Variations on a theorem by Alan Camina on conjugacy class sizes

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Journal of Algebra 296 (2006) 253–266

www.elsevier.com/locate/jalgebr

Variations on a theorem by Alan Caminaon conjugacy class sizes

Antonio Beltrána,∗, María José Felipeb

a Departamento de Matemáticas, Universidad Jaume I, 12071 Castellón, Spainb Departamento de Matemática Aplicada and IMPA-UPV, Universidad Politécnica de Valencia,

46022 Valencia, Spain

Received 1 January 2005

Available online 25 August 2005

Communicated by Paul Flavell

Abstract

Let G be a finite group. We extend Alan Camina’s theorem on conjugacy class sizes whichthat if the conjugacy class sizes ofG are exactly{1,pa, qb,paqb} for two primesp andq, thenG isnilpotent. If we assume thatG is solvable, we show that when the set of conjugacy class sizesG

is {1,m,n,mn} with m andn arbitrary positive integers such that(m,n) = 1, thenG is nilpotent andm = pa andn = qb for two primesp andq. 2005 Elsevier Inc. All rights reserved.

1. Introduction

We will assume that any group is finite. It is well known that there is a strong relbetween the structure of a group and the sizes of its conjugacy classes and there eeral results studying the solvability or the nilpotence of a group under some arithmconditions on its conjugacy class sizes. N. Itô shows in [12] that if the sizes of the cgacy classes of a groupG are{1,m}, thenG is nilpotent,m = pa for some primep andG = P × A, with P a Sylowp-subgroup ofG andA ⊆ Z(G). There exist other deepe

* Corresponding author.E-mail addresses:[email protected] (A. Beltrán), [email protected] (M.J. Felipe).

0021-8693/$ – see front matter 2005 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2005.06.031

254 A. Beltrán, M.J. Felipe / Journal of Algebra 296 (2006) 253–266

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results. For instance, in [13], Itô shows that if the conjugacy class sizes ofG are{1, n,m},thenG is solvable. D. Chillag and M. Herzog prove in [7] that if 4 does not divideconjugacy class size ofG, thenG is solvable. Later, A.R. Camina and R.D. Camina ga proof of that result independent of the Classification of Simple Finite Groups inOn the other hand, A.R. Camina proves in [6] that if the conjugacy class sizes ofG are{1,pa, qb,paqb}, with p andq two distinct primes, thenG is nilpotent. Notice that thehypotheses of Camina’s theorem imply the solvability ofG just by using Burnside’spaqb-theorem.

In the introduction of [6], Camina asserts that it seems extremely likely that a gwhose conjugacy class sizes satisfy the following property is solvable: Ifm and n arethe cardinals of two distinct conjugacy classes ofG with m � n, then eitherm dividesn

and (n/m,m) = 1, or (m,n) = 1 and there is a class of sizemn. One particular case othis property is when the set of such cardinals is exactly{1, n,m,nm} with (n,m) = 1,however it seems difficult to prove the solvability of such groups. In this paper, we pthe following.

Theorem A. Let G be a solvable group and suppose that the conjugacy class sizesG

are {1, n,m,nm} with (m,n) = 1. ThenG is nilpotent andn = pa andm = qb for somedistinct primesp andq.

In order to show Theorem A, we will first prove a particular case which is alsextension of Camina’s theorem. We also present a new proof of it, without making usome results due to I.M. Isaacs and D.S. Passman in [11] on primitive permutation gwhich appeared in the original proof. Such an extension is the following.

Theorem B. Let G be a solvable group and suppose that the conjugacy class sizesG

are {1,pa, n,pan} with (p,n) = 1 and a � 0. ThenG is nilpotent andn = qb for someprimeq.

Recently, there have appeared some papers analyzing thep-structure ofp-solvablegroups when some arithmetical conditions on the sizes of the conjugacy classesp′-elements are imposed (see, for instance, [2,3] or [14]). More precisely, in the proTheorems A and B we will use the main result of [3]. We believe that it is remarkablewe use these results related to local information of a group to obtain global informatithe structure of the group.

We will denote byxG the conjugacy class ofx in G and we call|xG| the index ofxin G. The rest of the notation is standard.

2. Preliminary results

We will need the following elementary results on conjugacy classes ofπ -elementswhereπ is an arbitrary set of primes.

Lemma 1. LetG be aπ -separable group.

A. Beltrán, M.J. Felipe / Journal of Algebra 296 (2006) 253–266 255

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(a) The conjugacy class size of anyπ ′-element ofG is a π -number if and only ifG hasabelian Hallπ ′-subgroups.

(b) The conjugacy class size of everyπ -element ofG is aπ -number if and only ifG = H ×K , whereH andK are a Hallπ -subgroup and aπ -complement ofG, respectively.

Proof. (a) is easy to prove by arguing on induction on the order ofG, and (b) is exactly[4, Lemma 8]. �

We stress that Lemma 1 implies that ifG is p-solvable andp does not divide anyconjugacy class size, thenG has a central Sylowp-subgroup. In fact, the hypothesisp-solvability is not needed, as the following result shows.

Lemma 2. Let G be a group. A primep does not divide any conjugacy class size ofG ifand only ifG has a central Sylowp-subgroup.

Proof. See, for instance, [9, Theorem 33.4].�We will use the following result due to N. Itô, which characterizes the structure of t

groups which possess only two conjugacy class sizes.

Theorem 3. Suppose that1 and m > 1 are the only lengths of conjugacy classes ogroupG. ThenG = P × A, whereP ∈ Sylp(G) andA is abelian. In particular, thenm isa power ofp.

Proof. See [9, Theorem 33.6].�The authors obtained in [3] the following generalization of Itô’s theorem forp-regular

conjugacy classes inp-solvable groups.

Theorem 4. Suppose thatG is a finitep-solvable group and that{1,m} are thep-regularconjugacy class sizes ofG. Thenm = paqb, with q a prime distinct fromp anda, b � 0.If b = 0 theG has abelianp-complement. Ifb �= 0, thenG = PQ×A, with P ∈ Sylp(G),Q ∈ Sylq(G) andA ⊆ Z(G). Furthermore, ifa = 0 thenG = P × Q × A.

Proof. This is exactly [3, Theorem A]. �We will also make use of the classic Thompson’sA × B-Lemma.

Theorem 5. Let AB be a finite group represented as a group of automorphismsp-group G with [A,B] = 1 = [A,CG(B)], B a p-group and A = Op(A). Then[A,G] = 1.

Proof. See, for instance, [1, 24.2].�We will prove the following result on conjugacy class sizes.


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Lemma 6. Suppose that the three smallest non-trivial indices of elements of a groupG area < b < c, with (a, b) = 1 and a2 < c. Then the set{g ∈ G: |gG| = 1 or a} is a normalsubgroup ofG.

Proof. Let C1,C2, . . . ,Cs be the distinct conjugacy classes ofG, and writeKi for theclass sum of the elements ofCi . It is well known that

KiKj =s∑

r=1

aijrKr

whereaijr is a non-negative integer for alli, j, r = 1, . . . , s. In addition, by [8, 87.4], forinstance, for 1� i, j, r � s there exists a non-negative integerl such that

aijr = |Cj ||Cr | l.

Now, assume thatCi andCj are two classes of sizea and notice that

a2 =s∑

r=1

aijr |Cr |.

Sincea2 < c, this shows that if|Cr | � c, thenaijr = 0. Moreover, if|Cr | = b thenaijr =al/b for somel � 0, so in particular,b dividesl anda dividesaijr . Thusab must divideaijr |Cr |, which forcesaijr = 0. From these facts we deduce that{g ∈ G: |gG| = 1 ora} isa (normal) subgroup ofG. �

Notice that if the solvability hypothesis of Theorem A is eliminated, then by Lemmafollows in the thesis of the theorem thatG is not simple. Finally, we will use the followinresult due to A. Camina.

Lemma 7. Let G be a group such thatpa is the highest power of the primep whichdivides the index of an element ofG. Assume that there is ap-element inG whose index ispreciselypa . ThenG has a normalp-complement.

Proof. This is [6, Theorem 1]. �

3. Proof of Theorem B

As we have pointed out in the introduction, we will also give a new proof of Camtheorem in the proof of Theorem B.

Proof of Theorem B. The proof has been divided into several steps.

Step 1. If G is p-nilpotent then the theorem is proved.


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Suppose thatG is p-nilpotent and letH be the normalp-complement ofG. For everyx ∈ H we have

|G :H |∣∣H :CH (x)∣∣ = ∣∣G :CG(x)

∣∣∣∣CG(x) :CH (x)∣∣.

If |xG| = 1 or pa , thenH ⊆ CG(x) and thus|xH | = 1. If |xG| = n or pan, then the aboveequality along with the fact that|xH | divides |xG| imply that |xH | = n. Therefore, anyconjugacy class ofp′-elements ofG has size 1 orn, and by Theorem 4, we have thn = pcqb for some primeq �= p. Since(n,p) = 1, thenn = qb and again by Theorem 4we conclude thatG is nilpotent, so the theorem is proved.

Step 2. We may assume that there are nop-elements of indexpa . Consequently, there exissomep′-element of indexpa .

If G has ap-element of indexpa then by Lemma 7,G is p-nilpotent and the theoremis proved by Step 1. In order to see the consequence in this step it is enough to consdecomposition of any element of indexpa as a product of ap-element by ap′-element.

Step 3. We may assume that there are nop′-elements of indexn. Consequently, there exip-elements of indexn.

Suppose thaty is a p′-element of indexn. Notice that the Sylowp-subgroups ofGcannot be central. Thus, we can choose some non-centralp-elementx ∈ CG(y) and thenCG(xy) = CG(x) ∩ CG(y) and|CG(y) :CG(x) ∩ CG(y)| must be equal to 1 orpa . Hence,any p-element ofCG(y) has index 1 orpa in CG(y). By Lemma 1(b), we can writCG(y) = Py ×Vy , with Py ∈ Sylp(G) andVy ap′-group. Now, chooseH ap-complementof G such thatVy ⊆ H . By Step 2, there exists somep′-element, sayt , of indexpa and upto conjugacy we may assume thatH ⊆ CG(t). Therefore,y ∈ Vy ⊆ CG(t), so t ∈ CG(y)

and thus,t ∈ Vy . In particular,Py ⊆ CG(t), contradicting the fact thatt has indexpa .The consequence in the statement follows as in the above step.

Step 4. If x is a p-element of indexpan, thenCG(x) = Px × Vx with Px a p-group andVx an abelianp′-group such thatVx �⊆ Z(G). If y is a p′-element of indexpan, thenCG(y) = Py × Vy with Py an abelianp-group such thatPy �⊆ Z(G) andVy a p′-group.

Let x be ap-element of indexpan and lety be anyp′-element ofCG(x). Notice thatCG(xy) = CG(x) ∩ CG(y) ⊆ CG(x) and sincepan is the largest class size ofG, thenCG(xy) = CG(x), soCG(x) ⊆ CG(y). This implies thaty ∈ Z(CG(x)), so we can writeCG(x) = Px × Vx with Px a p-group andVx an abelianp′-group. It remains to show thaVx cannot be central inG.

Suppose thatVx ⊆ Z(G), and notice that thenVx = Z(G)p′ and |G : Z(G)|p′ = n.Choosez a non-centralp-element, which must have indexn or pan by Step 2. In everycase, notice thatZ(G)p′ is ap-complement ofCG(z). This implies that if we choose annon-centralp′-elementw of G, then anyp-element ofCG(w) must be central inG.


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Thus Z(G)p is a Sylow p-subgroup ofCG(w). Sincew has indexpa or pan, then|G : Z(G)|p = pa . This yields

∣∣G : Z(G)∣∣ = ∣∣G : Z(G)

∣∣p

∣∣G : Z(G)∣∣p′ = pan,

which contradicts the existence inG of elements of indexpan. Thus, the first assertion othe step is proved.

The second part of this step can be proved by reasoning in a similar way with ap′-el-ement of indexpan.

Step 5. n = qb or n = qbrc for some primesq and r distinct fromp. Consequently, inthe first case we can assume thatG is a {p,q}-group, and in the second one thatG is a{p,q, r}-group.

By Step 2, we can choose ap′-element, sayy, of indexpa . Furthermore, if we considethe primary decomposition ofy as a product of elements of prime power order, it ismediate that we can assumey to be aq-element for some primeq �= p. Now if we take aq ′-elementw of CG(y), we haveCG(wy) = CG(w)∩CG(y) and|CG(y) :CG(w)∩CG(y)|must be 1 orn. This proves that anyq ′-element ofCG(y) has index 1 orn in CG(y), so wecan apply Theorem 4 to conclude thatn = qbrc, with b, c � 0 andr some prime distincfrom q andp (since(n,p) = 1). Therefore, the first assertion of this step follows. Tsecond assertion follows by applying Lemma 2.

Step 6. If pa > n, then the set

Lp := {x: x is p-element and

∣∣xG∣∣ = 1 or n

}

is an abelian normalp-subgroup ofG. If pa < n, then the set

Lp′ := {x: x is p′-element and|xG| = 1 or pa

}

is an abelian normalp′-subgroup ofG.

It is enough to apply Lemma 6 to obtain that ifpa > n then the setW := {x: |xG| = 1or n} is a normal subgroup ofG. Analogously, ifpa < n, then the setW ′ := {x: |xG| = 1or pa} is a normal subgroup ofG.

Now, if x is any element of indexn and factorizex = xpxp′ , with xp andxp′ a p-el-ement and ap′-element, respectively, it follows thatxp′ must be central by Step 2, whenx ∈ Lp ×Z(G)p′ . Therefore,W = Lp ×Z(G)p′ andLp is also a normalp-subgroup ofG.The argument forLp′ is similar.

Finally, we see for instance thatLp is abelian, as the argument forLp′ is the same. Ifwe take anyy ∈ Lp then|Lp :CLp(y)| divides(|Lp|, n) = 1. Consequently,Lp is abelian.

For the rest of the proof we fix the following notation. Ifpa < n, we define

Ls := {x: x is ans-element and

∣∣xG∣∣ = 1 orpa

}


t

e,

e

e,

for any primes dividing n. Notice thatLs is an abelian normal subgroup ofG by Step 6.Moreover, by Step 5, we haven = qb or n = qbrc for two primesq andr distinct fromp,so s ∈ {q, r}. We are going to distinguish three cases:pa > n, n = qb > pa and n =qbrc > pa with b, c > 0.

Step 7.

(7.1) If pa > n, thenLp is an abelian normal Sylowp-subgroup ofG.(7.2) If n = qb > pa , thenG is p-nilpotent and the theorem is proved.(7.3) If n = qbrc > pa with b, c > 0 and if Ls �⊆ Z(G) for somes ∈ {q, r} (observe tha

by Step2 such a primes must exist), thenLs is an abelian normal Sylows-subgroupof G.

(7.1) In order to prove thatLp is a Sylowp-subgroup ofG it is enough to show, bytaking into account Step 2, that there are nop-elements of indexpan. Suppose thatzis a p-element of indexpan and by Step 4, writeCG(z) = Pz × Vz, with Vz a non-central abelianp′-group andPz a p-group. If t ∈ Vz, it is clear thatCG(z) ⊆ CG(t), soin particularCLp(z) ⊆ CLp(t). By applying Theorem 5, we gett ∈ M := CG(Lp) andtherefore,Vz ⊆ M . On the other hand, by Step 3, we know thatt has indexpa or pan, so|CG(t) : CG(z)| must be equal to 1 orn. This proves thatLp ⊆ CG(z) and we concludethatLp centralizes everyp-element of indexpan. But on the other hand, anyp-element ofindexn trivially centralizesLp as it is abelian. Therefore, we conclude that anyp-elementof G lies in M , whence|G :M| is ap′-number. Furthermore, sinceLp ⊆ M ⊆ CG(k) foranyk non-central element ofLp, which has indexn, thenn must divide|G :M|. Now, ifwe consider the equality

|G :M||M :Vz| =∣∣G :CG(z)

∣∣∣∣CG(z) :Vz

∣∣,

then all the properties remarked above imply thatVz is ap-complement ofM .Let x be ap-element ofG, which we know lies inM . If x has index 1 orn, then it

certainly follows thatx ∈ Z(M). If x has indexpan, then by Step 4, we writeCG(x) =Px ×Vx with Vx a non-central abelianp′-group andPx ap-group. As we have seen abovVx is ap-complement ofM , and in particularVx ⊆ CM(x) and|M :CM(x)| is ap-number.Therefore, we have shown that the index of anyp-element ofM is ap-number. Thus, byapplying Lemma 1(b), we can factorM = P ×T , whereP ∈ Sylp(G) andT is ap′-group,which must be equal toVz. In particular,P is normal inG. But now, if we choose somnon-centraly ∈ Vz, thenP ⊆ CG(y), which contradicts Step 3.

(7.2) If n = qb > pa , we can argue as in case (7.1) to show thatLq is a normal Sylowq-subgroup ofG. But in this case we know thatG is a {p,q}-group by Step 5, soG isp-nilpotent and the theorem is proved by Step 1.

(7.3) In this case, by Step 5,G can be assumed to be a{p,q, r}-group. Moreover, wecan assume without loss of generality that the fixed primes of the statement is, for instancq and we will prove thatLq is a Sylowq-subgroup ofG.

To prove this, since we know thatLq is an abelian normal subgroup ofG, it is sufficientto show thatG does not possessq-elements of indexpan. Suppose thatw is such an


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h

element and write, by Step 4,CG(w) = Pw ×Vw , with Vw ap′-group andPw a non-centraabelianp-group. If u ∈ Pw, it is clear thatCG(w) ⊆ CG(u), so in particularCLq (w) ⊆CLq (u). By applying Theorem 5, we getu ∈ N := CG(Lq). Consequently,Lq centralizesanyp-element ofCG(w), that is,Pw ⊆ N . On the other hand, if we take some non-cenu ∈ Pw, then it has indexn or pan by Step 2, so|CG(u) :CG(w)| must be 1 orpa . Thisproves thatLq ⊆ CG(w) and hence,Lq centralizes everyq-element of indexpan. Butalso, anyq-element of indexpa trivially centralizesLq as it is abelian. Therefore, anyq-el-ement ofG lies in N , so in particular,N ⊆ CG(y) for anyy ∈ Lq andpa divides|G :N |.Now, if we consider the equality

|G :N ||N :Pw| = ∣∣G :CG(w)∣∣∣∣CG(w) :Pw

∣∣,

then all the above properties imply thatPw ∈ Sylp(N).We claim that the index inN of anyq-element (which lies inN ) is either 1 or a fixed

p′-number. Lety be aq-element ofG, which we know that has index 1,pa or pan. Ify has index 1 orpa , then certainlyy ∈ Z(N) and the claim is proved. Assume then thy has indexpan. As in the above paragraph, we can writeCG(y) = Py × Vy with Vy ap′-group andPy a non-central abelianp-group. However, we have seen thatPy ∈ Sylp(N),so in particular,Py ⊆ CN(y) and |N :CN(y)| is a p′-number. Lett be aq ′-element ofCG(y) and notice thatCG(yt) = CG(y) ∩ CG(t) ⊆ CG(y). Hence,CG(yt) = CG(y) andCG(y) ⊆ CG(t). Therefore,t ∈ Z(CG(y)) and we may writeCG(y) = Qy × Ty with Qy aq-group andTy an abelianq ′-group, which moreover cannot be central inG sincePy ⊆ Ty .In addition, if we choose any non-centralt ∈ Ty , we getCG(y) ⊆ CG(t). In particular,CLq (y) ⊆ CLq (t), and by Theorem 5, we obtain thatLq ⊆ CG(t), whenceTy ⊆ N . Sincewe know that anyq-element lies inN , we conclude thatCG(y) ⊆ N . Now the followingequality

|G :N |∣∣N :CG(y)∣∣ = pan,

together with the fact thatpa divides |G :N |, force |N :CG(y)| to be a fixedp′-number,m := pan/|G :N | for everyq-elementw of indexpan. Thus, the claim of this paragrapis proved.

Now, we will show that anyp-element ofN has also index 1 orm in N . Let x be a non-centralp-element ofN . Up to conjugacy, we can assume, for instance, thatx ∈ Pw wherePw ×Vw is the decomposition ofCG(w) andw is a fixedq-element of indexpan, given atthe beginning of this case. It is clear thatCG(w) ⊆ CG(x) and then|xG| = n or pan. Wealso know thatCG(w) ⊆ N by the above paragraph, soPw ⊆ CG(w) ⊆ CN(x). Also, asPw is a Sylowp-subgroup ofN , then|CN(x) :CG(w)| is ap′-number. If|xG| = n, thenthe following equalities

|G :N |∣∣N :CN(x)∣∣∣∣CN(x) :CG(w)

∣∣ = pan = ∣∣G :CG(x)∣∣∣∣CG(x) :CG(w)

∣∣

imply thatCG(w) = CN(x) and|N :CN(x)| = m. In the other case, that is, when|xG| =pan thenCG(w) = CN(x) and|N :CN(x)| = m, as we wanted to prove.


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Finally, we will show that any{p,q}-element ofN has also index 1 orm. Letx be a non-central{p,q}-element ofN and writex = xpxq , wherexp andxq are thep-part and theq-part ofx. We haveCG(x) = CG(xp) ∩ CG(xq) and we distinguish three possibilities fthe index ofxq in G. If xq is central inG, thenGG(x) = CG(xp) andCN(x) = CN(xp),so |xN | = |xN

p | = 1 or m according to the above paragraph. If|xGq | = pa thenxq ∈ Lq ,

so xq ∈ Z(N), so CN(x) = CN(xp) and again by the above paragraph we get|xN | = 1or m. Finally, if |xG

q | = pan, it follows thatCG(x) = CG(xq) andCN(x) = CN(xq), so|xN | = |xN

q | = 1 or m, since we have proved above that allq-elements inN also haveindex 1 orm in N .

Now we are able to apply Theorem 4 and obtain thatN = RQ × A, with R ∈ Sylr (N),Q ∈ Sylq(G) andA abelian. In particular, the non-centralp-elements ofN , which existbecausePw ⊆ N , have index not divisible byq, which is a contradiction with Step 2.

Step 8.

(8.1) If pa > n, then thep-complements ofG are abelian.(8.2) If n = qbrc > pa , with b, c > 0, then the Sylowp-subgroups ofG are abelian.

(8.1) LetH be ap-complement ofG and assume that it is not abelian. By Lemma 1and Step 3, there existp′-elements inH of index pan. Let w be any such element. BStep 4, we writeCG(w) = Pw × Vw with Pw an abelianp-group such thatPw �⊆ Z(G)

andVw a p′-group. We will prove thatVw is abelian too. We may choose a non-cenp-elementu ∈ CG(w), which certainly satisfiesCG(w) ⊆ CG(u). By (7.1), we know tha|uG| = n, so |CG(u) :CG(w)| = pa . Therefore,Vw is a p′-Hall subgroup ofCG(u). Onthe other hand, ifv is ap′-element ofCG(u), then|CG(u) :CG(uv)| = |CG(u) :CG(u) ∩CG(v)| is a power ofp. Thus, by Lemma 1(b),CG(u) has abelian Hallp′-subgroups. SoVw is abelian as we wanted to show and consequently,CG(w) is abelian too.

If Z(H) = Z(G)p′ , then there would not bep′-elements of indexpa , and this yieldsa contradiction with Step 2. Thus there exist non-central elements inZ(H). For any suchelement, sayy, note thaty ∈ CG(w) and asCG(w) is abelian, we haveCG(w) ⊆ CG(y) =CLp(y)H . Moreover, sinceLp � G, we haveCLp(y) � CG(y). SinceH ⊆ CG(y) andLp is abelian, it follows thatT := CLp(y) � G. Furthermore, as|CG(y) :CG(w)| = n, itfollows thatT is the Sylowp-subgroup ofCG(w), soT = Pw and in particular,T is notcentral inG. Notice that we have also proved thatT centralizes anyp′-element inH ofindexpan and any element inZ(H).

Now, if we takev ∈ H of indexpa , then there exists someg ∈ G such thatHg ⊆ CG(v),whencevg−1 ∈ Z(H). By the above paragraph,T ⊆ CG(vg−1

) and asT is normal inG,we get thatT also centralizesv. ThenT ⊆ CG(H) and asLp is abelian, we conclude thaT ⊆ Z(G), a contradiction.

(8.2) In this case, we know thatG is a {p,q, r}-group by Step 5. For one prime{q, r}, sayq, we can assume without loss thatLq is non-central inG, so by (7.3),Lq is anabelian normal Sylowq-subgroup ofG. If Lr is also non-central, thenLr is, again by (7.3)an abelian normal Sylowr-subgroup ofG. Consequently,Lp′ = Lq × Lr would be anabelian normalp-complement ofG, so by Step 1, the theorem is proved. Accordingly,


n.

is

h

e

may assume thatLr ⊆ Z(G), that is to assume that every non-centralr-element ofG hasindexpan.

Let P ∈ Sylp(G) and suppose thatP is not abelian. We will work to get a contradictioBy Lemma 1(a) and Step 2, there existp-elements of indexpan. Let z ∈ P be any suchelement. By Step 4, we writeCG(z) = Pz × Vz, whereVz is an abelianp′-group withVz �⊆ Z(G) andPz is ap-group. LetR0 be a Sylowr-subgroup ofCG(z) and letR0 ⊂ R

whereR is a Sylowr-subgroup ofG. If R0 is central, there can be nor-elements of indexpan as|R : R0| = rc. But thenR0 containsr-elements, sayw, of indexpan. Sincew ∈ Vz,thenCG(z) ⊆ CG(w), soCG(w) = CG(z). By applying Step 4, we conclude thatCG(z) isabelian. This is true for allz ∈ P of indexpan.

On the other hand, notice that there must existp-elements inZ(P )− Z(G)p , otherwisethere would not existp-elements of indexn, a contradiction with Step 3. For anyx ∈Z(P ) − Z(G)p and for anyz ∈ P of indexpan, we haveCG(z) ⊆ CG(x). SinceLq � G,then CLq (x) � CG(x) and as|CG(x) :CG(z)| = pa , it follows that T := CLq (x) is theSylow q-subgroup ofCG(z) for all z ∈ P of indexpan. Furthermore, we observe thatT

does not depend on the choice ofx.Now, let R be a Sylowr-subgroup ofG and note thatG = RPLq . Let y be a non-

central element ofR, which we know that has indexpan by the first paragraph (and thelement exists becauser dividesn). Again by Step 4, we writeCG(y) = Py × Vy , withPy an abelianp-group such thatPy �⊆ Z(G) andVy a p′-group. If we takek ∈ Py , thenCG(y) ⊆ CG(k). We distinguish two possibilities for the index ofk. If |kG| = pan, thenCG(y) = CG(k) and there existsg ∈ LqR such thatPy ⊆ P g . Also, by the above paragrapwe observe thatT g must be the Sylowq-subgroup ofCG(k) = CG(y). If |kG| = n, wemay chooseg ∈ LqR such thatPy ⊆ P g ⊆ CG(k). ThenCLq (k) is the Sylowq-subgroupof CG(k) and we know thatCLq (k) is the Sylowq-subgroup ofCG(u) for all u ∈ P g ofindexpan. Hence,CLq (k) = T g , for someg ∈ LqR. Since|CG(k) :CG(y)| = pa , we haveT g ⊆ CG(y). Therefore, we have proved that for any non-centraly ∈ R there exists somg ∈ LqR such thatT g is a Sylowq-subgroup ofCG(y). This yields

R ⊆⋃

g∈LqR

CLqR

(T g

)

and hence,

LqR =⋃

g∈LqR

CLqR

(T g

)Lq,

which implies thatLqR = CLqR(T )Lq and thenR ⊆ CG(T g) for someg ∈ LqR. We

defineH := Rg−1Lq and observe that sinceLq is abelian thenH ⊆ CG(T ).

We will prove now thatP ⊆ CG(T ). We have seen above thatT ⊆ CG(z) for all z ∈ P

of indexpan, so we only have to show thatT also centralizes any element inP of indexn.Let x ∈ P such that|xG| = n. Then there exists someg ∈ H such thatP g ⊆ CG(x),whence,xg−1 ∈ Z(P ). We know then thatCLq (x

g−1) is a Sylowq-subgroup ofCG(z) for

all z ∈ P of indexpan too, soT = CLp(xg−1). As g centralizesT , we obtainT = T g ⊆

CG(x). We conclude thatP ⊆ CG(T ), as required.


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The above paragraphs show thatT ⊆ Z(G). But now, if y is a non-central element oR of indexpan, then, as we have seen above, there exists someg ∈ G such thatT g is acentral Sylowq-subgroup ofCG(y). AsLq is non-central, we can take somev of indexpa

and we have thatCG(v) must contain some Sylowr-subgroup. This contradicts the aboassertion.

Step 9 (Conclusion). (9.1) Assume first thatpa > n. We claim that each prime divisors ofn satisfies|G : Z(G)|s = ns , so we will get|G : Z(G)|p′ = n. Suppose that this is proveLet z be an element of indexpan and writez = zpzp′ , with zp and zp′ the p-part andp′-part of z, respectively. Ifzp /∈ Z(G), then by Step 7,|zG

p | = n and Z(G)p′ is a Hallp′-subgroup ofCG(z), sozp′ ∈ Z(G), which is a contradiction sincez has indexpan. Ifzp ∈ Z(G), then|zG

p′ | = pan, sozp′ ∈ Z(G) and this is a contradiction too.We will prove the above claim. Lets be a prime divisor ofn and letS be a Sylow

s-subgroup ofG. By applying Brodkey’s theorem (see, for instance, [10, 5.28]) anding into account Step (8.1), we deduce that there exists somep-elementy ∈ Lp suchthat S ∩ Sy = Os(G). But notice that[Os(G),Lp] = 1 and as thep-complements ofGare abelian by (8.1), it follows thatOs(G) = Z(G)s . Furthermore,y cannot be centrain Lp, otherwiseS would be central inG contradicting the fact thats dividesn. Conse-quently,y must have indexn in G. If we choose ap-complementH of G with S ⊆ H , thenCG(y) = LpCH (y). If w is a s-element ofCH (y), thenw = y−1wy ∈ S ∩ Sy = Z(G)s .Thus, we deduce that everys-element ofCG(y) lies in Z(G) and hence,|G :CG(y)|s =|G : Z(G)|s = ns , as required.

(9.2) Suppose now thatpa < n = qbrc, with b, c > 0. Arguing the same as at thbeginning of (8.2), we can assume thatLq is non-central, and thus,Lq is an abeliannormal Sylowq-subgroup ofG. We can also assume thatLr ⊆ Z(G). Therefore, thereexists a non-centralr-element inG of index pan. Takey ∈ Lq of index pa and letwbe aq ′-element ofCG(y). Then |CG(y) :CG(yw)| = |CG(y) :Cg(y) ∩ CG(w)| is equalto 1 or n = qbrc. By applying Theorem 4, we obtain thatCG(y) = QR × A, whereQ

and R are q and r-Sylow subgroups ofG and A is an abelianp-subgroup. By (8.2)the Sylowp-subgroups ofG are abelian, so we haveA ⊆ Z(G) and as a consequencpa = |G : Z(G)|p . But if we takeg ∈ G an r-element of indexpan, then by Step 4CG(g) = Pg × Vg , with Pg a non-central abelianp-subgroup, and this is the final cotradiction. �

4. Proof of Theorem A

Proof of Theorem A. We will assume, for instance, thatm < n and we will denote byπthe set of primes dividingn. By Lemma 2, we can assume that the only primes divid|G| are the primes inπ and the prime divisors ofm.

Step 1. If G has a normal Hallπ -subgroup, then the theorem is proved.

Suppose thatH is a normal Hallπ -subgroup ofG. For everyx ∈ H we have

|G :H |∣∣H :CH (x)∣∣ = ∣∣G :CG(x)

∣∣∣∣CG(x) :CH (x)∣∣.


s

n

se

l

f

If |xG| = 1 or m, thenH ⊆ CG(x) and thus|xH | = 1. If |xG| = n or mn, then the aboveequality with the fact that|xH | divides |xG| imply that |xH | = 1 or m. Therefore, anyconjugacy class inH has size 1 orm, so by Theorem 3 we getm = pa for some primep.Then we can apply Theorem B to obtain thatG is nilpotent andn = qb, so the theorem iproved.

Step 2. We may assume that there are noπ -elements of indexn and that there are noπ ′-elements of indexm.

Suppose thatx is a π -element of indexn. By considering the primary decompositioof x we can assume without loss thatx is a p-element for some primep ∈ π . Now ify is a p′-element ofCG(x), then CG(xy) = CG(x) ∩ CG(y) ⊆ CG(x) and this forces|CG(x) :CG(x)∩CG(y)| = 1 orm. By applying Theorem 4, we obtain thatm = pcqb, butas(n,m) = 1, thenm = qb andG would be nilpotent by applying Theorem B. In this cawe have necessarilyn = pa for somea > 0.

The second assertion is also true since we can argue symmetrically withm andn.

Step 3. If x is a π -element of indexmn, thenCG(x) = Hx × Kx with Hx a π -groupandKx an abelianπ ′-group such thatKx �⊆ Z(G). Symmetrically, ify is a π ′-element ofindexmn, thenCG(y) = Hy × Ky with Hy an abelianπ -group such thatHy �⊆ Z(G) andKy a π ′-group.

This step follows arguing exactly as in Steps 4 and 5 of Theorem B.

Step 4. WriteLπ := {x: x is π -element and|xG| = 1 or m}. ThenLπ is an abelian normaπ -subgroup ofG.

By applying Lemma 6, we obtain that the setW := {x: |xG| = 1 orm} is a normalsubgroup ofG. Now, if x is any element of indexm and factorizex = xπxπ ′ , with xπ

andxπ ′ a π -element and aπ ′-element, respectively, it follows thatxπ ′ must be central byStep 2, whencex ∈ Lπ × Z(G)π ′ . Therefore,W = Lπ × Z(G)π ′ and consequently,Lπ isa normalπ -subgroup ofG.

Finally, if we take anyy ∈ Lπ then |Lπ :CLπ (y)| divides (|Lπ |,m) = 1, so Lπ isabelian.

Step 5. We may assume thatn = qbrc for some distinct primesq andr .

As a consequence of Step 2, we may choose aπ -element, sayx, of indexm. It is enoughto consider the decomposition of any element of indexm as a product of aπ -element bya π ′-element. In addition, if we consider the primary decomposition ofx as a product oelements of prime power order, we can assume without loss thatx is aq-element for someprimeq ∈ π . Now if we take aq ′-elementw ∈ CG(x), we haveCG(wx) = CG(w)∩CG(x)

and|CG(x) :CG(w) ∩ CG(x)| must be 1 orn. This proves that anyq ′-element ofCG(x)

has index 1 orn in CG(x), so we can apply Theorem 2 to conclude thatn = qbrc, with


is

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Theo-

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9

269.Math.

(1990)

cience,

998.

250.

b, c � 0, as wanted. Moreover, we can assumeb, c > 0 by Theorem B. Thus the stepproved.

We have seen that we can assumeπ = {q, r} and sinceLπ is abelian we can certainlwrite Lπ = Lq × Lr whereLq andLs are defined in the same way asLπ but for q andr-elements, respectively. Furthermore, as we know that there existπ -elements of indexm,we will assume without loss that one of these subgroups, sayLq , is non-central inG.

Step 6. Lq is a Sylowq-subgroup ofG.

This step can be proved by reasoning in the same way as in (7.3) of the proof ofrem B.

Step 7. G has abelian Hallπ ′-subgroups.

If we takeK a Hallπ ′-subgroup ofG, thenG = KRLq , with R ∈ Sylr (G) and one canprove, by following the same arguments as in (8.2) of the proof of Theorem B, thatK isabelian.

Step 8 (Conclusion). We can get a contradiction if we mimic the proof of (9.2) in Threm B. �

Acknowledgments

The authors thank the referee for all his/her comments and suggestions. This wopartially supported by grant Fundaciò Caixa Castellò P1 1A2003-06 and grant MTM06067-C02-02.

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Variations on a theorem by Alan Camina on conjugacy class sizes