

(a) H = 5{Sf) for a unique non-trivial filter 9; or(b) H is the almost stabilizer of a finite partition 9*.

In particular, this holds when H^S^ for some non-trivial filter CS.

In the next section we will address the question of which filters have stabilizerswhich are maximal subgroups of 5.

4. Almost stabilizers of sets and filters

Ball [2] proved that the setwise stabilizer of a finite set is a maximal subgroup,and that any transitive maximal subgroup contains the finitary symmetric group.Macpherson and Neumann [7, Observation 6.1] proved further that any maximalsubgroup which is not the stabilizer of a finite set contains the finitary symmetricgroup FS(Q). Since setwise stabilizers of infinite sets do not contain FS(Q), theyare not maximal subgroups. However, by introducing the concept of the almoststabilizer of a set, Ball generalized his result to infinite sets.

The almost stabilizer of an infinite set 2 is just

Moreover, AStab(2) = BkS{11) except when 2 is countable. Note that if 2 is finite,AStab(2) = 5{2:>, so Theorem 4.1 includes the finite case.

THEOREM 4.1 (Ball [2]). // 2 is a non-empty subset of Q with |2| = A < K thenAStab(2) is a maximal subgroup of S.

Proof. Let G = AStab(2), and choose any x e S — G. We wish to show that(G, x) = S. Since x $ G, we have |2 A 2*| = A. We may assume, by replacing xby x~l if required, that |2 - 2 r | = A. We show that we can find ye (G, x) suchthat 2y fl 2 = 0 . Then S(r) *£ G and S(xy) is conjugate to S(X) in (G, x) so by ourremarks at the end of Proposition 2.1, {G, x) = S.

It remains to find such a permutation y. Choose h stabilizing 2 such that thefollowing two conditions hold:

(a) (2* nifcH-r and(b) (r-I)"c(2U27.

Notice that (a) places a condition on the action of h within 2 while (b) restrictsthe action of h within 2C, so that the two conditions are not contradictory. Thefirst condition can be satisfied, since 12* D2| «£ A = |2 — 2*1, and the second issatisfiable since |(2 U2'r)c| = K. Thus h maps 2* into its complement, and soy =xhx~x maps 2 into 2C, as required.

Note that this result is the best possible, since if 2 is a moiety then G isproperly contained in the almost stabilizer of the partition Q = 2 U 2C, which byProposition 3.1 is maximal. In fact:

THEOREM 4.2. Let 2 be a moiety of Q and let H be the almost stabilizer of thefinite partition Q = 2U2C. Then H is the only proper subgroup of S properlycontaining AStab(2).

Proof. Let G be a group such that AStab(2)<G<S, and take g e G. LetA = 2*. Note firstly that G contains Sym(2), Sym(2c), Sym(A) and Sym(Ac). Wefirst deal with the case where both |A D2| = K and |A fl2c| = K. Then applyingProposition 2.1 to A and 2 yields Sym(AU2)=^G, and similarly we obtainSym(A U 2C) G. But then applying the proposition to A U 2 and A U 2C yieldsG = S. Therefore, we must have either A c*2 or A c*2C.

Assume first that A c ^ I Multiplying g by an element of AStab(2), we mayassume that A c 2. If |2 — A| = K then the proposition yields Sym(2 U Ac) = G.But 2UAC = Q, which is a contradiction. Therefore A=*2. Similarly, in thecase where A c " 2 c we obtain A=I<:2C. Thus any geG almost preserves thepartition, so G ^H.

To see that G is actually equal to H, we prove that (AStab(2), g) = H for anygeG - AStab(2). Take any permutation h eH - AStab(2). Then both g and halmost interchange 2 and 2C, so 2 ^ = *2. That is, hg e AStab(2), so h e<AStab(2),g>.

We will discuss almost stabilizers of partitions further in § 5.We saw in § 3 that an important class of maximal subgroups of infinite

symmetric groups consists of stabilizers of filters. We have also seen that almoststabilizers play an important role. We will now consider the almost stabilizer of afilter.

Recall that the almost stabilizer of a filter 9 of depth A is

AStab(^) = {geS: (VA e ^)(320, 2, e ^)(Ag =*20 A A*"' = AI,)}-

Observe that if 9 is the principal filter generated by A then AStab(^) =AStab(A). Then Theorem 4.1 says that the almost stabilizer of a principal filter ismaximal. We want to investigate under which conditions this is also true fornon-principal filters.

We define the closure of SF as

! = { A c Q : A=A2forsome2eSF}.

That is, the closure is obtained by adding to 9 all sets which are A-equal to sets in9. A filter 9 is closed if 9 = 9. A closed almost principal filter as defined in § 3is then the closure of a principal filter. A filter is almost principal if its closure is aclosed almost principal filter. Notice that if a filter is trivial, then its closure is theclosure of the principal filter {Q}, and it is therefore almost principal. Observealso that every closed filter 9 is either almost principal, or for every A e f thereare moieties of A which belong to 3F.

The following proposition shows that we can reduce the search for maximalalmost stabilizers of filters to the study of stabilizers of closed filters.

LEMMA 4.3. For any filter 8F of depth A:(a) 9 is a filter of depth A;

(b)

Proof, (a) Obviously, 0 $ 9, Q e l and 9 is closed under supersets. TakeA,, A2G 9. Then there exist 2, , 2 2 e 9 such that A, = A2i and A2=A22. Then2, D 22 e 9 and |(A, D A2) A (2, n 22)| < A, so A, n A2 e 9. Thus 9 is a filter,and clearly its depth is A.

(b) Let g e AStab(^) and A' e 9. Then A' =A A_for some A e 9 and As = A2for some 2 e 9. Thus (A')*=AA*=A2 so (A')* e 9. A similar result holds forg~\ so g G 5{#}. Conversely, take g e S{&} and A e f . Then A* e 9, so A* =A2for some 2 e 9. Again, g"1 is similar, so g e AStab(^).

THEOREM 4.4. Let 9 be a non-trivial filter. If 5{3P} is maximal then 9 is closed.

Proof. By Lemma 4.3(b), 5{SF} *£ S{&y Since 9 is non-trivial 5{^}<5, and soby maximality, S{S>f) = S{&). But then by Theorem 3.13, 9=9.

The next lemma tells us that if we have a group G which we know is thestabilizer of a closed non-trivial filter 9 then we may not be able to determine thefilter, but we can find its depth.

LEMMA 4.5. Let 9 be a non-trivial filter. Then

MARCUS BRAZIL ET AL.

Proof. Set A = A(^). Certainly Bk^ BkS{gf) =^5{#}. Suppose B^^S{^} forsome pi> X. Then S ^ contains all permutations with support of size A. Let g bea permutation with support of size A which interchanges a set A e SP such that|A| = A and |AC'| 5= A with a subset of Ac. Then g e S{^}. But then both A and Ac

belong to 9, which is a contradiction.

LEMMA 4.6. Let &c <3 be filters. Then A(5 ) = A(^) if and only if&c%.

Proof. (=>) This is trivial.«=) Assume f c t Since ^ c ^ , we have A(^)^A(^). Let k = k(&) and

suppose there exists F e <£ with |F| < A. Then for any A e 9 we have A - F =A A,and so A - F e ^ c t But then $ contains two disjoint sets, which is impossible.Hence we must have A(^) =

We can now strengthen Corollary 3.10 for the case of closed filters:

THEOREM 4.7. If S^ =s G <S for some non-trivial closed filter 2F then either:

(a) G^S{m <Sfor some closed filter <$^ & with k{<0) = k(&); or

(b) G H where H is the almost stabilizer of a partition & of Q. into exactly twoparts {, Oc}. Moreover, this case can only occur if cF is uniform and iscontained in either of the closed almost principal filters generated by 3>or <PC.

Proof. If Part (a) of Corollary 3.10 applies then we have a filter (S^&'. Sinceis closed, for every set A c Q with |A|<A(^) we must have Ac € 8F. Thus

= A(^) and <§ is closed.If Part (b) of Corollary 3.10 applies then we have G =s AStab(^) for some finite

partition <3>= {Qu Q2,..., &„} with |Q,| = K for all i. Moreover, we know thatany 2 in the dual ideal $> is /c-contained in, say, Q,. Assume n 5= 3. Then define apermutation g fixing Q, pointwise and interchanging a moiety of Q2 with a moietyof Q3. Then for any leJ> we have 2 = * 2 n Q , , and so Is = K(2nQl)

8 =SflQ,. Since 2 e 3>, it follows that 2 D Q, G ^, and then, since ^ is closed, it alsofollows that 2* 6 J>. Similarly, 2«~' G ^. Thus g e Sw = S w ^ AStab(^), whichcontradicts the fact that g interchanges moieties of Q2

ar |d ^3- Therefore, thepartition ^ must have only two parts.

We will show in § 6 that there are closed filters & which are not maximal. Wenow introduce a stronger notion of closure.

We define the superclosure 3<+ of a filter & of depth A which is not almostprincipal by

T e ^ + if and only if (31 e ^)(VA e #)(Z - A c A r ) .

It is easy to check that &+ is a filter, ^ c ^ + and A(^+) = A(^). We say that 9is superclosed if 9 = &+. That is, if whenever 2 e 9 and 2 - A c A r for all A e 9,then r e 9. Note that 9 c ^ c ^ + . For if F e ^, then F = 2 - O for some 2 6 ^and some set <J> with |<J>| < A, so F e 9*+. Consequently ^ + is closed.

The next lemma gives us an alternative characterization of the superclosure of afilter. A set is an almost intersection for SF if 4> c A A for all A e f . For example,

A is an almost intersection for the almost principal filter generated by A. Clearly,every set of cardinality less than A is an almost intersection, and every almostintersection has cardinality at most A. We call an almost intersection of cardinalityA a proper almost intersection. Note that the collection of almost intersections isclosed under finite unions and subsets, and in particular, intersections.

LEMMA 4.8. If SF is a filter which is not almost principal then

^ + = {2 - 4>: 2 e ^ and <& is an almost intersection for &}.

Proof. Let (&= {2 - 0 : 2 e ^ and is an almost intersection for SF}. We firstshow that ^ is a filter. Clearly, 0 $ c§, since ^ is not almost principal. Take 2 - 3>and 2 - < D ' in (§. Then their intersection is (2 D F ) - ( U <*>'). Clearly,2 D 2 ' e SF, and since the union of two almost intersections is an almostintersection, O U O ' is an almost intersection. Thus ^ is closed under intersec-tion. Now assume A 3 2 - <X>, where 2 e ^ and is an almost intersection. ThenA = (A U 4>) - (<D - A). But A U O 3 ( 2 - O ) U O ^ 2 : , so lies in 9, and - A Q 4>, so is an almost-intersection. Thus A e ^ and so ^ is closed undersupersets. Hence ^ is a filter.

We next show that ^ c 2F+. Let 2 e SF and let 4> be an almost intersection of9. Then O c A A for all A e ^ . Thus 2 - A c A 2 - 3> and so 2 - O e ^ + .

Finally, we show that SF+ c CS. Take F e SF+. Then for some 2 € SF we have2 - A c A F for all A e 9. Let O = 2 - F. Then 2 - A c A 2 n T = Z - <S>, and so c A A. Thus <J> is an almost intersection for ^ , so F 3 2 — O € ^.

COROLLARY 4.9. /I /i//er SF is superclosed if and only if it is closed and has noproper almost intersection.

Proof. (=» Assume SF is superclosed. We already know that SF is closed. Let2 e SF and let 3> be an almost intersection for SF. Then 2 — <£ e ^ and<DcA2-3>. Hence |4>|<A.

(<£:) Now assume SF is closed and has no proper almost intersection. Then ^ i snot almost principal. Take 2 e f and let be an almost intersection for SF. Since|4>|<A, we have 2 — O = A 2 e ^ , and because ^ is closed, this implies that2 - $ € SF. That is, SF+ c SF and it follows that 9 is superclosed.

LEMMA 4.10. Let SF be a filter which is not almost principal. Then S{&) «£ 5{gf+}.Hence, by Lemma 3.12, if S{&} is maximal then SF = SF = &+.

Proof. Take g e S{^} and 2 - $ e ^ + with 2 e & and O an almost intersection.Then (2 - 4>)s = 2 g - <!>*. Thus it suffices to show that 3>g is an almostintersection for SF. Now, for all A e SF we have 4> c A A, so in particular O c A Ag '.But then O*cAA, as required. The last assertion follows from Theorem 3.13.

The above result tells us that any filter whose stabilizer is maximal is eitheralmost principal or superclosed. However, in § 7 we will give an example whichshows that not every superclosed filter has maximal stabilizer.

LEMMA 4.11. Let 3*, <§ be filters which are not almost principal such that&<^<g<^&+. Then (§+^3:+.

In particular, &++= &+ and &+ = &+.

Proof. Take A e <^+. Then A contains some set 2 — 3> where 2 e ^ and is analmost intersection for (S. Then 2 € 3*+, and <E> is also an almost intersection forSF. But then 2 has the form 2 ' — <E>' for some 2 ' e 3* and some almost intersection4>' for 9. Then A 3 2 ' - (4> U <!>') e ^ + , so A G ^ + . Thus & c ^ + .

Putting ^ = S^+ gives ^_++ c ^ + . But also, &+ c ^ + + , so we have 3F++ = 5F+.Similarly, putting <£= ^ gives f + c f + , so it remains to show that &+ c ^ + .

Take 2 e ^ and let O be an almost intersection for ^ . Then l e f , so it willsuffice to show that <£ is an almost intersection for 3*. Take A e l . Then A = A Ffor some F e 3>. Now c A F , and so <£ c A A. Thus <J> is an almost intersection for9, and so 2 - O e &+.

In general, a maximal subgroup lying above the stabilizer of a filter need not bethe stabilizer of a filter. (For example, the stabilizer of the almost principal filtergenerated by a moiety A of Q is contained in the almost stabilizer of the partition{A, Ac}, which is not the stabilizer of a filter.) Fortunately, superclosed filters donot suffer from this affliction:

THEOREM 4.12. / / S w ^G <S for some superclosed filter 9* then G ^ S{<§) < Sfor some superclosed filter <§ 2 9.

In particular, if G is maximal then G = S{<s).

Proof. Assume this is not true. Then by Theorem 4.7, G $W, where H is thealmost stabilizer of a partition 9>= {<£, C} of Q into two moieties and 9 is auniform filter contained in the closed almost principal filter generated by, say, 4>.But then 4>c l f2 for all 2 e 9, and so 3> has a proper almost intersection,contradicting the assumption that 3* is superclosed. Thus every proper subgroupcontaining S^) must be contained in the stabilizer of a superclosed filter.

If 9 is a filter on Q and A e f , then 9 induces a filter on A, defined by9{^ = {2 fl A: 2 e 9). Similarly, a filter ^ on A is a filter base for a filter <§] onQ. Note that ^ | A | = 9 and «0f J,A = % that k(&) = A(^jA) and 9 is closed ifand only if ^\j,A is. Consideration of these restricted filters leads to the followingresults.

LEMMA 4.13. Let 3* be a non-trivial closed filter, let A e 9 and H = S^y Then

#{A} = Sym(A){3fiA} x Sym(Ac).

In particular, H^) = S{A) if and only if 3? is the closed almost principal filtergenerated by A.

Proof The groups Sym(A){Sf|A) and Sym(Ac) commute, since Sym(A){gf|A}acts on A and Sym(Ac) acts on Ac. It follows from Lemma 3.6 that Sym(Ac) <H{A). Take g e Sym(A){SF|A} and F e 3>. We need to show that F* e 9. It sufficesto consider F c A. Now since g e Sym(A) W A , , we have T8 e 3^[A c 9, and soV8 e 3*. Therefore g eH. But since g stabilizes A, it follows that g e H{A}.

Finally, //{A} = S{A) if and only if Sym(A) = Sym(A){35:|A}, which is the case ifand only if S |,A is trivial. However since & is closed, this is equivalent to sayingthat cF is the closed almost principal filter generated by A.

The next theorem tells us that in looking for maximal subgroups, we canrestrict our attention to uniform filters.

THEOREM 4.14. Let & be a filter on Q which is not almost principal, letH = S(&), and suppose A e ^ . Then H is maximal in Sym(Q) if and only ifSym(A){gF|A} is maximal in Sym(A).

Proof. Since SF is not almost principal, $\j,A is a non-trivial filter on A, soSym(A){sÂ} is a proper subgroup of Sym(A). Also if either 5{SF} or Sym(A){3Sr|A}

is maximal, then the corresponding filter ^ or ^ | A is superclosed, and henceboth the filters & and &\L are superclosed.

(4=) Assume Sym(A){SÂ} is maximal in Sym(A). By Proposition 3.3, there isa maximal subgroup G of 5 containing H. Then, by Theorem 4.12, G is thestabilizer of a filter ^ 3 & on Q. Now, H{A} = Sym(A){3FjA} x S(A) and G{A} =Sym(A){c9|A) x 5(A), so since H^G we have HlA) =£ G{A}, whenceSym(A){3f:jA} =s= Sym(A){cg|A} <Sym(A). But Sym(A){3F|A} is maximal in Sym(A)so Sym(A){3FjA) = Sym(A){Â}. Therefore by Theorem 3.13, ^ | A = ^JA andhence 8F = (S. Thus G = H and H is maximal in 5.

(^>) Assume H is maximal in 5. Again, by Proposition 3.3, there is a maximalsubgroup K of Sym(A) containing Sym(A){SFjA}. Then by Theorems 4.12 and4.10, K is the stabilizer of a superclosed filter 3^3 ^ | A . Since S i f ^ ^ A , itfollows that 3t\ 3 &. We claim that H^S{X^. Then, since H is maximal,H = S{X]] and so by Theorem 3.13, &=%]. It follows that &{t± = X, and thusSym(A){3Sr|A} = S{x) = K which is maximal in Sym(A).

We shall show that each element g e H lies in S{X^. Since 3F is not almostprincipal, there is a moiety 2 of A such that 2 e f . Then both |A — (2 n 2S~')| =jA| and |A - (I,8 fl 2)| = |A|, so there is a permutation h of A which agrees with gon SOS*"1. We show first that h e Sym(A){;nA}. Take T e ^ j A . We maysuppose (by replacing r by m s n ^ n Z * " ' ) that r c Z f l Z * DZ*"'. Thenr* = F e # , and r* c (2 n 2*"')* = 2 s n 2 c A, and so T'1 e S |A. Similarly,F'1 ' e ^ | A and hence /i 6 Sym(A){sf|A> ^ K. Now we can show that g stabilizesdC\. Let reSTf. Then since 2 HZ*"1 e ^ | A c X, we have r n 2 n2*"'e X So(r fl 2 fl 2«"')g = (r n 2 Pi 2s"')'1 G JC since /ie/C = 5{^}, and it follows that T8 e3t\. Thus g stabilizes Jtf.

THEOREM 4.15. Let SP be a closed filter which is not almost principal. Then S^îs transitive on moieties in &.

Proof. Let 2, and 22 be moieties in 9. If |2, - 2 2 | = | 2 2 - 2 i | , then, byLemma 3.6, there exists g e S^n^y^SiSF) interchanging 2 , - 2 2 and 22 —2^Hence we can assume \2.x — 22| < |22 - 2,|, and so |2, fl 22| = |2,| = K. Since 9 isnot almost principal, there is some A e ^ which is a moiety of 2 , f l22 . ByLemma 3.6, Sym(Ac)^5{gf}, and hence there exist g, h e5(A)^5{y} such that2f = 22 and 2j = 2,.

5. Almost stabilizers of partitions

We now return our attention to maximal subgroups containing the stabilizer ofa partition. Let 9 = {Qt: i el} be a partition of Q into /i = |/| parts each of size|Q,| = A. Then *: = A/*. When Q is finite, S{g>) is a maximal subgroup for anyp>l, A> 1. The case where Q is infinite is much more complicated. We show inthis section that in all cases a suitably defined almost stabilizer is maximal.

We have already seen (Proposition 3.1) that if 9 is a finite partition, that is, if/i is finite, then the almost stabilizer of 9* is maximal. Macpherson and Neumann[7, Observation 6.3] constructed a maximal subgroup containing S{9) in thespecial case where n = |/| is a regular infinite cardinal such that /i < cf(*-) (andtherefore A = K). For each g e S, let

We can think of this as the number of parts over which g spreads QJ} excludingparts which receive almost no elements. They defined a maximal subgroup H by

H = {g: ty(g) < \i and ^(g"1) < \x for all y e / } .

Our first result applies to arbitrary partitions with an infinite number of infiniteparts of the same size, and includes these subgroups of Macpherson andNeumann as a special case. Recall that if 9* = {Q,: / e /} is a partition of Q into \iparts, each of cardinality A, where A, ju are infinite, then

AStab(0>) = {geS: (Vi e I)(Bj, k e /)(Q? =AQ, A Qf ' =AQ*)}.

THEOREM 5.1. Let 9 = {Q,: / e /} be a partition of a set Q of cardinality K intofj, = \I\ parts, each of cardinality |Q,| = A, where both A and fj. are infinite. For anyS c Q , let

and define

Then 2F is a filter and S{3F} is a maximal subgroup of S = Sym(Q) containingAStab(^).

A set 2 lies in 2F if it 'almost contains almost all the parts of 9*\ First we provethat SF is a filter and establish a few of its properties.

LEMMA 5.2. The filter 9 is a non-trivial uniform filter on Q with

Proof It follows from the definition of & that Q e ^ and every superset of aset in 9 also lies in 9, and it is easy to see that 0 £ 9. Now choose 2, T e SF.Now, N(l H T) = N(l) U N(T) and both N(X) and N(T) have cardinality lessthan ju, whence 2 fl T e 9. Thus ^ is a filter.

Moreover, each 2 e ^ contains A points of each of /i parts of 9, and hence|2| = jtfA = *:. Therefore ^ is uniform. To show that 2F is non-trivial we constructa moiety in $*. Choose iel, and for all jel-{i} choose a point c} e Qy. Let2 = Q - (Q, U r) , where T = {c;: j ± /}. Then N(2) = {/}, so 2 e 9, and |2C| =max{A, JU} = K SO 2 is a moiety in the filter.

If g e AStab(^) then |N(2)| = |N(2*)|, so 2 e ^ if and only if 2* e &. Thusg e 5{SF}, and so AStab(^) ^ 5{Sf>.

Proof of Theorem 5.1. Suppose that S{g;}<M<S. Then by Theorem 3.7,M ^ 5(2} for some quasifilter S g f . Since S{gf} =£ 5{2}, ^ is a proper subset of 2..

LetHeQ-P. Since 2 $ ^ , the set N(2) = {/ e / : Q, f A2} has cardinality JU.Consider a permutation g which fixes each Q, setwise and acts as follows. For/ $ N(2), let g fix Q, pointwise. For / e N(2) the set Q, - 2 has cardinality A.Choose a partition Q, — 2 = Q/ U Q? of Q, — 2 into the disjoint union of twomoieties. Define g on Q, so that (Q, D 2)s c Q]. Note that

Since g fixes each Q, setwise, g e S{9>) ^S{^} <S{2.), and hence 2 s e l ButUi6^(2) ^? is contained in (2g)c, by definition of g, and is contained in 2C bydefinition of the Q?, and hence is contained in (2g)c D 2C = (2g U 2)c. Thus2gU2=£#fQ, and hence by the definition of a quasifilter, 2 g n S e l Now2 x f l 2 c U/«yv(z) &h a nd a s -2 is closed under supersets, T = U / ^ z ) &, e Si.

Now consider a permutation h with the following property. We choose apartition N(2) = NX\JN2 of N(2) (which has cardinality ju) into two disjointmoieties. Choose heS{9l) such that F* c Q-eA,( Q(.. Then to e S^} <5 { 2 } andhence T'1 e Si. On the one hand, TDTh=0, whence r n Th $ ± On the otherhand, U,eyv2 &, c r D (r*)c = ( r u T'X whence T u r ^ ^ ^ Q . This contradictsthe fact that 2. is a quasifilter, and it follows that 5{3F} is maximal in S.

In the special case of partitions into infinite parts of size A where A < JC and A isregular, the maximal subgroup found above can be defined in a rather differentway.

THEOREM 5.3. Let K, A be infinite cardinals, with X<K and A regular. Let Q bea set of cardinality K, and let & = {Q,: / € /} be a partition of Q of into K parts,each of cardinality |Q,| = A. For any 2 c Q, let

and define

Then $ is a trivial filter, and BKS{j,} is the maximal subgroup above AStab(^)given by Theorem 5.1. That is, BKS^) = S{&} where 2F is the non-trivial filterdefined in Theorem 5.1.

Of course this suggests that 'trivial' filters might not be so trivial after all. Theproof that $ is a filter is like the proof of Lemma 5.2. However, since thecomplement of each set in $ has cardinality at most A < K, $ is a trivial filter, andtherefore S{^} = S. We will prove Theorem 5.3 via the following four lemmas,assuming throughout that the assumptions of the theorem are in force. But firstwe need some notation. Analogous to jUy(g), define

to be the number of parts over which g spreads Qy. Noticing that v;(g) =\M((Qj)c)\, we can characterize 5{^} in a manner reminiscent of (but differentfrom) the Macpherson and Neumann groups mentioned above.

LEMMA 5.4. We have S w = {g e S: vy(g) < A and vy(g~') < A for all y e / } .That is, g e 5{Jf} if and only if g and g~l spread each part Qy over fewer than A

parts.

Proof. We have g e S{Jf) if and only if (VI c Q)(|M(2)| < A<=> \M(XS)\ < A), orequivalently,

(V2 c Q)(|M(2)| < A => |A#(2*)| < A and |Af (2*"')| < A).

In particular, |M(Qyc)|<A, so if geS{ J f } then taking 2 = Qy

c, we have vy(g) =|M((QJ)C)| < A and similarly, Vyfe"1) < A.

Conversely, suppose the set M(2) = {/ e /: Q, £ 2} has cardinality less than A.Then 2C c U/eM(2:) Q#, so> if g spreads each Q,, for / e M(2), over fewer than Aparts Qy, then

(2«) e s U QfsLJQyA/(2) V

where 7 is the union of fewer than A sets, each of cardinality less than A, andconsequently as A is regular, | / |<A. Thus |M(2S)|<A. Similarly, if g~l spreadseach Q, over fewer than A parts Qjt then |M(2S"')| < A.

From this lemma we can prove:

LEMMA 5.5. Let $ and cfi be as above. Then BKS{J>) =s S{9).

Proof. Since & is a non-trivial closed uniform filter, we have fl^S^), byLemma 4.5.

Now take g e 5{Jf) and 2 e 9. Then |N(2)| < K. Take i e N(2). Then |Q, - 2 | =A, and so |Qf —2S| = A. Now g spreads Q , - 2 over fewer than A parts, andA is regular, so there are at least one and at most A parts Qy such that\Qj H Qf - 2 g | = A. Then |Qy- - 1 8 \ = A, and so j e N(I8).

We will show that every / e ^(2*) is obtained in this way, and it then followsthat |W(Z*)|<*\ Assume | Q y - 2 g | = A. Then | Q f " ' - 2 | = A , and sinceg~x spreads Qy over fewer than A parts, there is some part Q, such that|Q, fi Qf"' - 2 | = A. Then / e N(X) and |Qf D Qy - 2*| = A.

Thus it remains to prove that S{3/;) ^ BKS{^,}.In view of Lemma 5.5 above, if x is a permutation which spreads each of K

parts Qy over A parts, then clearly x $ BKS{J,)f so BKS{^} is a proper subgroup ofS. Conversely, a property of this sort characterizes the permutations x $ BKS{J,y

LEMMA 5.6. Suppose the permutation x $ BKS^y Then there is some set J ^1with \J\ = K such that for all j eJ, either vy(jc) = A or vy(jc~') = A.

Proof. Let {Qy: jeJ) be the collection of all parts which are spread over Aparts by either x or x~\ That is,

Assume that \J\<K. Let Z denote the set of all integers, and let

r = U UQfjeJ reZ

be the union of all the cycles of x containing elements of any of these parts.Clearly |F| < K. Let y be the restriction of x to F, and let z be its restriction toQ - F. Then x = yz, where y e BK and z e 5{Jf}, which contradicts our hypothesis.

The proof of Theorem 5.3 is completed by:

LEMMA 5.7. S{3P) BKS{j,}.

Proof. Choose x eS^}, x $ BKS^}, and let H = {BKS^}, x). By Lemma 5.6,there are K parts {Qy: j eJ}, each of which is spread by either x or x~l over Aparts. We can assume, without loss of generality, that V,-(JC) = A for all j eJ. Wewill choose a subset K c J of cardinality K and sets {Fk: k e K} such that Tk is amoiety of Qk and such that for T = \^JkeKTk) its image Tx meets each Q, in atmost one element. Then taking 2! = Q — Yx we see that N(2) = 0 so 2 e 2F. ButHx ' = TC $ SP since N^T*) = /C which has cardinality *:, contradicting our assump-tion that x eS{Sp).

We now specify how to choose K and the sets Tk. Well-order J asJ = {ja'- <X<K) and proceed by transfinite recursion on oc. At stage a we willdefine disjoint sets Ka, K'a^.J. Ultimately we will take K = {Ja<K Ka and willhave Kc = K' where K' = {Ja<K K'a. To ensure that K U K' =J, we make certainat stage a that for k =ja, either k e Ka or k e K'a. To this end, if we have alreadyeliminated the possibility that k e K by putting k e K'p at some stage /? < a, thenwe put

Ka=\jKp, K'a=\jK'pP<a p<a

and proceed to stage a + 1. Otherwise, we put k into K by setting

Now suppose Q£ meets the A parts {Q,: i eL). Then we can choose a moiety Tk

of Q* such that Txk meets each part in at most one point, that is, so that

| r j n Q , | ^ l for all iel. We now eliminate from possible membership in K,every j = jp^k for which Qx meets any of the same parts as Q£. We achieve thisby putting

K'a ={j: j^k and Qxn Q, *0, for some i e L) U U K'p.P<a

It is clear inductively that no / e U ^ < f f ^ will be eliminated, so Ka and K'a aredisjoint. Also since \L\ = A and each Q,, for i e L, has cardinality A, at most Aindices j will be eliminated for each k put in K. Therefore after stage oc, at mostX\Ka\ of the jp will have been eliminated, so |K^|^A|Ka|. Consequently\K'\ A |/C|. But |K U /C'| = |/| = K, so it follows that \K\ = K.

Thus we have obtained K sets {Tk: k e K} such that P = [JkeK Yxk meets each

part Qh for / e /, in at most one point, as required.

A similar argument establishes the following lemma which we will use later.We first introduce some useful notation. For any collection {F,: i e /} of sets andany subset / c /, we let Fy = Uyey

rr In particular, Qy will be used to denote the

union of the parts whose indices lie in the subset /.

LEMMA 5.8. Let 9* = {Q,: i e 1} be a partition of a set Q of cardinality K into K

parts, each of cardinality |Q,| = A, where 1 =s A< K. For any infinite subset / c /with \J\ > A and any permutation x of Q, there is a subset L c / such that \L\ = \J\and for each j e L,

Proof. Let ju = |/|. We will define by transfinite induction two ascending chains

^n£^iC...cL^c...

andL I I I T I

o£L, c . . . cL/»c. . .such that L = U/?<^ Lp is the required subset and L' = U/s<M L'p is its complementin J. We construct L in stages; Lp is the set of indices added to L before the /3thstage. We need to ensure that for any j e L neither Qy - Qy nor Qy"' - Qy meetQL. To do this, we make sure that any index k =£y for which Q* meets either Qyor Q* ' is not in L. We do this by adding such indices to L'.

Choose an enumeration {y : /?<ju} of /. We will use the abbreviationAp = Qjfi. Start with Lo = 0 and LQ = 0. If jp e L'p then set Lp+l=Lp andL'p+X = L'p, otherwise set

/3 + l = Lp{jp},

If y is a limit ordinal then let Ly = U/3<y Lp and let L'Y = U/s<y L'p.First note that L U L' = J, since if jp $ L'p then jp e Lp+i. We show that \L\ = p.

Since |A^U AJ~'| = 2A, at most 2A points are added to L^ at any stage. Thus\L'p\ ^ 2A \Lp\ and so \L'\ 2A |L|. Hence as A < |/|, it follows that \L\ = |7|.

Next we prove by transfinite induction that Lp and L'p are disjoint for all ft < p.It then follows that L and L' are disjoint. Clearly, Lo and L'o are disjoint. AssumeLp and L'p are disjoint for all j3 < <5. If 6 is a limit ordinal then Ld = Up<«5 and^6 = Uf}<s L'p are disjoint. Now consider 6 = ft + 1. By our assumption, Lp andLJ( are disjoint. If jp e L'p then Lp+l = Lp and L^,+, = L'p are disjoint. If jp $ L'pthen AyDA^=0 and Ay" 'nA/ 3=0 for all y</3. But then A£PiA y =0 andA£~' n Ay = 0 , and so jy is not added to L'p+l at the /3th stage for any y < jS. Butevery index yy € L/3+1 has y p . Hence L^+1 and L^+1 are disjoint.

Finally, we show that for all j e L neither Qy - Qy nor Qy ' - Qy meet Q* forany k e L. If ;, k e L then / = jp and k = jy for some j8, y < p. But if y =£ /3 and(AJ U AJT') f l A r ^ 0 then yy 6 L and so k $ L. Thus (Q/ U Q/"') - Qy c QLc forall y e L, as required.

In the case not covered by either Proposition 3.1 or Theorem 5.1, that is thecase where A is finite, the filter & may still be defined as in Theorem 5.1. (Forfinite A, cA is just ordinary inclusion.) But & turns out to be the filter of allsubsets 2 such that |2c|<*r, and consequently S{g,} = S. We will now exhibit amaximal subgroup containing the stabilizer of such a partition. Note that in thiscase the almost stabilizer, as defined in § 2, is just the stabilizer.

The following theorem was proved with Dugald Macpherson.

THEOREM 5.9 (with Macpherson). Let 9>= {Q,: / e /} be a partition of a set Qof cardinality K into K parts, each of cardinality |Q,| = n, where 1< n < No.

Then BKS{gf>) is a maximal subgroup of S = Sym(Q).

We begin with a lemma which gives a characterization of the elements ofBKS{g,}, and tells us that it is a proper subgroup of 5.

LEMMA 5.10. Let 0> be as in Theorem 5.9. For any g eS define

Then

Thus BKS{0>) is a proper subgroup of S.

Proof. Let H = {geS: | / (g) |<j r} . Clearly, BK^H and S{9)^H, so^ S ^ j ^ H. So take /i e H and let J=J(h). Then |7| < #c and so |Qy| = n \J\ < K.Now it follows from the definition of J(h) that Qjr is a union of parts, and hencealso Qy is a union of parts, and |Qy| = |Qy| < ic. Define a permutation g e BK

fixing QyV pointwise and acting as h~l on Qy. Then /ig e S^} and so h e BKS{g,).Thus H^BKS{(9>).

To prove Theorem 5.9, we take any xoeS — BKS{9) and consider the groupH = (BKS{&), x<>). We will show that H = S. Let /„ = J(x0). Since x0 $ BKS{^]y wehave |/,,| = K.

The next lemma is crucial to the proof of the theorem.

LEMMA 5.11. Let 9* = {Q,: i e 1} be a partition of a set Q of cardinality K into K

parts, each of cardinality |Q,| = n, where \<n< N(). Take xoeS — BKS{^} and letH= {BKS{ap}, x{)). Let J0 = J(x0) and for each j eJ() choose an element a, e Qy.Then there is a moiety J of J{) and an element x e H such that, for all j e / ,

< = «» (1)

Qy no? = {<!,}, (2)

Qf-Qj^Qjr. (3)

That is, x fixes a, and sends the rest of Q, outside Qy.

Proof. We shall prove by induction on / that, for all 1 =? / =s= n - 1, there are amoiety 7, of 7,, and an element x, e H such that, for all j e /,,

a? = aj, (4)

\QjnQ?\^n-i. (5)

Then when i = n — \, we obtain a moiety /„_, of /„ and an element xn_ieHsatisfying conditions (1) and (2). We now apply Lemma 5.8 with / = /„_, andx = *„_, to produce / c / n _ , with \J\ = K such that condition (3) is also satisfied.

Now for the induction. First consider the case where / = 1. Choose a moiety 7,of /„ such that no two elements of the set {af\ j eJ\) lie in the same part of 9*.Then there is some h e S{&} such that a*"h = ay for all j eJ{. Let x} = xoh. Clearly,condition (4) is satisfied. Also, |Q, D Q/'| n - 1, since Q/° $ & for j e Jn.

If n = 2, we have nothing more to prove, so suppose that n s* 3. Suppose thatconditions (4) and (5) are satisfied for some / such that 1 ^ / =£ n — 2. ApplyLemma 5.8 to 7, to produce a subset L, c / ; with |L,| = K such that

(QfUQf)-Qj^QLr (6)

for all / e L,. Choose a moiety 7,+1 of L, and hence of 70. Since L, - 7 / + 1 and Lfare both moieties of Jc

i+X, there is an element y e S{Sfi} which fixes Qy,+I pointwiseand interchanges QLj-Ji+] and QLc Define a permutation 2 e 5 ( ^ as follows. Ifj e7-+, or j eJi+i and |Qy D Qf\ «£n — / — 1 then z fixes Qy pointwise. Otherwise,|Qy flQy'| = « - / 2 = 2 , so we can choose some bj^aj in Qy fl Qy'. Also|Qy D Qf'\ = n - i<n, so we can choose some cy € Qy - Qyr'. Let z interchange b,and cy and fix the rest of Qy pointwise.

Set xi+l=XjZyXj. We need to check that conditions (4) and (5) hold for alljeJi+i. Firstly, since xh y and z all fix ajy then a*l+l = aj. To show that|Qy D Qf+I\ ss n - i - 1, we show that

QynQf~+' c Q y D Q f ' (7)

and that the inclusion is proper if |Qy fl Qf) = n- i. Then

|Qy D Q^' l = |Qy n Q^'-l ^ n - i - 1,

as required.If a e Qy - Qy'' then ax' <£ Qy. Then by (6), aXi e QL.. Since z fixes QLr and y

sends it to QL _y.+l, we have ax'zy e Q* for some k e L, - / , + , . Then applying (6) toax'zy and QA.( we have either aXi+leQk or fl^'^1 e QLr. In either case, we haveflr'+l $ Qy, Thus a $ Qy n Qy

vr+I\ so the inclusion (7) holds.Now, suppose j eJi+i and |Qy D Qyr'| = n - i. Since b, e Qy fl Qy', we have

bj = df for some dj e Qy D Q^1'. Then

d?" = dx'zyx' = b)yx' = cf' = cf i Qy,

Thus d; $ Qj H Qf+> and hence |Qy D Qf-'| < |Qy D Qf ' | = n - / . Thus for alljeJi+l, we have |Qy H Qy

v+I| n — / — 1, and so condition (5) is satisfied for/ + 1. Then by induction, both conditions (4) and (5) are satisfied for all

In the proof of the following lemma we use the standard notation

Kz = {z~lyz: y e K}

for conjugates of a subgroup K.

LEMMA 5.12. Take x{)eS — BKS{&)y and let H = (BKS{^}, x{)).Then there is a moiety i c / such that H is full for Qy. That is, the setwise

stabilizer in H of Qy induces the full symmetric group on Qy.

Proof. Let 7() = J(x()) and for each j € /„ choose an element ay € Qy. By Lemma5.11 there are a moiety J of 7,, and x e H such that ax = ajt Qy fl Qy = {ay} andQy - Qy g Qy. for all j e J. Let F = {ay: j e7} and let K be the pointwise stabilizerof Q7< in 5 ( # } . Then K is full for F. Since Qx D Qy = F and Q j - F c Qy,, it followsthat K(r} fixes Qj setwise and fixes Q y - F pointwise. Thus if we denote the

symmetric group on Qy by 5", and let Ko be the subgroup of 5 ' induced on Qy byK, then Ko = Sym(r). (Here we are identifying Sym(r) with 5(rc) whereP = Qy — F, in violation of our usual practice of identifying it with 5(£2_r). Wewill adopt this modified convention for the rest of the proof of this lemma.)

Now consider the partition 9>x = {Q?: / € / } . Clearly S{9*}^H. Choose amoiety L of J and for each j € L choose a point bj e Qy - {ay}. Then H contains apermutation z which fixes Qy pointwise for j eJ - L, and fixes Qy setwise andinterchanges ay and bj for each j e L. Now (K{r))

z is full for Tz. Also, z fixes Qysetwise and so does K{r^; hence (K{r))

z fixes Qy setwise, and fixes Qy — Fz

pointwise. Thus if we let AT, be the subgroup of 5 ' induced on Qy by Kz, then

Now, since TC\rz = {dj\ jeJ — L] has cardinality K, we can deduce byapplying Proposition 2.1 to Qy that (Ko, /C,) = S y m ( r u P ) . Using a similarargument several times with different choices of the points 6y and the moiety L of/ , we obtain the full symmetric group 5 ' induced by H on Qy\ Then H is full forQy\ But since x e H, it follows that H is full for Qy.

We can now complete the proof of Theorem 5.9.

LEMMA 5.13. Let H be as in Lemma 5.12. Then H = S.

Proof. By Lemma 5.12 there is a moiety / of / such that H is full for Qy. Let Lbe a moiety of / such that I = JU L and \J D L\ = K. There is an element z eS{&>}

such that QL = Qj. Since zeH, H is also full for QL. So by Proposition 2.2,H = S.

We leave the following problem open:

Let 9* = {Q,: i e 1} be a partition of an infinite set Q into fj. = \I\ parts, each ofcardinality |Q,| = A. Characterize all the maximal subgroups of Sym(Q) whichcontain 5{go>.

6. Filters with a chain as a filter base

A fundamental problem which we have not solved is to find a set-theoreticcondition on a filter ^ which is necessary and sufficient for its stabilizer 5{3f} to bemaximal. Clearly the problem reduces to asking:

What are the maximal subgroups lying above the stabilizer S{3?} of a filter &?

In the present section we consider this question for filters cF which are 'simple'in that they have a chain as a filter base, that is, there are some linear ordering(/, < ) and filter base 38 = {A,: iel) for 8F such that A, z> Ay whenever i<j.Note that in particular, if a filter & has a countable filter base, then it has alinearly ordered filter base formed by taking finite intersections.

Let $* be a filter which is not almost principal (and therefore non-trivial), andlet Ogf = D^e3P A. Associated with cFwe have several filters, among them:

= {A: A U $ e ^ for some almost intersection of

Recall that the superclosure cF+ is closed and 5{3F} =sS{^} ^S{^+} <S. It is easilyseen that

and

If ^ V ^ J ^ , then the principal filter ^ with filter base {O^} will also have^ S ^* c &+ and satisfy 5{gf} S{3fm) ^ 5{i^}. The maximal subgroups lying above5 ^ } are given by taking cF = ^* in the following proposition which summarizesTheorems 4.1 and 4.2 about almost principal filters *&•.

PROPOSITION 6.1. Let |Q| = K. Suppose & is a non-trivial filter on Q and that SFis the closure of the principal filter with filter base {<!>}.

(a) / / |<J>| < ic then S{&} = AStab() is a maximal subgroup.(b) If\®\ = K then S{^ < AStab(^) where AStab(^) is the almost stabilizer of

the partition $P= {<!>, C}, and AStab(^) is the only maximal subgroup lyingabove S{§).

In the case where a filter $F is not almost principal and has a chain as a filterbase, the question of what maximal subgroups lie above S{3f) is settled by:

THEOREM 6.2. Let \Q\ = K and let & be a filter on Q which is not almostprincipal and has a chain as a filter base. Then the stabilizer 5{3F+} of thesuperclosure cF+ of cF is a maximal subgroup lying above 5{^}. Also the followinghold.

(a) / / <J>SF = 0 , there are no other maximal subgroups above S^).(b) / / p=t0 then S{Sp} lies below exactly one other maximal subgroup. If

|4>SF| < K then this is S ^ = AStab(4>3f). / / |<£>p| = if then it is the almost stabilizerAStab(^) of the partition &={&&, <&%}.

COROLLARY 6.3. If & is a filter which is not almost principal and has a chain as afilter base, then its stabilizer 5{3f} lies below at most two maximal subgroups.

Combining this with Proposition 6.1 gives:

COROLLARY 6.4. If SP is the closure of a non-trivial filter cF having a chain as afilter base, then S^ lies below at most two maximal subgroups.

As we saw in §4, for the stabilizer of a non-trivial filter to be maximal, thefilter must be closed. Our original motivation for this work was the conversequestion of whether the stabilizer of a closed filter which is not almost principal,is necessarily maximal. This it answers negatively. In fact, if the filter is theclosure 8F of some filter & having a chain as a filter base, then the possibilities(which, since 5{3f, ^5 { # } , must of course be among those given by Theorem 6.2)are summarized by:

THEOREM 6.5. Let \Q\ = K and let & be a filter on Q which is not almostprincipal, and has a chain as a filter base. Then one of the following holds:

(a) 5{^} is maximal;

(b) S{<pr} lies below exactly two maximal subgroups: if A(^} < K these are S(^)

and S(^i)t and if X(SF) = K they are 5 { ^ } and the almost stabilizerAStab(^) of the partition 9>= {&&, &%};

(c) S{fF) lies strictly below exactly one maximal subgroup, namely 5{3F+).

Each of these alternatives is possible, and we will give the conditions underwhich they occur. The content of Theorem 6.2_is not identical to that of Theorem6.5. For instance, as we will see, even if ^ c 5F*, it can happen that 5{gt} is not asubgroup of Sjs^}.

Some simplifying assumptionsWe now embark on the proof of these theorems. When dealing with a

filter cF which is not almost principal and has a linearly ordered filter base 38 ={A,: iel}, there are some convenient assumptions about the filter base whichmay be made without loss of generality. These, and the notation which wenow introduce, will be adopted without further comment whenever we considersuch a filter. Let fi be the cofinality of ( / , < ) . We will speak (a little inaccurately)of ju as the cofinality of the chain filter base 38. If ^ = 1 then & is principal, so wemay assume that \i is an infinite (regular) cardinal. By replacing / by a well-ordered cofinal subset, we may suppose that / is the set of ordinal numbers / ={a: a< ii}. Let A(^) = A. Since | Aa| = A for all sufficiently large a, we may alsoassume that |Aa| = A for all a. It is then clear that (i A. Let Qa = Aa — Aff+1 fora< /i and Q^ = D«<^ Aa = <P&. Since 9 is not almost principal, for each y < /ithere exists 8 < /i such that |Uy«a<6 Q J = A, and therefore since \{a: <x < 6}\ <fj, =s A, there exist arbitrarily large <*<ju for which |Qa | = A. Consequently wemay assume that IQQ.1 = A for all a</j,. (This might be a good time to draw apicture of the situation under these standard assumptions for reference during thecourse of the proof. That produced by one of the authors is given in Fig. 1.)

The filters associated with ^ c a n be characterized in terms of the filter base 38:2F* is the filter with filter base 38* = {Aa - (~)p<fi Ap: a<fx}; and the super-closure ^+ of 9 has filter base

38+ = {Aa — 3>: oc < /i, |<X>| =s A and 4> c=A A^ for all j3 < \i).

If \i =£ cf(A) (which is the case if A is regular, and in particular, since \i is regular,if \i = A), then the superclosure can also be represented in the form

r = { I c Q : |N(Z)|<f*} where N(2) = {ore/: Q ^ c ^ } ,

suggesting (correctly) a connection with Theorem 5.1. (To justify this lastcharacterization of SF+ notice that 2 e 9*+ if and only if there exists /? such that = A^ - 2 = U/s=sy«^ S2y - 2 is an almost intersection of ^ . Obviously 3> c A Aa

for all a =s p. For a el with oc > /?, we see that $ c A A t f = Ua«ysM &y if and onlyif IU/s«y<o'^y ~ 2 | < A. So is an almost intersection of 2F if and only if|LJ/}=sy«r^y ~ 2 | < A for all a with fi< a<fi. But if (j, =£cf(A), this is equivalentto the condition that \Qa — 2 | < A for all a with /S =s a < fx. So )3 exists if and onlyif Qa c

A 2 for all sufficiently large a < ii, that is, appealing to the regularity of \i,if and only if |W(I)|<f*.)

By Corollary 3.8 and Theorem 4.4 we know that when 9 is a filter withA(^) < K, the maximal subgroups lying above S{9) (which include, but as we willsee later are not necessarily limited to, those above S{§}) are all of the form S{(S)

FIG. 1

where ^ is a closed filter containing 9. An analogous, though somewhat lesselegant, result holds when A = K. Thus it turns out that to find all such maximalsubgroups it suffices to investigate which closed filters ^ 3 ^satisfy S[9)^S{V).

THEOREM 6.6. Suppose 9 is a filter which is not almost principal and has a chainas a filter base. Let <S^_9 be a closed filtej^satisfying S{&}^S{V}. Then the onlypossibilities are <S = 9, 9*, 9+, 9*, 9+ H9*.

The reader should be warned _that_these filters need not _be distinct. Forexample, if 9 is superclosed then 9=9* = 9+. Also 9*, 9^n9* may not existas filters (when &<? = 0 ) , and even if it is a filter, 9+ D 9* may not be closed(when 0< |O, |<A) . The proof of the above theorem will be accomplished bymeans of the following three lemmas, the third of which (Lemma 6.9) is just arestatement of Theorem 6.6 as a sequence of propositions. The first of theselemmas holds in a more general setting:

LEMMA 6.7. Let & be a filter with \®9\ = X{9) = A. Suppose^ is a closed filtercontaining & with S{3?} *£ S m and A(^) = A. Then either <$^&*or&*Q %

Proof. Suppose Te^-IF*. Since r$9*, it is not the case that $ , c A r .Therefore m $ y has cardinality A.

Case I: | r D Oy| < A. Since 9 £ % for all A e 9 we must have A f l T e ^ . But

A n r c ( A - $ , ) u (r n <=A A -

Therefore &* c <8, so A(^*) = A as A = \{&) ^ k{&*) s* A(^) = A, and since <S isclosed with A(^) = A(^*), it follows that &* c <£

Case II: | r n ^ > ^ | = A. Let g be a permutation which fixes Q - Ogc pointwiseand interchanges P fl Ogc and F D <&&. Then g fixes each A e f setwise sog e S{gP] =s 5{cg}. Since F e <£ and A e f c ^ , we see that r n A € CS, so A ^ J y 3( r D A) fl ( r n A)* € <S for all A e ^ . Therefore ^ * c « and consequently 3^* £ <gagain.

LEMMA 6.8. Suppose & is a filter which is not almost principal and has a chainas a filter base. Let <£^ 8F be a closed filter satisfying 5{3F} *s S{<g}.

Then either ^ c ^ + , or <P&^0 and <S=&*.

Proof. Suppose *# is a closed filter containing *&, and let A = A(^). Assumealso that S{3f) ^ S{<3). For each F € % let F^ = Q a D F and F^ = Qa D Fc for alla =s JI. There are two cases to consider.

Case I: for every F e ^ , there is some y<ju such that for every 6<ju,IUy«a<6 F^.| < A. Then taking 2 = Ay we see that for every A e ^ there is some6<fi such that A a c A and therefore

2-AcAy-A,= U rau u r;=A u rffSr.y«cr<6 y=so'<d y«a-<6

Hence F e + . But this is true for every F e (3, so c &+.Case II: there is some F e with the property that for every y < ju, there exists

6 < jU such that |Uy=s£r<<5 F^.| = A. Define a sequence of ordinals 5a by transfiniterecursion as follows:

(a) <50 is the least 6 such that |U/*<<!> T^| = A,

(b) 6^+, is the least 6 such that |U<5a«/3«5 r y = A,

(c) 6a = sup{<5p: f}< a}, if or is a limit ordinal.

If 6a is defined and 6t t .</i, then by our assumption da+l is also defined.Therefore the least a for which da+l is undefined has da = fi and {dp: P<a}cofinal in fi. Since ju is regular, this must be a = p.

As {6a: <x<ii} is cofinal in fi, we can replace the filter base 39 by the 'cofinal'subset {A^: a < //} which also generates 2F. Renaming this new filter base as{Aa: ff</i}, we see that for all a < \i the corresponding sets Ya, Va (defined interms of the new Qa) satisfy |F a | ^ A = |F^|. Consider any permutation g whichfor all a<(i, exchanges F a for a subset of F^ of equal cardinality, and fixes allother points. Now g fixes Aa setwise for every a, and so g e S{&} ^S^. On theother hand, let 2 = Ao D F. Then 2 e and therefore 2 g D 2 e (S. But 2 g n 2 = F^so FM e CS. It follows that FM = 0 , and consequently as F^ c QM = <&&, thatOaf # 0 and QM e ^,_whence % c ^.

To show that (S=9'*, it suffices (by closure of <$) to show that ^ c . SupposeS e ' S - ^ . Since Q^e^, we may assume (by replacing 2 by I f l Q J that2 c QM, and hence that 2 c: QM, since 2 £ ^*.

Let v = IQJ. If v is finite, then no proper subset 2 a QM can be in (3. We showthis by induction on |2 | . Since ^ is a filter, |2 | = 0 is impossible. So suppose1 *£ |2 | < v and 2 e CS, but no subset of QM with cardinality less than |2 | is in (S.Consider any permutation h which fixes all points in Q^ and takes some element

of 2 to an element of QM—2, so 2* =£2. Since h fixes each Aa setwise,h e S w < S{<S} so 2* n 2 e <& But |2'' D 2 | < |2 | and therefore 2* H 2 <£ <£ by ourinductive assumption.

If v is infinite then we must show that 2 = V Q M , that is, | 2 ' | < v where2 ' = Q / i - 2 . So suppose that | 2 ' | = v2=|2|. Consider any permutation h whichexchanges 2 for a subset of 2 ' of the same cardinality, while fixing all otherpoints. Clearly h e S{9) ^ 5{^} and therefore 2* D 2 e CS. But 2* D 2 = 0 , whichcontradicts ^ being a filter.

LEMMA 6.9. Suppose t-F is a filter which is not almost principal and has a chainas a filter base. Let ^ 3 & be a closed filter satisfying S{3f)m S{<s). Then

(i) (g = ¥lor(S^&+n¥~<or¥*^(g^&+,(ii) iftf^^nWtthen <$= § or <#= &+ n¥^,(iii) / / " ^ * c ^ c ^ + then <£ = ¥* or cg=&+.

Proof of {iy Suppose ^ ^ * . Then by Lemma 6.8, ^ c ^ ^ T h u s X{(S) = kwhere A = A(^) = A(^+). If |<D,|<A then as <S is closed, &* = 8F_^(g^9+

as required. So suppose |4>y| = A. Then by Lemma 6.7 either ^ e ^ * so <£c&+ D ^ , or ^ * c ^ so ^ * c ^ c ^ + .

Proof of (ii). Suppose ^ c ^ + n ^ . Let A = A(^). As any_filter containing theclosed filter ^ must also have depth A, we see that A = A(^) and A = A(^+) =A_( ) = A(^). Since ^ ^ ^ and ^ is closed, it follows that ^ c % Now suppose^ c ^ and let T e <§- &. Consider any 2 e &+ n ^ . We will show that there issome g eS{3p)^ S^ and some y < ju such that

and consequently since ^ is closed, 2 e <S.For all a < ju let Ta = Q a D T, 2 a = Q a D 2 and T^ = Q a D r , 2 ^ = Qa n 2C.

Since 2 e ^ + , there is some y < jU such that |2^,| < A for all a- with y a< /i. Tosimplify notation, renumber the Aa so that y = 0.

Since F $ &, for every <5 < /i we have |Ufi«/3<M F^| = A, and thus for everya<fi there is some y < ^ such that |Ud=s0<yF^| ^ |2^|. Define a sequence ofordinals da satisfying a ^ da < n by transfinite recursion according to

(a) <5() = 0,

(b) da+x is the least y>da such that I L k - ^ y rj, | ^ |2^|,

(c) 6a = sup{<5 ,: jS < a}, if or is a limit ordinal.

Arguing as in the proof of Lemma 6.8, we see that da is defined for all a < ju,{6a: a< <u} is cofinal in //, and {A6a: oc< fi} is a filter base for $*. Now considera permutation g which fixes QM pointwise, and for each oc<\i maps Q a ontoL)da*£t}<6a+l Qp taking 2^. to a subset of Uda«/3<6a+1 T^. (Recall that since 9 is notalmost principal, we are allowed to assume that both Qa and {J6o^p<6n+I ^p havecardinality A, so by (b) above, g exists.) Now A^,= A6o so both g and g~' merelymap one filter base for 9 onto another, and therefore geS{3f}. ButLJ6a«/)<6o+l Tp c 2^., so taking unions over all a< /i yields

( r n A 0 ) - Q M c F .

Also Q M c A 2 a s I e ^ , s o Q M = Q^cA2s. Thus Ffl AocA2g, as required.

Proof of (in). The argument is almost identical to the proof of (ii) but easier!Suppose r e ^ - ^ * . Consider any 2 e ^ + . One shows as above that there issome g e S{3F) and some y < ju such that T D (Ay - Q M ) c 2 * , so 2 e CS.

For the filters SF we are interested in here, the precise connection between themaximal subgroups lying above 5{SF} and closed filters ^ 3 ^ i s given by:

LEMMA 6.10. Let ^ be a filter which is not almost principal and has a chain as afilter base. If G is a maximal subgroup with S{Sf}^G then either G = S{<s) for someclosed filter ^ D ^ , or G = AStab(^) for the partition & = {O^, O^}.

Proof. Let G be a maximal subgroup containing S{^} and suppose that there isno (by Theorem 4.4 necessarily closed) filter ^ 3 2F for which G = S{<g}. Then byCorollary 3.8, A(^) = K, while by Corollary 3.10, G = AStab(^) for some finitepartition 0>= {4>,, ... , <!>„} with n ss2 and |<J>,-| = K for i = 1, ... , n. Observe firstthat <Ps?^K<&i f°r some /, say for / = 1. If |<J>gf| < K, this is trivial, so suppose\Q>&\ = K. Any permutation h which fixes <&% pointwise is in 5{SFj. But if some part<!>, intersects Og? in a moiety of <£.? then h can be chosen to map that moiety ontoany given moiety of $>&. So we can find h $ AStab(^), contrary to ourassumption that 5{3f} *£ AStab(^).

Observe next that in fact the proof of Corollary 3.10 (via Theorem 3.9) showsthat G = 5{2}, where .2= {2: S c ' O , for some /} is a quasi-ideal, and thecorresponding quasifilter of sets whose complements are in .2 contains $>. Using(for the first time in the proof) our assumption that $F has a chain as a filter base,we see this means that Ac

ae Q. for all oc < /i, and hence for each a < ju there issome / such that A^.c "<!>,. But since A . can be assumed to have cardinality K,and A . c A£ whenever a =s /3, it follows from the fact that 9* is a partition intodisjoint parts that this / is the same for every <*</i. Therefore there is some /such that Qa c * , for all a< p.

We show that $>% = Ua<M ^ a ^ ^ V F°r suppose this is not so. Let r = 4> ,Ta = Qa n T = Qa PI 0> , and T'a = Qa D P: = Qa n O,, Then U«<, T^ = D <Df,so \{Ja<nâ\ = K- Since & is not almost principal, one of our standardobservations is that we may assume |Qa,| = K, SO as Q ^ c ^ , , we have \Ya\ < K,\r'a\ = K for all <*< JU. Consider any permutation g which fixes Q^ pointwise, andfor each a<ju, maps Qa onto Qa, taking r a onto a subset r^cr^.. ThenI C - r*| = K and thus |O, - (Of)*| = K. On the other hand, the subset (Ua<M Fa)

8

of (tyY has cardinality K and is contained in <£,. Therefore ( ^ ) s intersects , ina moiety so gÂStab(^). However g fixes each Aa setwise, so geS{9;}ÂStab(^), yielding a contradiction.

Thus <P^cK<Pi. But we have already shown that O y c ' O , , so since each ofthe disjoint parts of the partition ^ has cardinality K, we see that i = n = 2 andO.9 = *"<!>!, <J> = lf4>2, from which the lemma follows immediately.

In our preparation for the proof jrf Theorem 6.2 the main ingredient we stilllack is a means of eliminating &*+ n ^ from the list of filters ^ for which S{^} canbe maximal. By Theorem 4 . 4 ^ S^+n&i) is maximal, then ^ + n ^ must beclosed. But letting A = A ( ^ D &,) = A(^+) = A(^), we see that \<P9\ = A (since if|Oy| < A then O^$ ^ + D even though ^=AQ).

As a bonus, the following lemma also characterizes the intersection of the twomaximal subgroups above S{SP} given by Theorem 6.2 in the case where |<Dy| = A.

LEMMA 6.11. Let |Q| = K and suppose SF is a filter on Q which is not almostprincipal and has |4>sjr| = A(5F) = A. Then

(ii) ifX = K, then S^+nW.) = S{^+} n AStab(^) where 9> = {O^, 4>^}.

Proof of (i). Observing that 2 e &+ D W+ if and only if 2 = 2, U 22 for some2, e ^ + and 22 e #», it is obvious that

^_^ + c (^+ n ¥*y. For if E e ^ + then 2 U 4 y e ^ + D W*, so 2 e(^+ fl ^ ) + since 4>gf is an almost intersection of 3>+ D SF*. Therefore

f c f + n ^ c f + c (Sf+ n ~¥ly,

so by Lemma 4.11, it follows that &^+ = (SF+ n ^*) + . Hence 5{3f+n^;> =s= 5{3C+}.To prove that S^+nî^S(^), it suffices to show that for arbitrary ge

S{s?+n¥l) we have <t>gr A<ft . Now 4>ê^ + and therefore as g~'e5{3F+),(O^)«2e &+- Also O , 6 ^ * . Hence ( ^ f ' u ^ e f ' f l ^ . Thus <&Û<D£e^ + D ^ , so ^gjrCÔIc, as required.

Proof of (ii). Since 5{^;, «s AStab(^), we have

•S{y+} n S^ ^ 5{;?+} D AStab(^),

so in view of Part (i) it suffices to show that 5{3F+} n AStab(^) =^5{^7}.Suppose g € AStab(^) - S{^r}. Then (<&%)* = K Og,. But O^ e &+, so if geS{&+}

then 0)^fl (<D )« e S^+, which is impossible since |O^n (<D )g| < «r =

Proof of Theorem 6.2. Let be a filter with a chain as a filter base which is notalmost principal. Then S^)} S(^}, and_i£ SF+ D is a closed filter, 5{gi:+n^7)also, are subgroups of S{Sf+), while if 3F* is a filter, ^ + ^ ^+. Therefore byTheorem 4.12 and Theorem 6.6, S{3P+) is a maximal subgroup lying above S{3f).Alternatively in the case where /i^cf(A), one can prove maximality of 5{SF+} bycombining Theorem 5.1 and the lifting procedure of Theorem 4.14 using thecharacterization:

<|U} where N(2) = {a e I: Qac£A2}.

By Proposition 6.1, 5 { ^ } is a maximal subgroup above S{3P) if 0< |O^| < K,while if |<J>gf| = K then AStab(^) is a maximal subgroup with S{9)^S(^~}<AStab(^) for 9- {$>&, <b%). By Lemma 6.10 and Theorem 6.6 (again) these arethe only possibilities for maximal subgroups above 5{SF}. Finally, 5{3f+} and S(^~t)

are distinct (by Theorem 3.13) since these filters are obviously different. In thecase where |4>gp| = K, Lemma 6.11 tells us that 5{Sf+} n AStab(^) = S{Sf+) fl5{^;},so is a proper subgroup of AStab(^). Therefore S{<?+} and AStab(P) are distinct.

We now embark on a proof deducing Theorem 6.5 via a series of lemmas. Wewill appeal to the following combinatorial fact:

LEMMA 6.12. Suppose that p is an infinite regular cardinal, juÂ, and thatLJ*<Mra has cardinality A. // \i±cf(A) then there is some y<\i such thatILJa<y 1 = A.

Proof. Suppose |Ua<yrQ.| < A for all y < JU. If \Ja<llTa has cardinality A thenas it is the union of (j. sets each of cardinality less than A, we see that cf(A) =s= pi.

If A is regular, then cf(A) *£ fi =s A = cf(A) so // = cf(A), contrary to ourhypothesis. Thus we may assume A is singular, so JU < A and there exists asequence {rja: ar<cf(A)} of cardinals r]a<\ which is cofinal in A. For ar<cf(A)let ya be the least ordinal y such that |U/3<yr/3| s* r\a. (Note that yQ.<ju sinceotherwise \\Jp<fl

rpl ^M* <*•) L e t va = \(jp<Yarpl- T h e n 1a^v f f <Aso the va

must be cofinal in A. But since |U/3<yr^| < A for all y < JU, it follows that the ya

must be cofinal in n, so cf(A) 3= cf(/i) = fi. Hence JU = cf(A).

LEMMA 6.13. Let 3* be a filter with A(^) = A which is not almost principal, andhas a chain filter base with cofinality fx =£ cf(A). Then 9<+ = &* and consequently

Moreover if |4v| = A, then 5{^} =s S(W.\-

Proof. Recall that ^*cSf+. Suppose r e 9+ - ¥*. Let r ; = Q a n r for eacha<[i. Since T$&*, Ap-^^^T for each /S<^. That is, \Up*«<pT''oe\ = X.Therefore by Lemma 6.12 (applied to the sets r^+Q., <x<fi) there is some y<yisuch that |U^««<yr^| = A, and since \{a: )3=5 a< y}\ < JU =£ A, it follows thatIF I = A for some a^ f$. Thus |F^| = A for arbitrarily large a<pt, and thereforesince fi is regular, the set N(F) = {a< p: Qa ^

AT} = {a<n: \T'a\ = X} hascardinality p. But T e 2F+ (if and) only if |N(r)|<jU, which contradicts ourassumption that F e f + .

Now_suppose |<J>gtr| = A. To show that 5{^}=^5{^r;}, it suffices to prove thatOff 6 ^ for all g e 5{^>. Let g e 5{^} and for each or 3F=AAgyno3f = (a>in<D^)u u ra.

But this is a disjoint union, and the left-hand side does not depend on y. Socomparing y = 0 with arbitrary y, we see that |Ua<yrQ.|<A for all y<ju.Therefore it follows by Lemma 6.12 that_|Ua</«ra.| < A. Hence O^ = AO^n <£>&,that is, O s f C * ^ or equivalently

LEMMA 6.14. / / |<P,| <

Proo/. Clearly ^ c ^ * . Let A = A(^) and suppose 2 e l * . Then there is someAeS^such that A-4>3 FcA2. Thus if |<&y|<A then A c A S U $ f = ^ , so A c A 2and hence 2 e ^ .

LEMMA 6.15. Let 5F fee a filter with A(^) = A whichjs^not almost principal, andhas a chain filter base with cofinality yL = cf(A). Then &** a ^+.

Moreover, if | $ , | = A, then §a&+r\~¥~*.

Proof. We need only prove the containments are strict. Let {r)a: a<cf(A)}be a sequence of cardinals rja<X whose sum is A. For each a<fi choose asubset raczQa with \T'a\ = t]a. (As remarked earlier, since 9 is not almostprincipal, we can assume \Q.a\ = A, so this is possible.) Let F be the set with

complement V = {Ja<n C Then_/V(r) = {a<n: \T'a\ = A} = 0 , so T e &+. AlsoOycT, so if < l v * 0 then T e &*.

But (Aa - 4>3F) — r = Ua«/3<^ T'p has cardinality A (since U/3«*r/s is the unionof fewer than cf(A) sets of cardinality lessjhan A). Therefore there is no a < /J. forwhich Aw - 0>3»:CAr. Consequently r <£ ^* and therefore Y$&.

LEMMA 6.16. Let ^ be a filter with A(^) = A which is not almost principal, andhas a chain filter base £8= {Aa: <* < JU} with cofinality fj. = cf(A). Then there is afilter 9' with a chain filter base W = {A^: a<fi} such that <&9 =0andW=&.

Proof. Since I O ^ S S A and /i = cf(A), we may write ®& = {Ja<(1?>a where|ZJ<A for all OC<IJL. Let A* = Aa -\Jp«*2p. S i n ce a<n^k, it follows thatIU/3<^2^|<A so A'a=^Aa,_ and therefore the filter 9' with filter base 39'={A;: <X<\I) satisfies 9' = 9. Clearly, O^. = O3, - LU M 2 a = 0 .

Theorem 6.5 is included in the following theorem which gives the conditionsunder which each case occurs.

THEOREM 6.17. Let \Q\ = K. Suppose & is a filter on Q with A(5F) = A which isnot almost principal and has a chain filter base with cofinality /i.

(a) / / |4>SF| < A and n =£ cf(A), then S{§} is maximal.(b) / / |<J>SF| = A and jU^c^A), then S^ lies below exactly two maximal

subgroups. If X<K, these are S(^} and S(&~t). If A = K, they are 5{^> and thealmost stabilizer AStab(^) of the partition & = {O^, ®%).

(c) If n = cf(A), then 5{^} lies strictly below exactly one maximal subgroup,namely S^+).

Proof of (a). By Lemmas 6.13 and 6.14, &=cF+, so 5{#} is maximal byTheorem 6.2.

Proof of (b). By Lemma 6.13, &* = £F+ and S{§:) lies below both S{^} andS(Wt), so the claim follows immediately from Theorem 6.2.

Proof of (c). Let &'_ be a_filter given by Lemma 6.16 with W' = § andO». = 0 . Since $+ = (&)+ = (9')+ = (9')+, we have 5{SF-,^5{^}^5{SF+}, andby Theorem 6.2 applied to y , S^} is the only maximal subgroup above5{iF} = £{£}. By Lemma 6.15, §:^2P+, so by the uniqueness of ^ + (guaranteedby Theorem 3.13) 5{^} <5{3f+}.

Let ^ be a filter which is not almost principal and has a chain as a filter base.The results above essentially determine the relationship between the stabilizergroups of all the closed filters $ 3 & satisfying S{3f} ^S{.3). We will illustrate thisby sketching the argument in one of the more interesting cases.

An example. Let |Q| = K and choose an infinite subset of Q of cardinality A. (IfA = K, this subset may be Q itself.) Partition this subset into (j, = cf(A) disjointparts {Qa: a^/j.} where |QJ = A for each a. Consider the filter ^ with chainfilter base {Aa: a<p} where Aa = U ^ / } « M ^ - Note that Q>p = Qfi which hascardinality _A_. The closed filters $ 3 ^ satisfying 5{3f) ^5{c§} are§, &*, &+, &*, &+ D %. By Theorem 6.17(c), S{§) lies strictly below only one

maximal subgroup, namely S{&+), and by Lemma 6.11, 5{3F+n^;} = 5{3f+} H5{^;}

is the intersection of the two maximal subgroups above S{S/f). (These are S{Sf+)

and S{^} if A<*r; 5{3F+} and AStab({<g>.?, &%}) if A = *r.) Also let ^ ' be a filtergiven by Lemma 6.16. Then

and f-F* is not one of the filters given by applying Theorem 6.6 to 2F', so

Putting this together with results explicitly stated above, we see immediatelythat in this example, there are the following strict containments between thestabilizers:

and these are the only containments between these groups.The only containments (all strict) between the corresponding filters are:

+ n

Notice for example, that even though & a ^ and S{3F) < S^. j , the groups S{^}

and S(^rm} are incomparable.

7. A superclosed filter with non-maximal stabilizer

We know from Lemma 4.10 that every non-trivial filter whose stabilizer ismaximal is either almost principal or superclosed. In this section we give anegative answer to the question:

Is the stabilizer of every non-trivial superclosed filter maximal?Consider a set Q of cardinality K and a regular cardinal A satisfying

a)<A<cf(tf). Consider a descending cu-chain I 0 3 2 , D 2 2 3 . . . such that

(a) 2* = n«<a»2, is a moiety of Q;(b) |Z, - 2/ + 1 | = K for all i < w.

Consider a descending A-chain Fo=»rj 3 ^ 3 . . . =>r^=>... of subsets of 2*satisfying

(a)a<Ar; = 0;(b) |r;-r;+1 | = /cfora

Choose a moiety V* disjoint from 2* meeting every 2, in a moiety and define

r =r'ur*Let & be the filter with filter base {2 , -n iV i < (o, a<X} and let ^ be the

filter generated by {T'a: a<\}. Note that k(&) = k{<0) = K. Then A e l(respectively A e *$) if and only if A ic-contains some 2,- D Tp (respectively T'p).

LEMMA 7.1. Let 3* and <& be as just defined. Then

(a) 2F is super closed;

(b) S{^^S{rSY

Proof, (a) By Corollary 4.9, it is enough to show that SF has no proper almostintersection. So assume <1> is an almost intersection for §>. Then

since |<J> - 2, | < K for all / < co and a> < cf(#c). So c * 2Also,

a<X

since |<J>- Fn\ < K for all a< A and A<cf(*r). So O c T * .But since 2* and F* are disjoint, we must have |4>| < K. Thus there is no proper

almost intersection, and so ^ i s superclosed.(b)_ To show that S{^} ^S{$}, we take g e 5{^} and show that for any element

A e § there is some ft < A such that A* =)" F^. But A D T ^ for some a < A, so itsuffices to show that ( Q S D T ^ for some )3 < A. Since geS{§;}, we haver ^ 3 * 2 , fl Fy 3 2* fl Fy = Fy for some i<w, y<A. Now consider (2*)s =n,<«, 2f. For each / < cu, we have 2f 3v2y- D Ya. 3 2* D r a . = Y'a. for some y, < coand cr, < A. Let r] = sup,<w OCJ. Since w < cf(A) = A and each &/ < A then 77 < A.Thus (2* )*2*2*nr , , = r ; . Then, letting /? = max{y, ry}, we have (Y'a)

s^*Yp,as required.

COROLLARY 7.2. S{^} is not maximal.

Proof We have proved that S{§) S{^}, so if 5{^} is maximal then S{&) = S{^}.But then by Lemma 3.12, 3* = iS, which gives a contradiction.

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Marcus Brazil Tim PenttilaJacinta Covington Cheryl E. PraegerDepartment of Mathematics Alan R. Woods

University of Melbourne Department of MathematicsParkville University of Western Australia

Victoria 3052 NedlandsAustralia Western Australia 6009

Australia

maximal subgroups of infinite symmetric groups

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