INDUCED MODULES AND SOME ARITHMETIC INVARIANTS OF THE FINITARY SYMMETRIC GROUPS … › ... ›...

Nova Journal of Algebra and GeometryVol. 2, No.1, pp. 89-105, 1993

ISSN 1060-9881@ 1993 Nova Science Publishers, Inc.

INDUCED MODULES AND SOME ARITHMETICINVARIANTS OF THE FINITARY SYMMETRIC GROUPS

S. K. SEHGALDepartment of Mathematics

University of AlbertaEdmonton, Alberta, Canada T6G 2G 1

A. E. ZALESSKIIInstitute of Mathematics

Academy of Sciences of Byelorussian SSR220072 Minsk, USSR

The finitary symmetric group S(Q) is the group of all finitary per-

mutations of an infinite set 11. It appears in a natural way in numerous

applications. In particular, it plays a crucial role in the theory of poly-

nomial identities of algebras. However we view this group as an important

90 S.K. Sehgal and A.E. Zalesskii

member of the class of locally finite groups. There are some general and

somewhat mysterious conjectures on locally finite groups which would be

good to verify first for most important. examples.

Let P be a field and let C denote the field of complex numbers. For

a group G we denote by PG the group ring of Gover P. If X is a

subgroup of G, then Ix denotes the trivial PX-module and Mx = 1~denotes the induced module. By Ix we denote the annihilator of Mx in

PG.

For what subgroups Xc G is the ideal Ix not zero? This question

seems to be difficult for finite G and we do not expect that any good

description of such X exists. However there are some indications that such

X for infinite G has very specific nature and possibly plays some essential

role from the group theoretical point of view. In order to clarify this we

consider the following generalization of the notion of a normal subgroup:

DEFINITION 1. Let G be a group and H C G a subgroup. We say

that H is enormous, if there exists a finite subgroup F C G such that

g-lFgnH =I {l} for all 9 E G.

CONJECTURE 1. Let G be a simple locally finite group and X a subgroup.

Let Ix C PG be the annihilator of Mx. Then Ix ~ {OJ if and only if .

X is enormous.

Induced Modules and Some Arithmetic Invariants... 91

PROPOSITION1. Let P be an arbitrary field and X C G a subgroup. If

Ix i= {O}, then X is enormous.

However, it seems to be difficult in general to check Conjecture 1 in the

opposite direction. We show that this conjecture is true for the alternating

finitary group. Furthermore, we give a complete description of enormous

subgroups of the finitary symmetric and the finitary alternating group.

THEOREM1. Let n be an infinite set and G = B(n) or G = A(n). Let

X C G be a subgroup and Ix the annihilator in PG of the induced

module l~. Then the following conditions are equivalent:

(i) Ix =f {O};

(ii) X is enormous;

(ill) X contains a Young subgroup Y of finite index;

(iv) the set of overgroups of X is finite.

REMARK. The term "Young subgroup" in this paper is slightly different

from the standard one, see Definition 3 below. According to Formanek and

Lawrence [2] every ideal of CG is described in terms of Young diagrams;

however, the dependence between X and the collection of Young diagrams

describing Ix is rather complicated.


l(n) or S(n) contains the center Z of l(n); then Ix contains the

augmentation id~ of P Z and, in particular, Ix =f:{O}. This indicates

that Conjecture 1 is true for a larger group class than the class of locally

finite simple groups.

In order to understand better the nature of enormous subgroups of 10-

ca.llyfinite groups we make an attempt to find a quantitative characterization

of them. Let G be a loca.llyfinite group. Let .c = {G,\} '\EA be a system of

finite subgroups of G such that G>., Gp. E.c implies (G,\, Gp.) S;Gv E .c

and G = U,\EA G,\, that is, .c is a local system of G. We write ,\ < J.'

if G,\ c Gp. so A is a directed set. Let X be a subgroup of G. For

h E G let Cl (h) denote the conjugacy class of G containing h. For

x EX) 9 E G we define

h(Xtg) = IG.\n Cl(x )//(1 + IG,\n gXg-1 n CI (x)!).

We introduce the following:

DEFINITION 2. Let H be subgroup of a locally finite group G. We say

that H is cobounded if there exists a local system .c of finite subgroups

of G such that the set {b,\(x,g) I ,\ E At x E Ht 9 E G} is bounded.

CONJECTURE2. Let G be a simple locally finite group and 1 =f: X c G a

subgroup. Then X is enormous if and only if X is cobounded.

Induced Modules and Some Arithmetic: Invariants... 93

assumptions.

PROPOSITION2. Let G be a locally finite group without finite normal sub-

groups different:&om {I}. Let Xc G be a subgroup. If X is cobounded,

then X is enormous.

The converse is proved only for the finitary alternating groups.

THEOREM3. Let G be a finitary alternating group. Let X be a subgroup

of G. If X is enormous, then X is cobounded.

§l. Classification of enormous subgroups

Let n be an infinite set. We recall that a finitary permutation is one

that act nontrivially only on finitely many elements of n. We denote by

s(n) the group of all finitary permutations of n and call it the finitary

symmetric group of n. Furthermore A(!l) is its subgroup of even permu-

tations; it is usually called a finitary alternating group. If M ~ S(n) be a

subset, then supp (M) is the set of I.IJE!l such that MI.IJ:f=UJ. If 01 C n,

then we keep the symbol S(Od (resp. A(~I)) for the group of all the

permutations s E s(n) (resp. s E A(!l)) such that suppes) S;;; 01.

DEFINITION3. Let Y c S(n) be a subgroup. We say that Y is a Young


is finite.

Let k = 10-(01 U.. .uOn)l. It is clear that the pair k, n is an impor-

tant invariant of a Young subgroup (for countable 0 pairs k, n determine

Young subgroups up to conjugacy in the group of all (not necessarily fini-

tary) permutations of 0). We shall often write Y(k, n) in order to stress

the dependence of Y on k and n.

LEMMA 1. Let X, F be subgroups of G with F finite. The restric-

tion (l~)IF is a direct sum of modules (lFngxg-l)F where 9 runs over

a transversal oE the (F,X)-double cosets oE G. Moreover, Ix n PF =

ngEG Ann (lFngxg-l )F.

This lemma follows from standard facts on induced modules (see, for

instance, (1, Theorem 44.2]). As usual, Ann denotes the annihilator of a

module.

PROOF OF PROPOSITION 1: Let 0 1:- a E Ix; write a E L: aigi with

gi E G, 0 =I ai E P. Take for F the group generated by these g/s. Then

Ann(Mx/PF) = Ix n PF =I o. By Lemma 1, MxlPF is a direct sum of

modules (lFngxg-l)F where 9 runs over a transversal of (F,X)-double

cosets of G. By way of contradiction, suppose that X is not enormous.


contains a regular submodule. Therefore the annihilator of MxlPF is {O},

a contradiction. 0

LEMMA2 (WIELANDT[5, SATZ 9.4]). Every primitive subgroup of S(n)

contains A(n).

We recall some facts of the representation theory of the symmetric and

alternating groups, see [3, 4]. Let SI: and AI: denote the symmetric and

the alternating group on Ie symbols respectively.

(I) The irreducible representations of SI: are parameterized by Young

diagrams. Let .A = (.Al,...,.Ar) be a set of natural numbers such that

.AI+... +.Ar = k and .AI ;::: ... ;:::.Ar. The Young diagram corresponding

. to.A is the picture

with r rows where .Ai is the number of boxes in the i-th row. There

exists a 1-1 correspondence between irreducible representations of SI: and

Young diagrams with kboxes. For each tfJ E IrrSk the restriction tfJlSI:-l

is multiplicity free and its irreducible constituents are described by Young

I I I II


~E Irr Sk-b the irreducible constituents of the induced representation ~Slo

are those whose Young diagrams are obtained from the diagram of ~ by

adding one single box.

(II) To ~ is associated another diagram obtained by changing rows

into columns. We say ~ is symmetric if it is identical with its associate. The

restriction <pIAkis irreducible provided the diagram of <P is not symmetric.

Otherwise <PIAl is a direct sum of just 2 nonequivalent representations

which are conjugate by an element of S k. Moreover, for <PI, <P2E Irr S t

their restrictions to At are equivalent if and only if their diagrams are

associate. As ~ runs over the representations of Sk with diagrams whose

first row is not shorter than the first column, the irreducible constituents of

~IAk run over all irreducible representations of At.

(III) Littlewood-Richardson rule (see [3, 4]). Let <P and c.p be irre-

ducible representations of Sk and Sl with diagrams ~ and 1-', respec-

tively. Then the irreducible representations of Sk+l which appear in the

induced representation «p 0 tp)S'+l correspond to the diagrams which can

be obtained by the following procedure:

Add to ~ the boxes of the first row of 1-'. These may be added to

one row or divided into any number of sats, preserving their order, the first

set being added to one row of ~, the second to a subsequent row, the third

to a row subsequent to this and so on. After the addition no row of the


row of J.L, according to the same roles, followed by the remaining rows in

succession until all the boxes of J.L have been used. These additions must

be such that each box of J.L shall appear in a later row of the compound

diagram than the row containing the box immediately above it in J.L.

LEMMA3. Let S = S(r) where Irl = r < 00. Let Q ~ r be a subset and

let Q = Ql U... UQt be a disjoint partition. Let T = S(Ql) X ... x S(Qt)

and R=A(Qdx...xA(Qt). Let d=lr\QI. Then (i) the Youngdiagram

of any irreducible component of If contains at most d boxes outside the

first trows. (ii) The Young dia.gram of any irreducible component of I ~contains at most d boxes outside both the first t rows and the first t

columns.

Proof follows easily from the Littlewood-Richardson role. Observe that

l~ is a direct sum of pS where p runs over the I-dimensional represent a-

tions of T with peT) E {:f:l}.

PROOF OF THEOREM 1: (i) ==> (ii) is just Proposition l.

(ii) ==> (iii): Let F be a finite subgroup of G such that g-1 Fgn

x -# {I} for all 9 E G. Observe that the number of orbits of X is finite. (If

not then let R be a set of representatives of these orbits; let T = supp (F).

Then there exists a map T -+ R which can be extended to an element


XIL ~ A(L) or XIL is imprimitive. In the latter case XIL has at most

ITI - 1 blocks of imprimitivity. Indeed, if not then choose a transversal

R of them; as above there exists an element s E G such that sT C R.

Then SFS-lnX = {I} as SFS-l fixes all but one element in each block of

imprimitivity. Let L1"'" Lt be the set of the blocks of imprimitivity of L

with maximal t and let XL, be the stablizer of Li in X. It is well-known

that XL.ILi is primitive. By Lemma 2 we have XL,IL;::> A(n;).

Let !h,..., nn be the total set of blocks of imprimitivity in L where

L runs over the infinite orbits of X. Put A/(nj) = {g E G: glnj = Id}.

Then A/(nj) ~ A(nj). We have shown in the previous paragraph that

Ujlnj ~ A(nj) where Uj is the stabilizer of nj in X. As sjT C nj

for a suitable Sj E G we have SjFS;l C A/(nj). It follows that the

normal closure of SjFsjl in A'(nj) coincides with A/enj). Therefore X

contain A/(nj) for j = 1,..., n. Then Y = A/(nd x ... x A/(nn) is a

normal subgroup of X and (X : Y) < 00.

(iii) =* (i): Let Y = Y(k, d) be a Young subgroup of G con-

tained in X and let N(Y) be the normalizer of Y in G. Let n1,..., nd

be the infinite orbits of Y and K = {n - (nl u... u nd)} so that k = IKI.

lt is easy to see that (N(Y) : Y) < 2dd! k! < 00. Obviously, Iy ~ Ix

since Mx is a quotient of My. Hence it suffices to prove that Iy i= {O}.

Suppose that P = C, the field of complex numbers.


want to show that IynCF:f: {O}. By Lemma 1, My IF = E9 (lFngyg-l)FgET

where T is a transversal of the (F, Y)-double cosets of G. It is clear that

FngYg-l =A(AI)X",xA(Ar) where Al,...,Ar arethenonempty

intersections f2; n Q. By Lemma 3 the Young diagrams of the irreducible

components of (1Fngyg-l)F contain at most k boxes outside both the

first d rows and the first d columns. It follows that some irreducible

representation <p of F does not occur as a component of (1Fngyg-l)F for

all 9 E G. Therefore the annihilator of the module E9 (lFngYg-l)F is notgET

{O} proving our assertion for the case G = B(f2).

Since IQI > (d + k)2, we can choose tjJ such that the first row of

its Young diagram is longer than the first column. Let F' stand for the

commutator subgroup of F so F'~ A(Q). It follows from (II) that the ir-

reducible components of tjJlF' do not occur as components of (1F'ngYg-l )F'

A(11) . 5(11)for all 9 E G. Observe that ly IS a component of. 1y . Hence the

irreducible components of tjJlF' do not occur as components of 1:<11). It

follows that the anninHator in CF' of 1:<fl) is not {O} so the annihilator

of l:<fl) in CA(f2) is not {O}. This means that (m) => (i) is true for

G = A(f2) and P = C as desired.

Let now P be an arbitrary field and p = char (P). We have shown

that the annihilator in CF' of l:(fl) is not {O}; hence the dimension

d of the image of CF' in the representation associated with the module..i{O\


d, < d. This implies immediately that the annihilator in P F' of l:<fi)

is not {OJ, as desired. Let fI"", fd" be a basis of the image of P F'.

As l:<fi) IF' is a direct sum of modules (IF'ngF'g-l)F' with 9 E A(n),

we can find a finite subsum r of these modules such that It Ir, . . . ,fapIrare linearly independent. Put hi = fi Ir (i = 1,. .. ,dp). Observethat Ii

and hi are matrices with 0, 1 entries under the basis {gY}gET. Let If

and h~ be the same matrices as Ii and hi where, however, 0,1 entries

are regarded as complex numbers. Obviously, it suffices to show that h~

are linearly independent over C. Suppose that L: aih~ = o with ai E C

such that all ai's are not zero's. Observe that a system of linear equations

with integer coefficients has a solution in (J provided it has a solution in

C. Hence all ai's can be chosen to be rational and then to be integral.

Moreover, having divided ai by a common multiplier we can assume that

a's are not all multiples of p = char (P). Then 0 =L: (tihi (mod p)Ii is

a nontrivial relation with the coefficients ai (mod p) E P. This contradicts

the choice of hi'S and proves the desired implication.

The next lemma establishes equivalence of (iii) and (iv) and completes

the proof of the theorem.

LEMMA 4. The set of overgroups of a subgroup X of B(n) is finite if and

only if X contains a Young subgroup as a subgroup of finite index.


overgroups. Let R -# Y be an overgroup of Y and L an orbit of R. It

is clear that L is a union of some orbits of Y. H R normalizes Y, then

(R : Y) < 00. H not, R contains a cycle c of length 3 which does not

normalize y: The group (c, Y) is primitive on each of its orbits and all its

elements are even permutations. It follows that (c, Y) is a Young subgroup

and the number of its orbits is less than those of Y. Therefore the number

of overgroups of Y is finite. This proves the part "if".

Now let the set of overgroups of a subgroup X of S(O) be finite.

Then, obviously, the set of orbits of X is finite; let L be one of them. Then

the set L\ of the blocks of imprimitivity of X on L is finite. Indeed, if not,

then let us choose an infinite set Ll' L2"" of L\. For a natural number k,

put L(k) = {L},..., L2.} and Tk = {S(Lk),X}. It is clear that Tk £;Tk+1

and the sequence Tl £; T2 £; . .. consists of distinct members. As X C Tl

then X has infinitely many overgroups in S(O); as (S(il) : A(O)) = 2

then the same is true for G = A(il). This contradiction shows that L\ is

finite. Then it follows from Lemma 2 that X contains a Young subgroup

of finite index. 0

We consider now enormous subgroups of the covering group G where

G is S( il) or A( 0). Recall that the center Z of G is of order two,

G ~ GjZ and the commutator subgroup of G is .4(0). IT 01 C il with

3 < lil11 < 00, then the pullback of A(ild c A(il) in G is not split over

Z. The following theorem shows that describing enormous subgroups of G


is reduced to the same problem for G.

THEOREM2. Let X c a be an enormous subgroup. Tben X containS Z.

PROOF: There exists a finite subgroup Fe a such that FngXg-I -# {I}

for any g E G. By replacing F by a larger subgroup we can assume that

F::: S(n) with n > 4. Denote by XI, gl and Fl the image of X, g, F,

respectively under the projection a -t r;/Z = G. H X does not contain

Z then FI n hXlh-1 -# {I} for any hE G. By Theorem 1, Xl contains

a Young subgroup 1": Therefore for a suitable g E a, supp (FI ng1XIg1I)

contains at least five elements so FI n glX1g11 contains the alternating

group A(5) where pullback in a is isomorphic to SL(2,5). IT X n Z = 1

then the preimage of A(5) in X is isomorphic to A(5). It follows that

SL(2,5) has a subgroup of index 2 which is not true. Hence xnz -# {I}.

0

§2. An arithmetical description ofenormous subgroups of s(n)

PROOF OF PROPOSITION2: We are given that X is cobounded. Let .c

be a local system of finite subgroups of G as in Definition 2. Suppose that

X is not enormous. Then for each G~ there exists g~ E G such that

G~ng~Xgi"I={l}. Hence IG~ng~Xgi"lnCl(x)l~l for all xEG. On

the other hand, as Cl(x) is infinite for 1 -# x E G, then IG~ n CI(x)1 is

not bounded. So b~(x,g~) is not bounded as desired. 0


PROOF OF THEOREM 3: Take £, = {G)..J where G).. = A(i)..). Let X

be an enormous subgroup of G. By Theorem 1, (X : Y) < 00 for an

appropriate Young subgroup Y = Y( r, n). Let !2I,...,!2n be infinite or-

bits of Y and R =!2 - (!2J,...,!2n) with r = IRI. As IG)..ngYg-l n

Cl(x)1 :'5 IG)..n gXg-1 n Cl(x)l, it suffices to estimate {h(x,g)} for

X = Y and all x E G. Denote by CG(x) the centralizer of x in G.

We have IG)..n CI(x)1 < 2IG)..I/ICG.\(x)1 because G>. = A(i)..) and ele-

ments of A(i)..) conjugate in A(11) are conjugate in S(i)..). Furthermore,

CG.\(x) is a subgroup of index 2 of CS(k)(X) x S(i>'-k)' It follows that

IG>.n CI(x)1 :'5 2i).!/(i).. - k)!lCS(k)(x)l. Choose A such that x E G).,

(l/n) -. (r/i>.n) > (l/n + 1) and (i). - r)/n > k. (Observe that only

finitely many i>. do not satisfy these conditions.) Put e = [(i). - r)ln];

here [ ] denotes the greatest integer function. Fix any 9 E G. Then

! supp (G)..)ng11jol > e for a suitable jo E {l,...,n} depending on g.

Put Dj = supp (G>.)n g11j and dj = IDjl,j = 1,..., n. We have

G).n gYg-1 n CI(x) ;;2 CI(x) n A(Djo)' Therefore,

IG>. n gY g-1 n CI(x)1 ~ IA(Djo)//ICs(k)(x)IIS(djo - k)1

~ !CS(k)(x)I-1djo!/(djo - k)!.

Since djo ~ e, then djo!/(djo - k)! > d/(e - k + I)!, Thus

IG>. n CI(x)/lIG>.n gYg-l n CI(x)/


= 2i>.(i,\-l)(i>. - 2)... (i>.- k + l)/(e!/(e - k)!).

Hence for any g.E G we have

b>.(x,g)= IG,\n Cl(x)l/(l + IG,\n gYg-l n Cl(x)1)

$IG>.nCl(x)I/IG>.ngYg-1 nCl(x)1

< 2i>.(i>.- 1)(i>.- 2) .. . (i>. - k + 1)/ e(e - 1) . .. (e - k + 1)

< 2i~/(e - k + l)k = 2«e- k + l)/i>.)-k

< 2«1/n + 1)- «k - l)/i>.»-k .

as (l/n) - (r/i>.n) > (l/n + 1). Observe now that h(x,g) depends on i>.

rather than on ..\. As we have only finitely many i>. which do not satisfy

the conditions of our choice of ..\ above the theorem follows. 0

Acknowledgement

Most of this work was done when the second author was a guest of the

University of Alberta. He wishes to thank the Department of Mathematics

for its warm hospitality.

References

[1] C. W. Curtis and 1. Reiner, "Representation theory of finite groups andassociative algebras," John Wiley & Sons, New York-London, 1962.

[2] E. Formanek and J. Lawrence, The group algebra of the infinite 3ym-metric group, Israel J. Math. 23 (1976), 325-331.


[3} G. James and A. Kerber, "The representations of the symmetric group,"Addison-Wesley, London, 1981.

[4J G. de B. Robinson, "Representation theory of the symmetric group,"Edinburgh Univ. Press, 1961.

[5J H. Wielandt, Unendlich Permutationsgruppen. Vorlesungen, Tubin-gen, 1959-1960.

Received August 20, 1991

Revised September 21, 1992

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