George M. Bergman and Saharon Shelah- Closed subgroups of the infinite symmetric group

8/3/2019 George M. Bergman and Saharon Shelah- Closed subgroups of the infinite symmetric group

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2 GEORGE M. BERGMAN AND SAHARON SHELAH

for arbitrary subgroups; but we will show, for countable, that if our subgroupis closed in the function topology on S, then one element suffices if and only if G

satisfies condition (i) of the above abstract. The method of proof generalizes togive the four-way classification of closed subgroups asserted there.

(The four conditions of that classification could be stated more succinctly, if notso transparently to the non-set-theorist, by writing for the least cardinal suchthat for some finite subset , all orbits of G() in have cardinality < .Then the conditions are (i) = 1, (ii) = 0, (iii) 3 < 0 and (iv) = 2.But we shall express them below in the more mundane style of the abstract.)

The proofs of the above results will occupy 28 of this note. In 912 we notesome related observations, questions, and possible directions for further investiga-tion.

The authors are indebted to Peter Biryukov, Peter Neumann, and ZacharyMesyan for corrections to earlier drafts of this note.

2. Definitions, conventions, and basic observations.

As usual, appearing before the symbol for a group means is a subgroupof, and ... denotes the subgroup generated by.

We take our notation on permutation groups from [5]. Thus, if is a set,Sym() will denote the group of all permutations of , and such permutationswill be written to the right of their arguments. Given a subgroup G Sym()and a subset , the symbol G() will denote the subgroup of elements ofG that stabilize pointwise, and G{} the larger subgroup {f G : f = }.Elements of will generally be denoted , , . . . .

The cardinality of a set U will be denoted |U|. Each cardinal is understood tobe the least ordinal of its cardinality, and each ordinal to be the set of all smallerordinals. The successor cardinal of a cardinal is denoted +.

Let us now define, in greater generality than we did in the abstract, the relationswe will be studying.

Definition 1. If S is a group, an infinite cardinal, and G1, G2 subgroups of S,we shall write G1 ,S G2 if there exists a subset U S of cardinality < suchthat G1 G2 U. If G1 ,S G2 and G2 ,S G1, we shall write G1 ,S G2,while if G1 ,S G2 and G2 ,S G1, we shall write G1 ,S G2.

We will generally omit the subscript S, and often as well, when their valuesare clear from the context.

Clearly ,S is a preorder on subgroups of S, hence ,S is an equivalencerelation, equivalent to the assertion that there exists U S of cardinality


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CLOSED SUBGROUPS OF THE INFINITE SYMMETRIC GROUP 3

but G1 G2 = {1}, which is not 0-equivalent to G1 and G2, since the lattergroups are uncountable, hence not finitely generated.)

We noteLemma 2. Let be an infinite set, G a subgroup of S = Sym(), and a subset such that |||| || (e.g., a finite subset). Then G() ||+ G.Proof. Elements of G in distinct right cosets of G() have distinct behaviors on

, hence if R is a set of representatives of these right cosets, |R | |||| ||.Clearly, G() R = G = G R, so G() ||+ G, as claimed.

Two results from the literature have important consequences for these relations:

Lemma 3 ([4], [5]). Let be an infinite set. Then on subgroups of S = Sym(),

(i) The binary relation 0 coincides with 1 (hence 0 coincides with 1).(ii) The unary relation

0

S coincides with

||

+ S.

Proof. (i) follows from [4, Theorem 3.3], which says that every countably generatedsubgroup of S is contained in a 2-generator subgroup.

We claim that (ii) is a consequence of [5, Theorem 1.1] (= [1, Theorem 5])which, as recalled in 1, says that any chain of proper subgroups of S having S asunion must have > || terms. For if G ||+ S, then among subsets U S ofcardinality || such that G U = S, we can choose one of least cardinality.Index this set U as {gi : i |U|}. Then the subgroups Gi = G {gj : j < i}(i |U|) form a chain of || proper subgroups of S, which if |U| were infinitewould have union S, contradicting the result of [5] quoted. So U is finite, soG 0 S.

We have not introduced the symbols and

for finite because in gen-

eral these relations are not transitive; rather, one has G1 m G2 n G3 =G1 m+n1 G3. However, the proof of (i) above shows that in groups of the formSym(), 0 is equivalent to 3 . Thus, in such groups, we do have transitivityof and when 3 < 0, but have no need for these symbols. Theseobservations are also the reason why in 57 we wont make stronger assertionsthan 0 , though some of our constructions will lead to sets U of explicit finitecardinalities.

(Incidentally, [4, Theorem 5.7] shows that on Sym(), the unary relation 0 Sis even equivalent to 2 S.)

3. Generalized metrics.

In this section we will prove a result which, for countable, will imply thatclosed subgroups of Sym() falling under different cases of the classification de-scribed in the abstract are indeed 0-inequivalent.

To motivate our approach, let us sketch a quick proof that if is any setsuch that || is an infinite regular cardinal (e.g., 0), and if G is a subgroupof S = Sym() such that every orbit of under G has cardinality < ||, then


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G || S. Given U S of cardinality < ||, let us define the distance betweenelements , to be the length of the shortest group word in the elements ofG U that carries to , or to be if there is no such word. It is not hardto see from our assumptions on the orbits of G and the cardinality of U that foreach and each positive integer n, there are < || elements of withindistance n of , hence this distance function on has no finite bound. Now ifan element of G U is expressible as a word of length n in elements of G U,it will move each element of a distance n. Thus, taking f Sym() whichmoves points by unbounded distances, we have f / G U; so G U = S.

The next definition gives a name to the concept, used in the above proof, ofa metric under which points may have distance , and introduces some relatedterminology and notation.

Definition 4. In this definition, P will denote {r R : r 0}{}, ordered inthe obvious way.

A generalized metric on a set will mean a function d: P, satisfyingthe usual definition of a metric, except for this generalization of its value-set.

If d is a generalized metric on , then for r P, , we will write Bd(, r) for the open ball of radius r about , { : d(, ) < r}. If is an infinitecardinal, we will call the generalized metric space (, d) -uncrowded if for every and r < , the ball Bd(, r) has cardinality < . We will call (, d)uniformly -uncrowded if for every r < , there exists a < such that for all , |Bd(, r)| . (For brevity we will, in these situations, also often refer tothe metric d as being -uncrowded or uniformly -uncrowded.)

Given two generalized metrics d and d on , we shall write d d if d(, ) d(, ) for all , .

For g Sym() and d a generalized metric on , we define

||g||d = sup d(,g) P.

If ||g||d < , then the permutation g will be called bounded under d.The case of these concepts that we will be most concerned with in subsequent

sections is that in which || = = 0.Observe that if a group G Sym() has all orbits of cardinality < ||, then

giving the generalized metric under which distinct points in the same orbit havedistance 1 and points in different orbits have distance , we get a ||-uncrowdedgeneralized metric with respect to which G acts by bounded permutations thisis the U = case of the construction sketched in the second paragraph of thissection. Thus, if we can change the hypothesis of that construction from G has

orbits of cardinality < || to G acts by bounded permutations with respect to a||-uncrowded generalized metric (the condition we deduced held for GU), wewill have a stronger statement. This is done in the next theorem. If the cardinalitiesof the orbits of G have a common bound < ||, then the above constructiongives a uniformly ||-uncrowded generalized metric. The theorem will generalizethat case as well.


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Note that if a generalized metric space (, d) is ||-uncrowded, the function dis necessarily unbounded, i.e., takes on values exceeding every positive real number.

Theorem 5. Suppose is an infinite set, a regular cardinal ||, d a -uncrowded (respectively a uniformly -uncrowded) generalized metric on , andG a subgroup of S = Sym() consisting of elements bounded with respect to d.

Then for any subset U Sym() of cardinality < , there exists a generalizedmetric d d on , again -uncrowded (respectively, uniformly -uncrowded),such that every element of U, and hence every element of GU, is bounded withrespect to d.

Thus, for subgroups G S, the property that there exists a (uniformly) -uncrowded generalized metric on with respect to which every element of G actsby bounded permutations is preserved under passing to groups H G, and inparticular, under passing to groups H G.Proof. Given , , d as in the first sentence and U as in the second, let us defined(, ) for , , to be the infimum, over all finite sequences of the form

(1) (0, 1, . . . , 2n+1) where = 0, 2n+1 = , and where for iodd, i+1 i (U U1),

of the sum

(2) d(0, 1) + 1 + d(2, 3) + 1 + . . . + 1 + d(2n, 2n+1).

This infimum is d(, ) because the set of sequences over which it is taken in-cludes the sequence (, ). We also see from (2) that whenever d(, ) = d(, ),we have d(, ) 1, hence d(, ) is nonzero for = . Symmetry and the trian-gle inequality are immediate from the definition, and each g U satisfies ||g||d 1because elements and g are connected by the sequence (,,g,g). Thus, d

is a generalized metric d, with respect to which every element of U is bounded;

it remains to show that d

is again (uniformly) -uncrowded.So consider a ball of finite radius, Bd(, r). An element lies in this ball ifand only if there is a sequence (1) for which the sum (2) is < r. But (2) has nsummands equal to 1, so given r < , there are only finitely many values ofn that need to be considered; so it suffices to show that for fixed n and , thenumber of sequences (1) making (2) less than r is < , and in the uniform casehas a bound < depending only on r. Now for 0 i < 2n + 1, if we are giveni, then the number of possibilities for i+1 consistent with (2) being less thanr is |Bd(i, r)| < if i is even, while it is | U U1| < if i is odd. Bythe regularity of , it follows that the number of possible sequences (1) of length2n + 1 making (2) less than r and starting with a given 0 is < . Moreover, ifwe have a bound on |Bd(i, r)| independent of i, we also clearly get a bound onthe above cardinal independent of , as required.

The final assertions, concerning the relations and , clearly follow. Remark. What if in the above theorem we weaken the assumption that U has

cardinality < to say that it has cardinality ?If = 0, we get exactly the same conclusions, since the result of [4] cited in the

proof Lemma 3(i) lets us replace any countable U by a set of cardinality 2. Can


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we see this stronger assertion without calling on [4]? Yes, by a slight modificationof the proof of our theorem: We write the countable set U as {f1, f2, . . .}, andreplace each of the 1s in (2) by a value N such that i+1 = if

1N . Thus in the

final step of the proof, the number of choices of i+1 that can follow i for i oddwill still satisfy a bound below 0, namely 2r.

For an uncountable regular cardinal , the sort of generalized metric we haveintroduced is not really the best tool. (Indeed, if (, d) is -uncrowded, then forevery , most elements of must be at distance from , and themain import of d lies in the equivalence relation of having distance < ; so sucha metric is hardly a significant generalization of an equivalence relation.) What ismore useful then is the concept of a {}-valued ultrametric, where the symbol is again taken as greater than all other values of the metric. Defining in theobvious way what it means for such an ultrametric to be (uniformly) -uncrowded,one can apply the same method as above when |U| = , with the summation (2)replaced by a supremum. Since we will not be looking at this situation, we leavethe details to the interested reader.

The above theorem, in generalizing the argument sketched at the beginning ofthis section, discarded the explicit connection with cardinalities of orbits. The nextresult records that connection.

Let us understand a partition of a set to mean a set A of disjoint nonemptysubsets of having as union. If A is a partition of and S = Sym(), wedefine

(3) S(A) = {f S : ( A) f = }.(This is an extension of the notation S() recalled in the preceding section.)

Theorem 6. Let be an infinite set, A a partition of , S = Sym(), G =S(A), and let be an infinite regular cardinal

|

|. Then

(a) If some member of A has cardinality , then there is no -uncrowdedgeneralized metric on with respect to which all members of G are bounded.

(b) If all members of A have cardinalities < , but there is no common bound < for those cardinalities, then there is a -uncrowded generalized metric withrespect to which all elements of G are bounded, but no uniformly -uncrowdedgeneralized metric with this property.

(c) If all members of A have cardinalities for some < , then there is auniformly -uncrowded generalized metric with respect to which all elements of Gare bounded.

Thus, by the last sentence of Theorem 5, for partitions A, B of falling underdistinct cases above, we have S(A) S(B). More precisely, if A falls under a latercase than B, then S(A)

S(B).

Proof. To show (a), let A have cardinality and let d be any -uncrowdedgeneralized metric on . To construct an element of G which is unbounded withrespect to d, let us choose elements j , j for each positive integer j asfollows: Assuming the elements with subscripts i < j have been chosen, take forj any element of distinct from all of these. Since Bd(j , j) has cardinality


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< ||, the set Bd(j, j) {1, . . . , j1, 1, . . . , j1} is nonempty;let j be any element thereof. Once all j and j are chosen, let f G = S(A)interchange j and j for all j, and fix all other elements of . Since d(j , j) jfor each j, f is unbounded with respect to d.

The final assertion of (b) is shown similarly: If d is uniformly -uncrowded, thenfor each positive integer n we can find a cardinal n < such that all balls ofradius n contain n elements. On the other hand, the assumption on A allowsus to choose for each n a set n A with more than n + (2n 2) elements.Assuming 1, . . . , n1, 1, . . . , n1 have been chosen, we take for n any elementof n {1, . . . , n1, 1, . . . , n1} and for n any element of n Bd(n, n) {1, . . . , n1, 1, . . . , n1}, and finish the argument as before.

To get the positive assertions of (b) and (c), we define a generalized metric dAon by letting dA(, ) = 1 if and are in the same member of A, and otherwise. This is clearly -uncrowded, respectively uniformly -uncrowded, andall elements of G are bounded by 1 under dA.

The conclusions of the final paragraph are straightforward.

The three cases of the above theorem will be used to separate the first three ofthe four situations described in the abstract. One may ask whether the remainingcase can be treated similarly. For parallelism, one might call a generalized metricabsolutely uncrowded if all balls of finite radius are singletons, i.e., if the distancebetween any two distinct points is , and then note that the trivial group isthe unique group of permutations whose elements are bounded with respect tothe absolutely uncrowded generalized metric. However, the property of acting bybounded permutations with respect to the unique absolutely uncrowded metric iscertainly not preserved under adjunction of finitely many elements, i.e., is not an0-invariant. Rather than any result of this sort, the property of countability willseparate this fourth equivalence class from the others.

4. The function topology.

If is an infinite set and we regard it as a discrete topological space, then theset of all functions becomes a topological space under the functiontopology. In this topology, a subbasis of open sets is given by the sets {f :f = } (, ). The closure of a set U consists of all maps f suchthat, for every finite subset , there exists an element of U agreeing withf at all members of . It is immediate that composition of maps is continuous inthis topology.

The group S = Sym() is not closed in in the function topology. For

instance when = , we see that the sequence of permutations (0, 1), (0, 1, 2), . . . ,(0, . . . , n), . . . (cycle notation) converges to the map n n + 1, which is notsurjective. Nevertheless, when restricted to S, this topology makes ( )1 as wellas composition continuous; indeed, {f S : f = }1 = {f S : f = }.

Given a subset U S, we shall write cl(U) for the closure of U in S (notin !) under the function topology. The fact that S is not closed in has


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the consequence that if one wants to prove the existence of an element f cl(U)behaving in some desired fashion, one cannot do this simply by finding elements

of U that show the desired behavior at more and more elements of , and sayingtake the limit; for the limit may be an element of which is not in S. However,there is a standard way of getting around this difficulty, the method of going backand forth. One constructs elements of U which not only agree on more and moreelements of , but whose inverses also agree on more and more elements. Takingthe limit, one thus gets a map and also an inverse to that map. (What is being usedis the fact that though S is not closed in , the set {(g, g1) : g S} is closedin .) Cf. [2, 9.2, 16.4] for examples of this method, and some discussion.The next result, a formalization of this idea, will be used at several points below.

Lemma 7. Suppose that = {0, 1, . . . } is a countably infinite set, and thatg0, g1, . . . S = Sym() and 0, 1, . . . are such that for all j > 0,

(4)

{0, . . . , j1

} {0 g

1j1, . . . , j1 g

1j1

} j ,

and

(5) gj S(j) gj1.Then the sequence (gj)j=0,1,... converges in S.

Proof. Let i 0.For all j > i, the conditions i j and (5) imply that i gj = i gj1. Thus,

the sequence (gj) is eventually constant on i.Likewise, (5) and the condition i g

1j1 j imply that i g1j = i g1j1; hence

the element of carried to i by gj is the same for all j > i.Since the first conclusion holds for all i , the sequence (gj)j=0,1,... converges

to an element of , which is one-to-one because all the gj are. Applying thesecond conclusion, we see that each i is in the range of g, so g Sym().

We note some elementary facts about closures of subgroups in the function topol-ogy.

Lemma 8. Suppose is a set and G a subgroup of S = Sym(). Then

(i) cl(G) is also a subgroup of S.

(ii) G and cl(G) have the same orbits in .

(iii) If is a finite subset of , then cl(G)() = cl(G()).

Proof. Statement (i) is an immediate consequence of the continuity of the groupoperations.

From the characterization of the closure of a set in our topology, we see that for, , the set cl(G) will contain elements carrying to if and only if Gdoes, from which (ii) is clear.

The direction in (iii) follows by applying (ii) to the orbits of elements of .(Finiteness of is not needed for this direction.) To get , assume f cl(G)().Since f cl(G), every neighborhood of f contains elements of G. But as f fixesall points of the finite set , every sufficiently small neighborhood of f consistsof elements which do the same, hence every such neighborhood contains points ofG(); so f cl(G()).


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The above lemma has the consequence that once we show (for countable)that the 0-class of a closed subgroup of Sym() is determined by which ofconditions (i)(iv) in our abstract hold, we can also say for an arbitrary subgroupG Sym() that the 0-class of cl(G) is determined in the same way by whichof those conditions G satisfies.

The subgroups of Sym() closed in the function topology are known to be pre-cisely the automorphism groups of the finitary relational structures on . (Indeed,one may take the n-ary relations in such a structure, for each n, to be all orbitsof n-tuples of elements of under the group.) But we shall not make use of thisfact here.

(Incidentally, Sym() is also not openin . It is easy to give a sequence of non-injective or non-surjective maps in which the failures of injectivity or surjectivitydrift off to infinity, so that the limit is a bijection, e.g., the identity.)

5. Infinite orbits.

In this and the next three sections (and with minor exceptions, in subsequentsections as well), we shall restrict attention to the case of countable . When anenumeration of its elements is required, we shall write

(6) = {i : i }.References to limits etc. in S = Sym() will always refer to the function topology;in particular, a closed subgroup of S will always mean one closed in S under thattopology. The symbols and will mean 0,S and 0,S respectively.

We shall show in this section that if G is a closed subgroup of S such that

(7) For every finite subset , the subgroup G() has at least oneinfinite orbit in ,

then G S. Our proof will make use of the following result of Macpherson andNeumann:

(8) [5, Lemma 2.4] (cf. [1, Lemma 3]): Suppose is an infinite set and Ha subgroup of Sym(), and suppose there exists a subset ofthe same cardinality as , such that H{} (i.e., {f H : f = })induces, under restriction to , the full permutation group of .Then there exists x Sym() such that Sym() = H {x}.

(This is stated in [5] and [1] for the case where is a moiety, i.e., a set of cardinality|| such that also has cardinality || . But if the hypothesis of (8) holdsfor some of cardinality ||, it clearly also holds for a subset of which is amoiety, so we may restate the result as above.)

We will also use the following fact. We suspect it is known, and would appreciate

learning of any reference. (A similar technique, but not this result, occurs in [9]and [10].)

Lemma 9. Let us call a permutation g of the set of natural numbers local if for every i there exists j > i in such that g carries {0, . . . , j 1} to itself.

Then every permutation f of is a product gh of two local permutations.


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Proof. Given f Sym(), let us choose integers 0 = a(0) < a(1) < a(2) < . . . re-cursively, by letting each a(i) be any value > a(i1) such that {0, . . . , a(i1)1} f {0, . . . , a(i1)1} f

1

{0, . . . , a(i)1}. Let 1 = , and for i 0 leti = {a(i), a(i)+1, . . . , a(i+1)1}. Thus the set A = {i : i 0} is a partitionof into finite subsets, such that for each i 0 one has if i1 i i+1.Note that for each i 0, the number of elements which are moved by f from i1into i is equal to the number that are moved from i into i1 (since these arethe elements of that are moved past a(i) 1/2 in the upward, respectivelythe downward direction).

We shall now construct a permutation g such that g carries each set 2i2i+1(i 0) into itself, and g1f carries each set 2i1 2i (i 0) into itself; thuseach of these permutations will be local, and they will have product f, as required.To do this let us, for each i, pair elements that f carries from 2i upward into2i+1 with elements that it carries from 2i+1 downward into 2i (having seenthat the numbers of such elements are equal), and let g exchange the members ofeach such pair, while fixing other elements. Clearly g preserves the sets 2i2i+1.It is not hard to verify that if we now look at as partitioned the other way, intothe intervals 2i1 2i, then the g we have constructed has the property thatfor every , the element g lies in the same interval 2i1 2i as does f.Hence g1f preserves each interval 2i1 2i, completing the proof.

We shall now prove a generalization of (8), assuming countable. To motivatethe statement, note that in the countable case of (8), if we enumerate the elementsof as 0, 1, . . . , then the hypothesis implies that we can choose elements g Hin ways that allow us infinitely many choices for 0 g, for each such choice infinitelymany choices for 1 g, etc.. But the hypothesis of (8) is much stronger than this,since it specifies that the set of choices for 0g include all the i, that the choices

for 1 g then include all i other than the one chosen to be 0 g, etc.. The nextresult says that we can get the same conclusion without such a strong form of thehypothesis.

Lemma 10. Let be a countably infinite set and G a subgroup of Sym(), andsuppose there exist a sequence (i)i of distinct elements, and a sequenceof nonempty subsets Di i (i ), such that(i) For each i and (0, . . . , i) Di+1, we have (0, . . . , i1) Di;(ii) For each i and (0, . . . , i1) Di, there exist infinitely many elements such that (0, . . . , i1, ) Di+1; and(iii) If (i)i has the property that (0, . . . , i1) Di for each i ,then there exists g G such that (i) = (i g) in .

Then G S.Proof. Let us note first that our hypotheses imply that for (0, . . . , i1) Di,the entries j are all distinct. For from (i) and (ii) we see that such an i-tuple canbe extended to an -tuple as in (iii), and by (iii) this -tuple is the image undera group element of the -tuple of distinct elements (i).


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We shall now construct recursively, for i = 0, 1, . . . , finite sets Ei Di. Foreach i, the elements of Ei will be denoted e(n0, . . . , nr; 1, . . . , r) with one such

element for each choice of a sequence of natural numbers 0 = n0 < n1 < .. . < nr =i and a sequence of permutations m Sym({nm1, . . . , nm1}) (1 m r).(Note that each e(n0, . . . , nr; 1, . . . , r), since it belongs to Di, is an i-tuple ofelements of , where i = nr; but we shall not often write it explicitly as a stringof elements. Nevertheless, we shall refer to the i elements of comprising thisi-tuple as its components.)

We start the recursion with E0 = D0, which is necessarily the singleton consist-ing of the unique length-0 sequence. Assuming E0, . . . , E i1 given, we choose anarbitrary order in which the finitely many i-tuples in Ei are to be chosen. Whenit comes time to choose the i-tuple e(n0, . . . , nr; 1, . . . , r) Ei, we define its ini-tial substring of length nr1 to be the nr1-tuple e(n0, . . . , nr1; 1, . . . , r1) Enr1 . We then extend this to an element of Dnr in any way such that its re-maining nr

nr1 components are distinct from all components of all elements of

E0 . . . Ei1, and from all components of those elements of Ei that have beenchosen so far. This is possible by nr nr1 applications of condition (ii) above: ateach step, when we extend a member of a set Dj to a member of the next set Dj+1(nr1 j < nr) we have infinitely many choices available for the last component,and only finitely many elements to avoid.

Once the sets Ei are chosen for all i, let us define an element s Sym() topermute, in the following way, those elements of that occur as components inthe members of

i Ei. (On the complementary subset of we let s behave in

any manner, e.g., as the identity.)

(9) For each (j)0j


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(11) e(n0) E0, e(n0, n1; 1) En1 , . . . ,e(n0, . . . , nm; 1, . . . , m) Enm, . . . .

Each of these tuples extends the preceding, so there is a sequence (i) ofwhich these tuples are all truncations. From (9) we see that the sequence (i) willsatisfy is = i for all i . Also, by our hypothesis (iii) and the conditionEi Di, there exists g G such that i = ig for all i. Substituting this intothe relation is = i, we get (10), as claimed.

Now (10) can be rewritten as saying that g s g1 acts on {i} by the mapi i. In view of Lemma 9, every permutation of {i} can be realized as therestriction to that set of a product of two such permutations, hence as g s g1h s h1

for some g, h G. Thus, the group H = G {s} satisfies the hypothesis of (8)with = {i}. Hence by (8) there exists x Sym() such that G {s, x} = S,completing the proof of the lemma.

We now consider a subgroup G Sym() satisfying (7). We shall show how toconstruct elements i and families Di i satisfying conditions (i) and (ii)of the above lemma, and such that if G is closed, condition (iii) thereof also holds,allowing us to apply that lemma.

We begin with another recursion, in which we will construct for each j 0 anelement j , and a finite subset Kj of G, indexed

(12) Kj = {g(k0, k1, . . . , kr1) : k0, k1, . . . , kr1, r , r+k0+. . .+kr1 = j}.To describe the recursion, assume inductively that i and Ki have been defined

for all nonnegative i < j , and let j denote the finite set consisting ofthe images of 0, . . . , j1 (cf. (6)) and of 0, . . . , j1 under the inverses of allelements of K0

. . .

Kj1. Let j be any element of having infinite orbit

under G(j) (cf. (7)). In choosing the elements g(k0, k1, . . . , kr1) comprising Kj,we consider two cases.

If j = 0, we have only one element, g(), to choose, and we take this to bethe identity element 1 G. A consequence of this choice is that for all larger j,we have 1 K0 . . . Kj1, hence the definition of j above guarantees that0, . . . , j1 and 0, . . . , j1 themselves lie in j .

If j > 0, we fix arbitrarily an order in which the elements of Kj are tobe constructed. When it is time to construct g(k0, k1, . . . , kr1), let us writeg = g(k0, k1, . . . , kr2), noting that this is a member of Kjkr11, hence al-ready defined. We will take for g(k0, k1, . . . , kr1) the result of left-multiplying g

by a certain element h G(r1). Note that whatever value in this group we choosefor h, the images of 0, . . . , r2 under hg

will be the same as their images un-

der g

, since elements of G(r1) fix 0, . . . , r2 r1. On the other hand, wemay choose h so that the image of r1 under hg is distinct from the imagesof r1 under the finitely many elements of K0 . . . Kj1, and also under theelements of Kj that have so far been constructed, since r1 has infinite orbitunder G(r1), and there are only finitely many elements that have to be avoided.So let g(k0, k1, . . . , kr1) = hg

be so chosen.


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In this way we successively construct the elements of each set Kj . Note thatthis gives us group elements g(k0, . . . , ki1 ) for all i, k0, . . . , ki1 . We canthus define, for each i ,(13) Di = {(0 g , . . . , i1 g) : g = g(k0, . . . , ki1) for some k0, . . . , ki1 }.

By construction, g(k0, . . . , ki) agrees with g(k0, . . . , ki1) on 0, . . . , i1, sochopping off the last component of an element of Di+1 gives an element of Di,establishing condition (i) of Lemma 10. Moreover, any two elements of the formg(k0, . . . , ki) Di+1 with indices k0, . . . , ki1 the same but different last indices kiact differently on i, so the sets Di satisfy condition (ii) of that lemma. Suppose,now, that (i) has the property that for every i the sequence (0, . . . , i1)is in Di. We see inductively that successive strings (), (0), . . . , (0, . . . , i1), . . .must arise from unique elements of the forms g(), g(k0), . . . , g(k0, . . . , ki1), . . . .Moreover, by construction each of these group elements g(k0, . . . , ki) is obtainedfrom the preceding element g(k0, . . . , ki1) by left multiplication by an element

of G(i), where i contains the elements 0, . . . , i1 and their inverse imagesunder all the preceding group elements. It follows by Lemma 7 that if G is closed,the above sequence converges to an element g G whose behavior on (i) is thelimit of the behaviors of these elements, i.e., which sends (i) to (i), establishingcondition (iii) of Lemma 10. Hence that lemma tells us that G S.

We can now easily obtain

Theorem 11. Let be a countably infinite set, and G a closed subgroup ofS = Sym(). Then G S (i.e., S is finitely generated over G) if and only if Gsatisfies (7).

Proof. We have just seen that (7) implies G S. On the other hand, if (7) doesnot hold, then for some finite , G() has only finite orbits. Letting A denotethe set of these orbits, we have G() S(A). But S(A) falls under case (b) or (c)of Theorem 6 (with = 0) while S falls under case (a), being determined by theimproper partition of . We thus get

(14) G G() S(A) S,where the first relation holds by Lemma 2 and Lemma 3(i), and the final strictinequality by the last sentence of Theorem 6. Thus G S.

Notes on the development of the above theorem: In the proof of Lemma 10,and again in the arguments following that proof, it might at first appear that ourhypotheses that certain subsets of were infinite (namely, in the former case,the set of next terms extending each member of Di, and in the latter, at leastone orbit of G() for each finite set ) could have been replaced by statementsthat those sets could be taken to have large enough finite cardinalities, since at

each step, we had to make only finitely many choices from these sets, and to avoidonly finitely many elements of . But closer inspection shows that we dipped intothese sets for additional elements infinitely many times. In the proof of Lemma 10,this is because for fixed n0, . . . , nr1 there are infinitely many possibilities fornr > nr1, and for each of these, the construction of Enr requires extending theelements e(n0, . . . , nr1; 1, . . . , r1) Dnr1 to elements of Dnr1+1. Likewise,


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in (12), note that r j, and each value of r comes up for infinitely many j,so that for each r we must choose, in the long run, elements of G(r1) having

infinitely many different effects on r1.This spreading out of the choices we made from each infinite set, into infinitely

many clumps of finitely many choices each, was necessary: If we had made infinitelymany choices at one time from one of our sets, we would have had infinitely manyobstructions to our choices from the next set, and could not have argued that thosechoices could be carried out as required.

Could the two very similar recursive constructions just referred to have beencarried out simultaneously? In an earlier draft of this note they were. That ar-rangement was more efficient (if less transparent as to what was being proved), andcould be considered preferable if one had no interest except in closed subgroups.However, the present development yields the intermediate result Lemma 10, whichcan be used to show the -equivalence to S of many non-closed subgroups forwhich, so far as we can see, Theorem 11 is of no help.

For example, consider a partition A of into a countably infinite family ofcountably infinite sets i, and let G be the group of permutations of that, foreach i, carry i, into itself, and move only finitely many elements of that set. Ifwe choose an arbitrary element (i)

i i, and let Di = 0 . . .i1 for each

i, then we see easily that the conditions of Lemma 10 hold, hence that G S.(The same argument works for the subgroup of the above G consisting of those

elements g for which there is a bound independent of i on the number of elementsof i moved by g.)

6. Finite orbits of unbounded size.

In this section, we again let be a countably infinite set, and will show that allclosed subgroups G S = Sym() for which

(15) There exists a finite subset such that all orbits of G() arefinite, but no such for which the cardinalities of these orbits havea common finite bound,

are mutually -equivalent. The approach will parallel that of the preceding section,but there are some complications.

First, there is not one natural subgroup that represents this equivalence class, asS represented the class considered in the previous section. Instead we will beginby defining a certain natural family of closed subgroups of S which we will prove-equivalent to one another. Second, we do not have a result from the literatureto serve in the role of (8). So we will prove such a result. The fact that a finitesymmetric group is not its own commutator subgroup will complicate the lattertask. (Cf. the proof of (8) as [1, Lemma 3], which uses the fact, due to Ore [7], thatevery element of an infinite symmetric group is a commutator.) So we will preparefor that proof by showing that certain infinite products of finite symmetric groupswithin S are -equivalent to the corresponding products of alternating groups.


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To define our set of representatives of the -equivalence class of subgroups ofS we are interested in, let

(16) P = {A : A is a partition of into finite subsets, and there is nocommon finite bound on the cardinalities of the members of A } .

For A P (and S(A) defined by (3)), we see that(17) S(A) =

A

Sym().

Note that if a partition A1 is the image of a partition A2 under a permutationf of , then S(A2) is the conjugate of S(A1) by f; in particular, S(A1) S(A2).Let us now show more; namely, that

(18) S(A) S(B) for all A, B P.We claim first that given A, B P, we can find two elements f, g S such

that

(19) For each B there exists A such that f or g.Indeed, write B as the disjoint union of any two infinite subsets B1 and B2. Wemay construct f by defining it on one member of A after another, making surethat each member of B1 ends up within the image f of some sufficiently large A. We map those members of A or subsets of members of A that are notused in this process into the infinite set

B2

, and we also make sure to includeevery element of

B2

in the range of f, so that f is indeed a permutation.We similarly construct g so that every member of B2 is contained in the imageg of some A. Condition (19) is thus satisfied.

For such f and g, we claim that

(20) S(B) (f1S(A)f)(g1S(A) g).Indeed, every element of S(B) can be written as the product of a member of S(B)

which moves only elements of

B1 and one which moves only elements ofB2

; and these can be seen to belong to f1S(A)f and to g1S(A) g respec-

tively.Thus S(B) S(A) {f, g}, so S(B) S(A). Since this works both ways, we

get (18), as desired.

We next prepare for the difficulties concerning alternating groups versus sym-metric groups. Let A be a partition belonging to P which contains infinitely manysingletons, and whose other members are all of cardinality at least 4, and let Seven(A)denote the subgroup of S(A) which acts by an even permutation on each memberof A. We shall show that

(21) Seven(A) S(A).

To do this, let us list the non-singleton members of A as 0, 1, . . . , and foreach i choose in i four distinct elements, which we name 4i, 4i+1, 4i+2, 4i+3.Let B denote the partition of (not belonging to P) whose only nonsingletonsubsets are the two-element sets {2j , 2j+1} (j 0). Thus S(B) can be identi-fied with (Z/2Z), and S(B) Seven(A) can be seen to correspond to the subgroup{(a0, a1, . . .) (Z/2Z) : ( i 0) a2i = a2i+1}.


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Let us now choose from the union of the singleton members of A infinitely manyelements, which we will denote i for i < 0, and let f S be any permutationsuch that if = i+2 for all i Z. Then we see that the conjugation mapg f1g f will carry S(B) into itself, by a homomorphism which, identifyingS(B) with (Z/2Z)

, takes the form (a0, a1, . . .) (0, a0, a1, . . .). Now it is nothard to see that every member of (Z/2Z) can be written (uniquely) as the sumof an element whose 2ith and 2i+1st coordinates are equal for each i, and anelement whose 0th coordinate is 0 and whose 2i+1st and 2i+2nd coordinates areequal for all i. Hence

(22) S(B) = (S(B) Seven(A) ) ( f1(S(B) Seven(A) )f) Seven(A) {f}.We also see that S(A) = S(B)S

even(A) . (For, given any h S(A) which we wish to

represent in this way, a factor in S(B) can be chosen which gives a permutation ofthe desired parity on each i A, and a factor in Seven(A) then turns this into thedesired permutation h.) Hence S(A)

Seven

(A) {f}

, so (21) holds.

We can now obtain our analog of (8). Suppose that

(23) H is a subgroup of S, and A an infinite family of disjoint nonemptysubsets of , of unbounded finite cardinalities, such that writing =

A , every member of Sym()(A) extends to an element of H{}.

That is, we assume we can find elements of H which give any specified familyof permutations of the sets comprising A but we dont assume that we cancontrol what they do off those sets. We claim that by adjoining to H one elementfrom S, we can get a group which contains a subgroup Seven(A) for some A

Psatisfying the conditions stated before (21) (infinitely many singletons, all othermembers having cardinality 4).

To do this let us split the set A of (23) into three infinite disjoint subsets,

A = A1 A2 A3, in any way such that A1 has members of unbounded finitecardinalities and no members of cardinality < 4. If we let 1 =

A1

, 2 =A2

, 3 = 1 2, we see that these sets each have cardinality 0 (thelast because it contains

A3

). Since 1 2 , it follows from (23) that(24) For every f Sym(1)(A1) there exists an f H which agrees with

f on 1, and acts as the identity on 2.

Now take any g S that interchanges 2 and 3, and fixes 1 pointwise.Conjugating (24) by g gives

(25) For every f Sym(1)(A1) there exists an f g1H g which agreeswith f on 1 and acts as the identity on 3.

Now if one forms the commutator of a permutation which acts as the identity on

2 and preserves 3 with a permutation which acts as the identity on 3 andpreserves 2, one gets an element which acts as the identity on 2 3. Hencefrom (24) and (25) we may conclude that H{g} contains elements which act asthe identity on 2 3 = 1, while acting on each A1 by any specifiedcommutator in Sym(). Moreover, in the symmetric group on a finite set , thecommutators are precisely the even permutations [7, Theorem 1]; so letting A be


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the partition of consisting of the members of A1 and all singleton subsets of 1, we have(26) H {g} Seven(A) .Combining with (21), we get our analog of (8), namely

(27) If H satisfies (23), then H S(A) for some (hence by (18), for all)A P.

With the help of (27) we can now prove a strengthening thereof, analogous toLemma 10 of the preceding section:

Lemma 12. Let be a countably infinite set and G a subgroup of Sym(), andsuppose there exist a sequence of distinct elements (i)i , an unboundedsequence of positive integers (Ni)i, and a sequence of sets Di i (i ),such that

(i) For each i

and each (0, . . . , i)

Di+1, we have (0, . . . , i1)

Di;

(ii) For each i and each (0, . . . , i1) Di, there exist at least Ni elements such that (0, . . . , i1, ) Di+1; and(iii) If (i)i has the property that (0, . . . , i1) Di for each i 0, thenthere exists g G such that (i) = (i g) in .

Then G S(A) for some, equivalently, for all A P.Proof. This will be similar to the proof of Lemma 10, but with two simplificationsand one complication. The simplifications are, first, that we will not need to han-dle simultaneously strings of permutations (1, . . . , r) for all decompositions of{0, . . . , i1} as {0, . . . , n11} . . . {nr1, . . . , nr1}, but only for a single de-composition, and, secondly, that we will not have infinite families of choices thathave to be spread out over successive rounds of the construction, as discussed at the

end of the last section. The complication is that in general not all of the Ni in ourhypothesis will be large enough for our immediate purposes; hence each time wemove to longer strings of indices, we will have to jump forward to a value i = i(j)such that Ni(j) is large enough.

We begin by fixing an arbitrary increasing sequence of natural numbers 0 = n0 1}.

Unlike the P of the preceding section, Q has, up to isomorphism, a naturaldistinguished member, namely a least isomorphism class with respect to refinement:

(35) We will denote by A0 an element of Q, unique up to isomorphism,which has infinitely many 1-element members, infinitely many 2-element members, and no others.

Clearly any A Q can be refined to a partition A0 isomorphic to A0, henceS(A) S(A0) S(A0), so S(A) S(A0). We claim that the reverse inequalityS(A) S(A0) also holds. To show this, let us draw a graph with the elementsof as vertices, and with edges making each member of our given partition Aa chain (in an arbitrary way), and no other edges. Now color the edges of eachsuch chain alternately red and green, subject to the condition that infinitely manychains have a terminal red edge and infinitely many have a terminal green edge.Clearly, the partition of whose non-singleton members are the pairs of pointslinked by red edges, all other points forming singletons, is isomorphic to A0; hence

the group of permutations whose general member acts by transposing an arbitrarysubset of the red-linked pairs of vertices and fixing everything else can be writtenf1S(A0)f for some f Sym(). Similarly, the group of permutations which actby transposing some pairs of green-linked vertices and fixing everything else canbe written g1S(A0) g. Moreover, for each A, any permutation of can beobtained by composing finitely many permutations, each of which acts either byinterchanging only red-linked pairs or by interchanging only green-linked pairs (thisis easiest to see by looking at permutations that interchange one such pair at a time);and the number of such factors needed can be bounded in terms of the cardinalityof . Since there is a common bound to the cardinalities of the sets A, wesee that every member of S(A) can be written as a finite product of members of

f1S(A0)f and g1S(A0)g, so S(A0){f, g} S(A), so S(A0) S(A). Combining

this with the observation at the start of this paragraph, we get S(A0)

S(A), so

(36) S(A) S(B) for all A, B Q.We obtain next the result that will play the role that (8) played in 5 and (27)

played in 6. The development will be similar to the latter case, though simpler.Suppose that


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(37) H is a subgroup of S, and A an infinite family of disjoint 2-elementsubsets of such that, writing =

A , every member of

Sym()(A) extends to an element of H{}.(Again we do not assume we have any control over the behavior of these elementsoutside of , though again our goal will be to get such control in an extendedsubgroup.) Let us index A by Z, writing A = {i : i Z}, and let h be anelement of S which for each i Z sends i bijectively to i+1, and which fixes allelements of . We claim that as f runs over all elements of H{} that extendelements of Sym()(A), the commutators h

1f1hf all fix pointwise, andtheir restrictions to give all elements of Sym() (A). The first fact holds because

h fixes pointwise. The second may be seen by looking at h1f1hf as(h1f1h)f, noting that both factors are members of Sym()(A), and examiningtheir behaviors on the general 2-element set i A. One sees that (h1f1h)facts by the trivial permutation on i if and only if f acts trivially either on both

of i1 and i, or on neither, while (h1f1h)f acts by the nonidentity elementof Sym(i) in the remaining cases. One easily deduces that by appropriate choiceof f one can get an arbitrary action on the family of subsets i.

Thus H {h} contains a subgroup conjugate in S to S(A0), proving(38) If H satisfies (37), then H S(A0).

The result analogous to Lemmas 10 and 12 will be quite simple to state andprove this time:

Lemma 14. Let be a countably infinite set and G a subgroup of Sym(), andsuppose there exist two disjoint sequences of distinct elements, (i), (i) ,such that for every element (i)

i{i, i} , there exists g G such

that (i) = (i g).

Then G S(A0).Proof. Let = {i : i }, let A be the partition of whose members are thetwo-element sets {2j , 2j+1} (j 0), and let s Sym() be any element whichfixes all the elements i and interchanges 2j and 2j+1 for all j 0. We claimthat every member of Sym()(A) extends to an element of G {s}.

Indeed, given f Sym()(A), define (i) by letting i = i if i ismoved by f (i.e., if it is transposed with the other member of its A-equivalenceclass) and i = i otherwise. By hypothesis we can find g G such that i = i gfor all i. It is now easy to see that g s g1 acts by f on .

Thus G {s} satisfies (37), so by (38), S(A0) G {s} G.

As the pattern of the two preceding sections suggests, we will now prove that

any closed subgroup G satisfying (33) satisfies the hypothesis of the above lemma.We begin with a reduction: Assuming (33), let M > 1 be the largest integer suchthat for every finite subset , the group G() has orbits of cardinality at leastM. Thus, there exists some finite such that G() has no orbits of cardinality> M . Since G() inherits from G the property (33), we may replace G by G()and so assume without loss of generality that


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(39) For every finite subset , the maximum of the cardinalities ofthe orbits of G() is M.

A consequence is that for any such , every orbit of G() of cardinality M is alsoan orbit of G (since the orbit of G containing it cannot have larger cardinality).Thus

(40) If is a finite subset of , and an element of such that| G() | = M, then for every g G we have G() g = G().

We shall now construct recursively, for each j 0, elements j , j and asubset Kj G, indexed as(41) Kj = {g(k0, k1, . . . , kj1) : (k0, k1, . . . , kj1) {0, 1}j}.

Again we start with K0 =

{g()

}=

{1

}. Assuming inductively for some j 0

that i, i have been defined for all i < j and Ki for all i j, we let j denote the set all of images of 0, . . . , j1, 0, . . . , j1, 0, . . . , j1 underinverses of elements of K0 . . . Kj. By assumption, G(j) has an M-elementorbit. Let j and j be any two distinct elements of such an orbit. (Note that jand j are distinct from all i, i for i < j, since the latter are fixed by G(j).)

For each (k0, . . . , kj1) {0, 1}j, we let g(k0, . . . , kj1, 0) and g(k0, . . . , kj1, 1)be elements of G obtained by left-multiplying g(k0, . . . , kj1) Kj by an elementh G(j), chosen so that j hg(k0, . . . , kj1) is j , respectively j . This ispossible by (40).

Given an infinite string (ki) of 0s and 1s, the elements g(k0, k1, . . . , kj1) willagain converge in S by Lemma 7. Assuming G closed, the limit belongs to G, andclearly gives us the hypothesis of Lemma 14, hence the conclusion that G S(A0).

Again we easily get the reverse inequality: Taking as in the first clause of (33)and letting A denote the partition of into orbits of G(), we have G G() S(A) S(A0) by (36).

We deduce

Theorem 15. Let be a countably infinite set, and Q the set of partitions of defined in (34). Then the subgroups S(A) S for A Q (which are clearlyclosed) are mutually -equivalent, and a closed subgroup G S belongs to theequivalence class of those subgroups if and only if it satisfies (33).

The members of this -equivalence class are the members of the equivalenceclass of Theorem 13.

Proof. This is obtained using the above results exactly as Theorem 13 was obtainedfrom the results of the preceding section, except that we need a different argumentto show that G does not belong to the -equivalence class in question if it does notsatisfy the final clause of (33), i.e., if there exists a finite subset such thatG() = {1}. In that situation, any subgroup G will be {1}, hence countable;but clearly S(A0) is uncountable, so S(A0) G.


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8. Countable subgroups.

The final step of our classification is now easy, and we even get a little extrainformation:

Theorem 16. The countable subgroups of S = Sym() form an equivalence classunder , and members of this class are the members of the equivalence class ofTheorem 15. Moreover, for G S, the following conditions are equivalent.

(i) G is countable and closed.

(ii) There exists a finite subset such that G() = {1}.(iii) G is discrete.

Proof. The countable subgroups are clearly the subgroups 1{1}, and as notedin Lemma 3, for subgroups of symmetric groups Sym(),

1

-equivalence is the

same as 0-equivalence, which is what we are calling -equivalence. This givesthe first assertion; the second is also immediate, since the trivial subgroup is allsubgroups, and is the subgroups of Theorem 15 by the only if assertion ofthat theorem.

To prove the equivalence of (i)(iii), we note first that (ii) and (iii) are equivalent,since a neighborhood basis of the identity in the function topology on G is given bythe subgroups G() for finite , so the identity element (and hence by translation,every element) is isolated in G if and only if some such subgroup is trivial.

To see that these equivalent conditions imply (i), observe that (ii) implies thatG G() = {1}, hence that G is countable, while (iii) implies that G is closed,by general properties of topological groups. (If G is a discrete subgroup of atopological group S, take a neighborhood U of 1 in S containing no nonidentity

element of G, and then a neighborhood V of 1 such that V V1

U. One findsthat for any x S, xV is a neighborhood of x containing at most one element ofG; so G has no limit points in S.)

Conversely, we have seen that any countable G is the members of the equiv-alence class of Theorem 15, hence does not belong to the equivalence class of anyof Theorems 11, 13 or 15. Hence if G is also closed, those theorems exclude allpossible behaviors of its subgroups G() (for finite) other than that there existsuch a with G() = {1}; so (i) implies (ii).

For convenience in subsequent discussion, let us name the four equivalence classesof subgroups of S = Sym() which we have shown to contain all closed subgroups:

(42) CS = the -equivalence class of S.CP = the -equivalence class to which S(A) belongs for all A P.CQ = the -equivalence class to which S(A) belongs for all A Q.C1 = the -equivalence class consisting of the countable subgroups

of S.


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9. Notes and questions on groups of bounded permutations.

It would be of interest to investigate the equivalence relation on classes ofsubgroups G Sym() other than the class of closed subgroups. One such classis implicit in the techniques used above: If is any set and d a generalized metricon , let us define the subgroup

(43) FN(, d) = {g Sym() : ||g||d < }.(We write FN, for finite norm, rather than B for bounded to avoid confusionwith the symbol for an open ball.) These subgroups are not in general closed.For instance, if d does not assume the value (i.e., if it is an ordinary metric)but is unbounded (say the standard distance metric on = R), then bythe former condition, FN(, d) contains all permutations of that move onlyfinitely many elements, which form a dense subgroup of S = Sym(), while bythe unboundedness of d, it is nevertheless a proper subgroup of S, hence it is not

closed.Here are some easy results about the relation on these subgroups. (Cf.

also [11].) Below, uncrowded means 0-uncrowded.Lemma 17. Let d be a generalized metric on a countably infinite set .

(i) If d is not uncrowded, then FN(, d) CS.(ii) If d is uncrowded but not uniformly uncrowded, then FN(, d) is the groupsin CP , but is CS.(iii) If d is uniformly uncrowded, but for some r < , infinitely many of the ballsBd(, r) contain more than one element, then FN(, d) is the groups in CQ,but the groups in CP .(iv) If d is uncrowded and for each r < all but finitely many balls Bd(, r) aresingletons, then FN(, d) C1.Proof. In situation (i), let Bd(, r) be a ball of finite radius containing infinitelymany elements. Then all g Sym()(Bd(,r)) satisfy ||g||d 2r, hence lie inFN(, d), and the conclusion follows by (8).

In cases (ii) and (iii) we can similarly find subgroups of FN(, d) of the formS(A) for A P, respectively A Q, while the last sentence of Theorem 5 givesthe negative statements for these two cases.

For d as in (iv), each set {g Sym() : ||g||d < n} (n ) is finite, so theirunion, FN(, d), is countable.

Thus, if a group FN(, d) belongs to one of the four -equivalence classesof (42), the above lemma determines precisely which class that must be.

Note that our definition (43) can be rewritten

(44) FN(, d) =n {g S : ||g||d < n}.

We claim that if the generalized metric d is uncrowded, then for each n the set{gS : ||g||d < n} is compact. Indeed, the condition ||g||d < n determines, foreach , a certain finite set of possibilities for g ; so {gS : ||g||d < n} isthe intersection of S with a certain compact subset of . But for each ,


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the condition ||g||d < n also limits us to finitely many possibilities for g1, fromwhich it can be deduced that any limit in of elements g S with ||g||d < nis again surjective, hence again belongs to S. So {gS : ||g||d < n} is closed ina compact subset of , hence, as claimed, is compact. This makes FN(, d) acountable union of compact sets, suggesting the second part of

Question 18. If d is an uncrowded generalized metric on a countably infinite set, must FN(, d) belong to one of the -equivalence classes of (42)?

More generally, does every subgroup of Sym() that is a union of countablymany compact subsets belong one of these classes? What about subgroups that are

unions of countably many closed subsets? What about Borel subgroups? Analytic

subgroups?

If the answer to any of these questions is negative, can one describe all the

-equivalence classes to which such subgroups belong?

Let us sketch a couple of cases where it is not hard to show that FN(, d) doesbelong to one of the equivalence classes of (42).Let d be the standard distance function on (inherited from R). To show that

FN(, d) CQ, let A1 be the partition of consisting of the subsets {2m, 2m+1}(m ), and A2 the partition consisting of the subsets {2m+1, 2m+2} and thesingleton {0}. Then A1, A2 Q, so S(A1) S(A2) CQ. This subgroup is clearlycontained in FN(, d); we claim that equality holds.

Indeed, given f FN(, d) with ||f||d = n, if we let i = {ni, ni+1, . . . ,n(i+1) 1} for i 0 a n d 1 = , then we see that for all i 0, if i1 i i+1. Letting B1 be the partition of into the subsets 2i 2i+1and B2 the partition into the subsets 2i1 2i (i 0), we see as in the secondparagraph of the proof of Lemma 9 that f S(B1)S(B2). On the other hand,it is easy to show that S(B1) and S(B2) are both contained in

S(A1)

S(A2)

,

using the fact that any permutation of a 2n-element string of integers 2i 2i+1or 2i1 2i can be written as a product of finitely many transpositions ofconsecutive terms, and that the number of transpositions needed can be boundedin terms of n (cf. end of paragraph preceding (36)). So f S(A1) S(A2), asclaimed, so FN(, d) CQ.

In the above example, the argument cited from the proof of Lemma 9 uses thefact that for any f FN(, d), the number of elements that f carries upward pasta given point is equal to the number that it carries downward past that point. Ifwe modify this example by replacing with Z, again with the standard metric,that property no longer holds, as shown by the translation function t : n n+1.It is not hard to see, however, that given f FN(Z, d), the difference between thenumber of elements that f moves upward and downward past a given point is the

same for all points, and that the function associating to f the common value of thisdifference is a homomorphism v : FN(Z, d) Z. If we let A1 denote the partitionof Z into sets {2m, 2m+1} and A2 the partition into sets {2m+1, 2m+2}, wesee that S(A1) and S(A2) lie in the kernel of v, while v(t) = 1. The argument ofthe preceding paragraph can be adapted to show that S(A1) S(A2) = ker(v),hence that S(A1) S(A2) {t} = FN(Z, d); so this group also belongs to CQ. (For


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some further properties of this example see Suchkov [9], [10], where FN(Z, d) andits subgroup ker(t) are called G and G respectively.)

An example that falls under case (ii) of Lemma 17 (so that if FN(, d) belongsto one of our four classes, that class is CP) is given by = {n : n } withthe metric induced from R. We suspect one can show that it does belong to CPby adapting the method we used for FN(, d), putting in the roles of A1 andA2 the partitions of arising from the integer-valued functions [/2] and [(+1)/2], where [] denotes the integer-part function.

Two cases that have some similarity to that of FN(Z, d) but seem less trivial,and might be worth examining, are those given by the vertex-sets of the Cayleygraphs of the free abelian group, respectively the free group, on two generators,with the path-length metric. An example of a different sort is the set with theultrametric under which d(, ) is the greatest n such that and differ in thenth digit of their base-2 expansions. From the fact that this d is an uncrowdedultrametric, it is easily deduced that FN(, d) is the union of a countable chain ofcompact subgroups. All three of these examples fall under case (iii) of Lemma 17,so that if they belong to any of the classes of (42) it is CQ.

10. Further questions about and .

It seems unlikely that one can in any reasonable sense describe all -equivalenceclasses of subgroups of the symmetric group on a countably infinite set . On theother hand, if one regards the set of such equivalence classes as a join-semilattice,with join operation induced by the map (G1, G2) G1 G2 on subgroups,one may ask about the properties of this semilattice. The cardinal |G | + 0 isan -invariant on subgroups G of Sym(), and induces a homomorphism fromthis join-semilattice onto the semilattice of cardinals between

0 and 2

0 under

the operation sup . Of our four classes, C1 maps to the bottom member of thischain, while the other three map to the top member. Although the operationof intersection on subgroups of S does not respect the relation , it is not clearwhether our join-semilattice may nonetheless be a lattice. The second author hopesto give in a forthcoming note further results about this semilattice, and in particular,on Question 18 above.

How much influence does the isomorphism class of a subgroup G Sym()have on its -equivalence class? It does not determine that class; for consider theabstract group G = (Z/pZ) for p a prime. If for each i we let i be aregular Z/pZ-set (hence of cardinality p) on which we let G act via the projectionon its ith coordinate, and we take for a disjoint union of the i, then we get arepresentation of G as a compact subgroup of Sym() belonging to CQ.

On the other hand, we may identify G with the direct product

i0 (Z

/pZ

)i

,and let be a disjoint union of regular representations of the factors in thisproduct, getting a representation of G in Sym(), also compact in the functiontopology, but belonging to CP . Finally, observe that if V is a vector space ofdimension 0 over the field of p elements, and we also regard G = (Z/pZ) asa vector space over this field, then G and V , both having the cardinality of


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the continuum, are both continuum-dimensional, hence isomorphic. Performingthe same construction as before on this product expression G =

V, we get a

representation of G as a group of permutations of a countable set (with Gagain closed in the function topology, but no longer compact), which Lemma 10(with i a representative of the ith orbit, and Di the product of the first i orbits)shows belongs to CS.

Of course, membership of a subgroup in the class C1 is determined by its cardi-nality, hence by its isomorphism class. But to any isomorphism class I of groupsof continuum cardinality, we may associate the subset of {CS, CP , CQ} consistingof those -equivalence classes (if any) that contain members of I. Which subsetsof {CS, CP , CQ} arise in this way (or in various related ways; for instance, by as-sociating to an isomorphism class I the set of closed subgroups that belonging toI), we do not know.

If we take account of the topological structure of a subgroup G S, this canimpose restrictions on its

-equivalence class:

Lemma 19. If is an infinite set, then a subgroup G S = Sym() is compactin the function topology if and only if it is closed and has finite orbits.

Proof. If G is closed and the members of the partition A given by the orbits of Gare all finite, then G is a closed subgroup of S(A) =

ASym(). It is not hard

to see that this isomorphism is also a homeomorphism, hence as the above productof finite discrete groups is compact, so is G.

Conversely, if G is compact, it is closed in S by general topology, and for each the orbit G, being an image of the compact group G under a continuousmap to the discrete space , is finite.

So for || = 0, a compact subgroup of S cannot belong to CS. Note also thatif G is a closed subgroup of S not in

CS, then by Theorem 11 there exists a finite

set such that G() has finite orbits, so by the above lemma G() is compact.Thus, though G itself need not be compact, it will be a countable extension of acompact subgroup that is open-closed in it, and thus will be locally compact.

11. Some finiteness results.

This section assumes only the notation recalled in the first two paragraphs of2,and the contents of 4 (the definition of the function topology, and Lemma 7. Atone point we will call on a result of a later section, but our use of that result willsubsequently be superseded by a more general argument.) We begin with a resultthat we will prove directly from the definitions.

Lemma 20. Suppose is a set, and G a subgroup of S = Sym() which isdiscrete in the function topology on S, and has the property that each member ofG moves only finitely many elements of . Then G is finite.

Proof. The statement that G is discrete means that there is some neighborhood of1 containing no other element of G. Since a neighborhood basis of 1 in S is givenby the subsets S() for finite , there is a finite such that G() = {1}.


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Take such a , and assuming by way of contradiction that G is infinite, let0 be maximal for the property that G(0) is infinite, and let be anyelement of 0. Then G(0) inherits the properties that we wish to show lead toa contradiction; so, replacing G with this subgroup, we may assume that for some , G({}), unlike G, is finite, say of order n. Then the orbit G must beinfinite, so let g0, . . . , gn be n+1 distinct elements of that orbit. By hypothesis,each of g0, . . . , gn moves only finitely many elements of , so the infinite set Gmust contain an element g not moved by any of them. Hence G({g}) containsthe n+1 elements g0, . . . , gn, contradicting the fact that, as a conjugate of G({}),it must have order n.

If we generalize the hypothesis of this lemma by letting be a subset of andG a discrete subgroup of Sym(){} each member of which moves only finitelymany elements of , it does not follow that G induces a finite subgroup of Sym().For example, partition an infinite set into two sets and

of the same

cardinality, and let v be a homomorphism from a free group F of rank || ontothe group of those permutations of that move only finitely many elements. Takea regular representation of F on , and consider the representation of F on = ( ) given by the graph of v, i.e., the set of elements of Sym(){}that act on by an element a F, and on by v(a). This subgroup isdiscrete because for each we have G({}) = {1}; but it does not inducea finite group of permutations on .

On the other hand, the proof of Lemma 20 easily generalizes to show that if G isa discrete subgroup of S = Sym(), and is the union of a family of G-invariantsubsets i such that every element of G moves only finitely many members ofeach i, then G is finite. (Incidentally, note that throughout this section, whenwe refer to families of subsets i or i of , there is no disjointness assumption.)In a different direction, if is countable we can formally strengthen Lemma 20 byweakening the hypothesis discrete to closed; for a subgroup of Sym() whosemembers each move only finitely many elements must be countable, and we saw inTheorem 16 that a countable closed subgroup of Sym() is discrete.

Now suppose that for countable we combine the above two weakenings of thehypothesis of Lemma 20, and consider a closed subgroup G < Sym() such that forsome expression =

Ii of as a union of G-invariant subsets, each element

of G moves only finitely many members of each i. We would like to concludethat G induces a finite group of permutations of each i; but we cannot argue asabove, for now G need not be countable, making Theorem 16 inapplicable.

In an earlier version of this preprint we asked whether this conclusion nonethelessheld. Greg Hjorth has shown us a proof, which, with his permission, we give below.We will use

Lemma 21. Let be a countable set, G a closed subgroup of S = Sym(), and(i)iI a family of subsets of . Then either

(i) there exists a finite set such that G() S{i} for all but finitely manyi I, or(ii) there exists an element g G such that g / S{i} for infinitely many i I.


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Moreover, if all i are finite, then in (i) we can strengthen all but finitelymany to all.

Proof. As in Lemma 7, let = {0, 1, . . . }. Assuming (i) does not hold, we shallconstruct g0, g1, . . . G which converge, by that lemma, to an element g withthe property asserted in (ii).

Suppose inductively that for some j 0 we have chosen 1 = g1, g0, g1, . . . ,gj1 G, and also distinct indices i0, . . . , ij1 I and elements 0, . . . , j1 , such that for k = 0, . . . , j1, gk moves k either out of or into ik , and suchthat defining, for 0 k j,

(45) k = {0, . . . , k1 } {0 g1k1, . . . , k1 g1k1} {0, . . . , k1 } {0 g1k1, . . . , k1 g1k1} (cf. (4)),

we have

(46) gk G(k) gk1 for 0 k < j (cf. (5)).Since j is finite, our assumption that (i) fails tells us that there are infinitelymany i such that G(j) fails to preserve i. It follows that by multiplying gj1on the left by a member of G(j) if necessary, we can insure that for some indexother than i0, i1, . . . , ij1, which we may call ij, the resulting product gj fails topreserve ij , i.e., moves an element ij into or out of ij . Also, (46) shows thatgj retains the properties of the preceding elements gk of moving k into or outof ik (0 k < j). Applying Lemma 7, we get a limit element g which clearlypreserves none of i0 , i1 , . . . .

To get the final assertion, observe that if all i are finite and (i) holds, we maytake a as in (i) and then adjoin to it the elements of the finitely many sets inot preserved by G().

Applying the above lemma (in particular the final sentence) in the case where

the i are the singleton subsets of a set , we getCorollary 22. Let be a countable set, G a closed subgroup of S = Sym(),and a subset of . Then either

(i) there exists a finite set such that G() S(), or(ii) there exists an element g G which moves infinitely many members of .

We shall now get our desired result by an argument similar to the proof ofLemma 20, with the above corollary replacing our use of discreteness.

Theorem 23 (G. Hjorth, personal communication). Let be a countable set, Ga closed subgroup of S = Sym(), and {i | i I} a family of G-invariant subsetsof such that each element of G moves only finitely many elements of each i,and

Ii = .

Then G acts on each i as a finite group of permutations; equivalently, G fixesall but a finite subset of each i.

Proof. The equivalence of the two forms of the conclusion follows from the hypoth-esis that each member of G moves only finitely many elements of each i. Toprove the first form of that conclusion, suppose, on the contrary, that G induces


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an infinite group of permutations on j for some j I. Applying the preced-ing corollary to j , and noting that, by hypothesis, case (ii) of that corollary is

excluded, we get a finite such that G() S(j ).As in the proof of Lemma 20, let 0 be a maximal subset of such that G(0)

induces an infinite group of permutations of j, and any element of 0.Replacing G by G(0), which clearly inherits the hypotheses of the theorem, wehave that G({}), unlike G, induces a finite group of permutations of j, sayof order n. Thus for any g G, the group G({g}) likewise induces a group ofpermutations of j of order n.

As before, must have infinite orbit G. Now applying to some i that con-tains the hypothesis that each element of G moves only finitely many elementsof i, we see that each element of G lies in G({g}) for all but finitely many dis-tinct g G. It follows that every finitely generated subgroup of G is likewisecontained in G({g}) for all but finitely many distinct g . Since G induces aninfinite group of permutations of j , we can find a finitely generated subgroup ofG that induces a group of > n permutations of that set. But by the last sentenceof the preceding paragraph, a group of order > n cant be contained in any of thesubgroups G({g}), let alone in all but finitely many of them. This contradictioncompletes the proof of the theorem.

We remark that the analog of Lemma 20 (and hence of Theorem 23) fails forsubmonoids of . For instance, let = , and for i, j define fi(j) =max(i, j). Then G = {fi | i } satisfies the hypotheses of Lemma 20 withsubmonoid of in place of subgroup of Sym(), but does not satisfy theconclusion.

12. Other preorderings, and further directions for investigation.

In the arguments of5-7, when we obtained a relation G H, we often did thisby showing that G lay in the subgroup of S generated by finitely many conjugatesof H. This suggests

Definition 24. If S is a group, an infinite cardinal, and G1, G2 subgroups ofS, let us write G1

cj,S G2 if there exists a subset U S of cardinality < such

that G1 fU f

1G2f.As with ,S, we may omit the subscripts and S from

cj,S when their

values are clear from context, and we will write cj,S or cj for the inducedequivalence relation. For the remainder of this discussion, will be 0 and S willbe Sym() for a countably infinite set , and these subscripts will not be shown.

In general, cj

and cj

are finer relations than and . Since not all thearguments in 5-7 were based on combining conjugates of the given subgroup G (inparticular, some were based on conjugating a carefully constructed element s Sbyelements of G), it is not obvious whether those results can be strengthened to saythat the classes of subgroups that we proved -equivalent are in fact cj-equivalent.Let us show that the answer is almost.


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Recall (cf. [2, p.51, Theorem 6.3]) that since is countably infinite, the onlyproper nontrivial normal subgroups of S are the group of permutations that move

only finitely many points, which we shall denote Sfinite

, and the subgroup of evenpermutations in Sfinite, which we shall denote Seven.

Lemma 25. Let be a countably infinite set, and G a subgroup of S = Sym()not contained in Sfinite. Then the unary relations cj G and G on the set ofsubgroups of S coincide. Hence if G is uncountable, the unary relations cj Gand G on that set also coincide.Proof. In the first assertion, the nontrivial direction is to show that H G impliesH cj G. The former condition says that H G U for some finite U S.Now since G is not contained in the largest proper normal subgroup of S, thenormal closure of G is S. Hence each element of U S is a product of finitelymany conjugates of elements of G. The desired conclusion follows immediately.

To see the second assertion, note that if G is uncountable, so is any H

G,

hence we also have H Sfinite, and the preceding result can be applied to H aswell as to G, giving H cj G.

We can now get

Proposition 26. Let be a countably infinite set, and S = Sym(). Then all-equivalence classes of subgroups of S other than C1 are also cj-equivalenceclasses. The class C1 decomposes into the following six cj-equivalence classes:

(i) The set of countable (finite or infinite) subgroups not contained in Sfinite,(ii) The set of infinite (necessarily countable) subgroups of Sfinite not contained

in Seven.(iii) The set of infinite (again countable) subgroups of Seven.(iv) The set of finite subgroups contained in Sfinite but not in Seven.

(v) The set of finite nontrivial subgroups of Seven

.(vi) The set containing only the trivial subgroup.

Proof. The first assertion follows immediately from the preceding lemma. In thesecond, it is not hard to see that the sets (i)-(vi) partition C1, and that a subgroupin one of these sets cannot be cj-equivalent to one not in that set, so it remainsonly to show that any two groups in the same set in our list are cj-equivalent.

That this is true of (i) follows from the first assertion of the preceding lemma,and the fact that all members of C1 are -equivalent.

Skipping to (iii), if G is in that class, then Lemma 20 shows that G is non-discrete. From a sequence of nonidentity elements of G approaching 1, we canextract an infinite subsequence consisting of elements whose supports, supp(g) ={ | g = }, are pairwise disjoint. If we take s S whose support has single-ton intersection with each of those supports, we find that each of the correspondingcommutators s1g1sg is a 3-cycle [2, p.51, Exercise 6(i)]. These 3-cycles lie inGs1Gs, and no point belongs to the support of more than two of them, so wecan find an infinite set of 3-cycles in that group with disjoint supports. By droppingsome of these, we may assume that the complement in of the union of their sup-ports is infinite. Hence we may assume without loss of generality that = Z and


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that we have gotten the 3-cycles (4n, 4n+1, 4n+2) for all n Z. Three more con- jugations now give us all 3-cycles of the form (k, k +1, k + 2) (k Z), and thesegenerate S

even

. Hence G cj

Seven

. So all subgroups in (iii) are cj

-equivalent tothat subgroup, hence to each other, as required.

For G in class (ii), the above result shows that finitely many conjugates ofG Seven generate Seven, and since G also contains an odd permutation, the cor-responding conjugates of G generate Sfinite. So all such groups are cj-equivalentto Sfinite, and so again, to each other. That the members of each of (iv), (v),and (vi) are mutually cj-equivalent is easily deduced from standard results aboutfinite symmetric groups [2, 2.4].

We note that the ordering on these sets induced by the relation cj on subgroupsis

(i) cj (ii) cj {(iii), (iv)} cj (v) cj (vi),

with (iii) and (iv) incomparable.There is another family of preorders also implicit in the methods we have used.

Given subgroups G, H Sym() and a cardinal , let us write

(47) G fix H if for some with || < we have G() H,and let us write G fix H for the conjunction of G fix H and Hfix G.

Lemma 2 yields an implication between these relations and those studied in thisnote:

(48) G fix0

H = G ||+ H.The relations fix and fix tend to be quite fine-grained. For instance, given

partitions A1 and A2 of , it is not hard to see that S(A1) fix S(A2) if and onlyif A1 and A2 disagree at < elements, meaning that one can be obtained from

the other by redistributing < elements of .

In a different direction, one might define on abstract groups (rather than sub-groups of a fixed group) a preordering analogous to , by letting G1

emb G2

mean that G1 admits an embedding in a group H which is generated over G2 by< elements.

In our study of symmetric groups in this note, we have considered only countable, except when no additional work or distraction was entailed by allowing greatergenerality. It would be of interest to know what can be said about -equivalenceclasses of closed subgroups of Sym() for general and ; in particular, whetherthere are simple criteria for a closed subgroup G Sym() to be 0-equivalent(equivalently, ||+-equivalent) to Sym().

A related topic which has been studied extensively (e.g., [8], [12]) is the cofinalityof groups Sym(), defined as the least cardinal such that Sym() can bewritten as the union of a chain of < proper subgroups. If S = Sym() isof cofinality , then our unary relation 0,S S is equivalent to ,S S (cf.proof of Lemma 3(ii) above); though the converse fails under some set-theoreticassumptions.


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Still another direction for study would be to consider questions of the sort studiedhere, but for structures other than groups. Mesyan [6] gets some results of this sort

for the ring of endomorphisms of the -fold direct sum of copies of a module.

References

[1] George M. Bergman, Generating infinite symmetric groups, to appear, J. London Math.Soc.. Preprint version, 8 pp.: http://math.berkeley.edu/ gbergman/papers/Sym Omega:1.{tex,dvi} , arXiv:math.GR/0401304 .

[2] Meenaxi Bhattacharjee, Dugald Macpherson, Rognvaldur G. Moller and Peter M.Neumann, Notes on infinite permutation groups, Texts and Readings in Mathematics, 12,Hindustan Book Agency, New Delhi, and Lecture Notes in Mathematics, 1698,Springer-Verlag, 1997. MR 99e:20003.

[3] Stephen Bigelow, Supplements of bounded permutation groups, J. Symbolic Logic 63 (1998)89102. MR 99b:20007.

[4] Fred Galvin, Generating counta

George M. Bergman and Saharon Shelah- Closed subgroups of the infinite symmetric group

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