some engel conditions on finite subsets of certain groups

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HOUSTON JOURNAL OF MATHEMATICS© 2001 University of Houston

Volume 27, No.3, 2001

SOME ENGEL CONDITIONS ON FINITE SUBSETS OFCERTAIN GROUPS

ALIREZA ABDOLLAHI

Communicated by Bernhard H. Neumann

ABSTRACT.Let nand k be positive integers. We say that a group G satisfiesthe condition £(n) (respectively, £dn)) if and only if any set with n + 1elements of G contains two distinct elements x, y such that [x,t y] = 1 forsome positive integer t = t(x, y) (respectively, [X,k y] = 1). Here we studycertain groups satisfying these conditions. We prove that if G is a finitegroup satisfying the condition £(n), then G is nilpotent if n < 3 and G issoluble if n < 16. If G is a finitely generated soluble group satisfying thecondition £(2), then G is nilpotent. If k and n are positive integers and Gis a finitely generated residually finite group satisfying the condition £k(n),then G is nilpotent if n < 3 and G is polycyclic if n < 16. In particular,there is a positive integer c depending only on k such that G/ Zc (G) is finite,where Zc(G) is the (c + 1)-th term of the upper central series of G. Alsothese bounds cannot be improved.

1. INTRODUCTION

Paul Erdos posed the followingquestion [16]:Let G be an infinite group. If there is no infinite subset of G whose elements donot mutually commute, is there then a finite bound on the cardinality of eachsuch set of elements?The affirmative answer to this question was obtained by B. H. Neumann whoproved in [16] that a group is centre-by-finite if and only if every infinite subsetof the group contains two different commuting elements.Further questions of a similar nature, with slightly different aspects, have beenstudied by many people (see [1]-[3], [5]-[10], [12], [13]).

1991 Mathematics Subject Classification. 20F45j 20F99.Key words and phrases. Engel conditions, finite subsets.

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512 ALIREZA ABDOLLAHI

Let k be a positive integer, n a positive integer or infinity (denoted 00), Nk theclass of nilpotent groups of class at most k, F the class of finite groups and Nthe class of all nilpotent groups. We denote by (N,n) ((Nk,n), respectively)the class of all groups in which every subset of cardinality n + 1 contains twodistinct elements x, y such that (x, y) is nilpotent (nilpotent of class at most k,respectively), if n is infinity then n + 1 is so. We also denote by £k(n) (£(n),respectively) the class of all groups in which every subset of cardinality n + 1contains two distinct elements x, y such that [X,k y] = 1 ([x,t y] = 1 for somepositive integer t depending on x, y, respectively), where [x, y] = x-1y-1xy and[x,m+1 y] = [[x,m y], y] for all positive integers m. For a group G, we denote byZt(G) and ft(G), respectively, the (t + 1)-th term of the upper central series andthe t-th term of the lower central series of G.There are some results about the groups in the classes £ (00) and (N, 00), (£k (00)and (Nk, 00)) which show that such groups have the same behavior on certainclasses of groups such as the Class of finitely generated soluble groups. We havelisted some of these results in the following. Lennox and Wiegold in [12]showedthat if G is a finitely generated soluble group then G E (N, 00) if and only ifG is FN. Longobardi and Maj in [13] proved that if G is a finitely generatedsoluble group then G E £(00) if and only if G is FN. Therefore if G is a finitelygenerated soluble group then G E (N,oo) if and only if G E £(00). Delizia in [5]showed that if G is a finitely generated soluble group then G E (N2, 00) if andonly if G/Z2(G) is finite. Abdollahi in [2]proved that if G is a finitely generatedsoluble group then G E £2 (00) if and only if G/ Z2 (G) is finite. Thus if G is afinitely generated soluble group then G E (N2, 00) if and only if G E £2 (00). Also,Abdollahi in [1] proved that G E (N3,00) if and only if G E £3(00). Recently,Delizia, Rhemtulla and Smith in [7] have proved that if G is a residually finitegroup satisfying the condition (Nk, 00) then G is FN. In particular, there is apositive integer c, depending only on k such that G/Zc (G) is finite. We have beenunable to prove a result similar to that of [7]about the finitely generated resid-ually finite £k (00 )-groups. But we have obtained some results which are similarto that of [7]about the finitely generated residually finite £k (n)-groups, where nis a positive integer. Here we study certain groups satisfying the conditions £(n)and £k(n), where n < 00. Our main results are the following.

Theorem 1.1. Let n < 3 be a positive integer. Then every finite group satis-fying the condition £ (n) is nilpotent. Also, every finitely generated soluble group

ENGEL CONDITIONS FOR CERTAIN GROUPS 513

satisfying the condition £ (n) is nilpotent. The symmetric group of degree 3, 83,satisfies the condition £(3).

Recall that a group G is an Engel group (respectively, a k-Engel group) if foreach ordered pair (x, y) of elements in G there is a,positive integer m = m(x, y)such that [X,m y] = 1 (respectively, [X,k y] = 1). In 1936 Zorn [24] proved thatevery finite Engel group is nilpotent. Obviously, every Engel group satisfies thecondition £(n), for all positive integers n; therefore Theorem 1.1 says that theconverse is also true only for n < 3.

Theorem 1.2. Let n < 16 be a positive integer. Then every finite group whichsatisfies the condition £(n) is soluble. The alternating group of degree 5, A5satisfies the condition £(16).

Theorems 1.1 and 1.2 are similar to the results in [9]. In [9], Endimioni studiedfinite groups satisfying the condition (N, n) and proved that a group G satisfyingthe condition (N, n) is nilpotent, if n ~ 3, G is soluble, if n ~ 20 and these boundscannot be improved. Also about the groups in the class (N, 00), Tomkinson in[22], among other things, proved that if G is a finitely generated soluble group

4

satisfying (N, n) then IGj Z* (G) I ~ nn , where Z* (G) is the hypercentre of G.

Theorem 1.3. Let G be a finitely generated residually finite group satisfyingthe condition £k(n). Then G is finite-by-nilpotent. In particular, there exists apositive integer c depending only on k such that GjZc(G) is finite.

2. PROOFS

We use the following lemma in the proofs of some results; its proof is straight-forward and one can easily check it by some commutator calculations.

Lemma 2.1. Let G be a group and A a normal abelian subgroup of G. Then forall a, b E A, x E G and any positive integer k we have:i) [ax,k bx] = [a,k x]X ([b,k XJ-1t .ii) if [ax,k x] = 1 or [X,k ax] = 1 then [a,k x] = 1.iii) if [xa,k x] = 1 or [X,k xa] = 1 then [a,k x] = 1.iv) if [ax,k xa] = 1 or [xa,k xa] = 1 then [a,k+1 x] = 1.v) [ai'k x] = [a,k x]i for all integers i.vi) [a,k x] = axk (aXk-1 )-C(k,1) (aXk-2)C(k,2) ... a( _1)k = 1, where C(k, r) = r!(:~r)!

for 0 ~ r ~ k.

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PROOF OF THEOREM 1.1. Assume that the first part of Theorem 1.1 is falseand choose a counterexample G of smallest order. Thus G is a finite minimalnon-nilpotent group. By a result of of Schmidt (see Theorem 9.1.9 of [17])G is a{p, q}-group, where p, q are unequal primes and G has a normal Sylowp-subgroupP. Let Z be the centre of P. Thus Z is a non-trivial normal abelian subgroup ofG. Now,weprove that Z is a subgroup ofR(G), the set of all right Engel elementsof G. For this we must show that for all a E Z and x E G, there exists a positiveinteger k such that [a,k x] = 1. Suppose, for a contradiction, that there existselements a E Z and x E G such that for all positive integers k, [a,k x] :I 1. Thusthe elements ax, xa and x are pairwise distinct. Now, by the condition £(2) andLemma 2.1 (ii), (iii) and (iv), we get' a contradiction. Therefore Z ~ R(G). Nowby a result of Baer (see Theorem 12.3.7 of [17]) Z is a subgroup of Zm(G), forsome mEN. Hence Zm(G) is non-trivial and by the minimality of G, GjZm(G)is nilpotent and so G is nilpotent, a contradiction.By a result of Robinson and Wehrfritz (see Theorem 15.5.3 of [17]) and the firstpart of Theorem A, every finitely generated soluble group satisfying the condition£(2), is nilpotent.Since 53 has an abelian normal subgroup A of order 3, and 1531 = 6 thus everysubset of 53 which contains 4 elements, certainly contains an element x of A andso if y is an arbitrary element of the subset, we have [y, x, x] = 1. Therefore 53satisfies the condition £(3). 0

Lemma 2.2. Let G be a finite group satisfying the condition £(n) and let p be aprime greater than n. Then G is p-nilpotent.

PROOF. Suppose that G is a counterexampl~ of minimal order. Thus G is aminimal non p-nilpotent group and so by a result of Ito (see Theorem 10.3.3 of[17]), G is a minimal non-nilpotent group and

pG has a normal Sylowp-subgroup

P such that IG : PI is a power of a prime q :I p. Let Z be the centre of P,a E Z be non-trivial and x E G. Consider the elements x, ax, ... aP-1x. Sincen + 1 ~ p, by the condition £(n) there exist distinct integers i,j E {O, ... ,p - I}and a positive integer k such that [aix'k ajx] = 1. Thus by Lemma 2.1 (i) and(v), [a,k X]i-j = 1 and so [a,k x] = 1, since [a,k x] E Z and gcd(i - j,p) = 1.Hence a E R( G). Therefore Z ~ R( G), and we get a contradiction as in Theorem1.1. 0

Therefore we have:

o

------------------------------------------

ENGELCONDITIONSFORCERTAINGROUPS 515

Corollary 2.3. Let G be a finite group satisfying the condition &(n), then theset of all 7rn -elements of G forms a 7rn -group, where 7rn is the set of all primesp "5. n.

Theorem 2.4. Let G be a non-abelian finite simple group satisfying the condition2

&(n), (n < 00), then IGI < cn for some constant c.

PROOF. By Lemma 2.2, G has no prime divisor greater than n. Now Remark 5.5of [4] yields that IGI < cn2 for some constant c. D

Now, we are ready to prove Theorem 1.2.

PROOF OF THEOREM 1.2. Assume that the first part of the theorem is false andsuppose that G is a counterexample of minimal order. Thus G is a minimal sim-ple group and so by Thompson's classification of minimal simple groups [21], Gis isomorphic to one of the following simple groups:1) A5 the alternating group of degree 5,2) PSL(2, 2m), where m is an odd prime,3) PSL(2, 3m), where m is an odd prime,4) PSL(2,p), where 5 < p is prime and p == ±2 (mod 5),5) PSL(3, 3), and6) Sz(2P), where p is an odd prime.

We note that if a > 1 is an integer and m an odd prime then every primedivisor of am - 1 is of the form 2km + 1 (k E N) or is a divisor of a - 1, also forall prime powers q and positive integer n, IPSL(n, q)1 = (qn _ 1)(qn _ q) ... (qn _qn-2)qn-1 /gcd(n, q -1). Therefore, by Lemma 2.2, G is isomorphic to one of thefollowing groups:A5, PSL(2, 7), PSL(2, 8), PSL(2, 13), PSL(2, 27), Sz(8).Our task is to find a 16-element subset in each of the above groups, such that forall pairs {x, y} in these subsets and for all positive integers k, [X,k y] =I- 1. Thefollowing list of such subsets was found by computer search. For the projectivespecial linear groups, we represent by matrices in the corresponding special lineargroups which must then be reduced modulo the centre of the corresponding speciallinear groups. Also, in the following, GF(q) is the finite field with prime powercardinality q and z is always a generator of the cyclic group GF(q)\{O}.(a) A5;

{(3,4,5),(2,3,4), (2,3,5), (2,4,5),(1,2,3),(1,2,5),(1,3,4),(1,3,5),(1,4,2),(1,4,5),(1,2,4,3,5),(1,3,2,4,5),(1,5,3,2,4),(1,5,4,3,2),

o

516

(1,5,2,4,3),(1,2,4,5,3)}.

(b) PSL(2,7);

ALIREZA ABDOLLAHI

(c) PSL(2,8);

(e) PSL(3,3);

{[~ ; ~], [~ ~ ~], [~ ~ ~], [~ ~ ;],[; ~ ~], [~ ~ ~],zll OZZ OZZ OZZ OZZ OZZ

[~ ~ ~],[~ ~ ~],[~ ~ ~],[~ ; ~],[~ ~ ~],[~ ~ ~],OZZ Ozz 011 011 011 zOO

------------------------------_._~-

ENGEL CONDITIONS FOR CERTAIN GROUPS

[

1 Z 1

] [

1 Z Z

] [

1 Z 0

] [

z 1 1

]}zzz,1z0,Oz1,011.zOO zOO zOO z 1 0

In the following cases, we do not use a computer search.

(f) PSL(2,27)j

S17-II

One may consider the following set {[r~ ~] I x E GF(27) } , where r is an arbi-

trary fixed element of GF(27)\{0, 1}. We note that for all x,y E GF(27) and for

all kEN: [[r~ ~]'k[r~ ~]]=[(r-l-1~k(X-Y) ~].

(g) Sz(8)j

Sz(8) has a subgroup H = (a) (b) were (a) is a normal cyclic subgroup of H oforder 13, I (b) I = 4 and for all x E (a), xb = x8, (see Theorem 3.10 in Chapter XIof [11]). We consider the following 16-element subset of H:

{b-1 7 b-1 6 b-1 4 b2 2 b b b S b 7 b 12}a, a, a, a" a, ... , a , a , ... , a .

Note that for all kEN, t, i E {O, 1,2, ... , 12} and s, j E {4, 6, 7} we have:[b2a2'k bai] = a7k(2-9i), [bai'k b2a2] = a63k-l.7(9i-2), [b-1aj,k b2a2] = a63k-l.8(3j-l),[b2a2'k b-1aj] = aHk(1-3j), [b-1aj,k b-1aS] = a4k.(j-S), [bai'k bat] = a7k'9(i-t).Therefore, in any case, we get a contradiction, and so the proof of the first partis complete.

The alternating group of degree 5 satisfies the condition £(16). Suppose, for acontradiction, that M is a subset of As with 17 elements, such that for all pairsof elements {x, y} of M and positive integers k, [X'k y] f:. 1. If M has no elementof order 2, then M is covered by 16 Sylow p-subgroups where p E {3,5}, and sowe get a contradiction. Now, assume that M has an element a of order 2 and putM' := M\{a}. Thus a E Stab(i) for some i E {1, ... ,5}. If bE M'nStab(i), then[b, a, a] = 1, since a belongs to the unique normal abelian subgroup of Stab(i).Thus

M' C (~(StabU» \Stab(i») U C~ R,.)


--

where R1, ... , R6 are the Sylow 5-subgroups of G. On the other hand,

;~, (Stab(j)\Stab(i)) = C~ p,) uC~,Q,)for some Sylow 2-subgroups PI, ... ,P4 and Sylow 3-subgroups Ql, ... , Q6' ThusM' has 4 elements of order 2 and 6 elements of order 3. But every Pt is theunique abelian normal subgroup of Stab(j) for some j -I- i,1 :::;j :::;5. There-fore there exist elements c and d in M' of orders 2 and 3, respectively, such thatc, dE Stab(j) for some j and so [d, c, c]= 1, a contradiction. 0

For example we do the checking of the 16-element subset of PSL(2,13) inTheorem 1.2 by the followingGAP program [19].Note that in a finite group H with n elements for all a, bE H we have:

Vk E N [a,k b] -I- 1 ¢::::::> 1 ~ {[a, b], [a,2 b], ... , [a,n b]}.

E:=function(L) local o,z,i,j,r,k,c,b,n,I,M,h,V,W; n:=Length(L)jI:=IdentityMat(2,GF(13)); z:=Z(13)j 0:=O*Z(13)jh:=[[z-6,0],[0,z-6]]j i:=lj M:=Oj while i<=n-l and M=O do j:=i+=1jwhile j<=n and M=O do k:=l; c:=L[i]; b:=L[j] j W:=[]j whileComm(c,b)<> I and Comm(c,b)<> h and Size(W)=k-l doW:=Union(W,[Comm(c,b)])j c:=Comm(c,b)j k:=k+lj odj if Size(W)=k-lthen M:=1; fi;r:=lj c:=L[j] j b:=L[i]; V:=[];

while Comm(c,b)<> I and Comm(c,b)<> h and Size(W)<k-l andSize(V)=r-l do V:=Union(V, [Comm(c,b)]); c:=Comm(c,b)j r:=r+lj odjif Size(V)=r-l then M:=l; fi; j:=j+lj odj i:=i+l; odj if M=lthen Print("There exist two distinct elements x, y","\n", "in Land a positive integer k such that, [x,_k y]","\n"," belongs to the centre of SL(2,13)","\n"); elsePrint("For all x, y in L and for all positive", "\n","integers k: [x,_k y] does not belong to the", "\n", "centre ofSL(2,13)","\n"); fij end;; I:=IdentityMat(2,GF(13))j z:=Z(13)j0:=O*Z(13)j t:=z-O; a:=[[o,t], [0,0]] ;d:=[[o,o] ,[t,o]] jG:=Group(I+z*a,I+z-2*a,I+z-3*a,I+z-4*a,I+z-5*a,I+a,I+d,I+zA6*a,I+zA7*a,I+zA8*a,I+zAg*a,I+z-(10)*a,I+z-(11)*a~I+z-(12)*a,I+z-6*d,I+zA7*d,I+zA8*d,I+zA9*d,I+zA(10)*d,I+z-(11)*d,I+zA(12)*d,I+z*d,I+zA2*d,I+z-3*d,I+z-4*d,I+z-5*d); L:=[

---- -------------~------------- --,.~-_._----- --~- -

- '"

ENGEL CONDITIONS FOR CERTAIN GROUPS

[[z·8,z·4J,[0,z·4JJ,[[z·7,z·10J,[t,tJJ,[[z·8,oJ ,[z·3,z·4JJ,[[z·6,z·6J ,[t,oJJ ,[[z·3,z·5],[z·4,z·2JJ,[[z·4,oJ,[z·ll,z·8JJ,[[z·4,z·5J,[0,z·8JJ,[[z,z·3J, [z·2,z·lOJJ, [[z·7,z·6J, [z·4,tJJ,[[0,z·9J,[z·9,z·6JJ,[[0,z·3J,[z·3,z·6JJ,[[z·3,z·10J, [z·ll,z·2JJ, [[z·8,z·9J ,[0,z·4JJ,[[z·9,z·3J ,[t,z·IIJJ ,[[z·ll,z·10J,[z·9,z·7JJ, [[z·4,oJ ,[z·4,z·8JJJ ;

519

o

Now, we investigate finitely generated residually finite groups in Ek(n), wheren < 00. We want to use Theorem 4 of [23] which states that if G is a finitelygenerated residually finite group and N is a positive integer such that G has nosections isomorphic to groups EtwrLH, with H finite and cyclic, E an elementaryabelian group acted on faithfully and irreducibly by L, and IH : LI > N, then G isvirtually a soluble minimax group. For the definition of twisted wreath products,we refer readers to Neumann [15] and Meldrum [14].

Lemma 2.5. Let nand k be positive integers. Let A be a non-trivial abeliangroup, B a cyclic group, C a subgroup of B, and () a homomorphism from Cinto A ut(A). If the twisted wreath product W of A with B for (C, ()), satisfies thecondition Ek(n), then t = IB : CI ~ k + n.

PROOF. Let B = (b) so that a right transversI to C in B can be chosen tobe Y = {I, b, ... , bt-1}. Suppose, for a contradiction, that t > k + n. Fix anarbitrary non-trivial element a of A and define the functions fi from Y to A, fori = 0,1, ... ,71, as follows:

..{I if j '"i'v' J E {O, 1, ... , t - I} fi (fil) = ...a IfJ=z

Now, consider the following n + 1 elements: bfa, bh, ... , bfn. By the conditionEk (n), there exist two distinct integers i, j E {O, 1, ... , n} such that [bfi,k bfJ] = 1.Since, Ii, fJ E AY and AY is a normal abelian subgroup of W, [fdj-

1,k b] = 1, byLemma 2.1 (i). Therefore by Lemma 2.1 (vi)

h:= lk (lk-l)-C(k,I)(lk-2)C(k,2) ... g(_I)k = 1,

o


where g = fdj-1. If i > j, then for all l E {O, 1, ... , k}, 1 ~ i + l ~ n + k ~ t - 1,

and so bi+l E Y. Therefore

l' (bi) = g(bl+i) = fi(bl+i)(iJ(bl+i»-1 = Ji(bl+i) = {I ~f l ¥: 0 .a If l=O

Hence h(bi) = a( _1)k = 1, a contradiction. Similarly, if j > i, then h(b1) =a(-I)k+l = 1, a contradiction. This completes the proof. 0

Corollary 2.6. Let nand k be positive integers. If G is a finitely generatedresidually finite group satisfying the condition £k(n), then G is soluble-by-finite.

PROOF. It follows from Theorem 4 of [23] and Lemma 2.5. o

D

Lemma 2.7. Let G be a finitely generated nilpotent-by-finite £(oo)-group. ThenG is finite-by-nilpotent.

PROOF. Suppose, for a contradiction, that G ~ G II is not finite-by-nilpotent.Since G is finitely generated nilpotent-by-finite, G satisfies the maximal condi-tion on subgroups. Therefore there exists among the normal subgroups of Gone, N say, which is maximal subject to GIN is not finite-by-nilpotent. Nowthe group G = GIN is a finitely generated nilpotent-by-finite £k (00 )-group, andevery proper quotient group of G is finite-by-nilpotent. Since G is not finite-by-nilpotent, G is not finite. There exists a normal nilpotent subgroup H of finiteindex in G, since G is nilpotent-by-finite. Thus H is an infinite finitely generatednilpotent group and so H has an infinite free abelian characteristic subgroup A(see Exercise 9 of page 17 of [20]). Now we claim that A is a subset of the setof all right Engel elements G, R(G). Suppose that the claim is true, then A is asubgroup of the hypercentre of G by Theorem 12.3.7 of [17], so that in particularthe centre Z(G) is non-trivial. Thus GIZ(G) is finite-by-nilpotent. Now Theorem4.25 of [18] yields that G is finite-by-nilpotent, a contradiction.Now we prove the claim. Let 1 ¥: y and x be any elements of A and G respectively,and consider the infinite sequence

By hypothesis there exits two different positive integers m, n such that the re-peated commutator

[ynx'k ymx] = 1.

Since A is a normal abelian subgroup of G, [y,k x]n-m = 1. Thus [y,k x] = 1 sinceA is torsion-free. Hence y E R(G). 0

ENGEL CONDITIONS FOR CERTAIN GROUPS 521

Proof of Theorem 1.3. By Corollary 2.6, G is soluble-by-finite. Then thereexists a normal subgroup H of G such that H is soluble and G/ H is finite, so His also finitely generated. By Theorem 3 of [1], th.ere exists a positive integer cdepending only on k, such that H/Zc(H) is finite and so is G/Zc(H). Since Zc(H)is nilpotent, G is a finitely generated nilpotent-by-finite group. Therefore G isfinite-by-nilpotent, by Lemma 2.7. Thus there exists a finite normal subgroup Kof G such that G/ K is nilpotent. Since G/ K E £k (n) and G/ K is finitely gener-ated nilpotent, by Theorem 3 of [1],G/K/Zc(G/K) is finite. Then by Corollary2 to Theorem 4.21 of [18], rC+1(G/K) = rc+iG)K is finite, so rcH(G) is finite.Now it follows from Theorem 4.24 of [18],that G/Zc(G) is finite.

Corollary 2.8. Let G be a finitely generated residually finite group in £k(n). Ifn < 3 then G is nilpotent and if n < 16 then G is polycyclic.

PROOF. By Theorem 1.3, G is polycyclic-by-finite. If n < 16, then by Theorem1.2, G is polycyclic and so G is nilpotent by Theorem 1.1, if n < 3. 0

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Soc. 23 (1991), 239-248.[24] M. Zorn. Nilpotency of finite groups, Bull. Amer. Math. Soc. 42 (1936), 485-486.

Received March 3, 2000Revised version received July 18, 2000

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ISFAHAN, ISFAHAN 81744, IRAN.

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some engel conditions on finite subsets of certain groups

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