Cycles and bipartite graph on conjugacy class of groups

Cycles and Bipartite Graph on Conjugacy Class of Groups

BIJANTAERI(*)

ABSTRACT- LetGbe a finite non abelian group andB(G) be the bipartite divisor graph of a finite group related to the conjugacy classes ofG. We prove thatB(G) is a cycle if and only ifB(G) is a cycle of length6andGASL2(q), whereAis abelian, andq2 f4;8g. We also prove that ifG=Z(G) simple, whereZ(G) is the center ofG, thenB(G) has no cycle of length4if and only ifGASL2(q), whereq2 f4;8g.

Introduction and results.

Assigning a graph to a group is one of the interesting tools in in- vestigating the structure of groups. One of the extensively studied graphs is the graph related to conjugacy class sizes. In fact various graphs related to conjugacy class sizes have been defined and studied. For a survey on the subject see [15]. The bipartite divisor graph is recently introduced for a set of arbitrary integers in [11]. The bipartite graph divisor for the set of conjugacy classes is the object of [3].

Let us introduce some notation. Let Gbe a finite groupxbe any ele- ment of G. We denote by x^Gˆ fx^gjg2Gg, wherex^gˆg ¹xg, the con- jugacy class ofxinGand by cs(G)ˆ fjx^Gj jx2Ggthe set of the conjugacy class sizes of G. Let cs(G)ˆcs(G)n f1g be the set of sizes of the non- central classes of G. Recall that the prime vertex graphD:ˆD(G) is the graph with vertex set V(D)ˆr(G)ˆ S

n2cs(G)p(n), where p(n) is the set of prime divisors of n, and edge set E(D)ˆ ffp;qg jpq divides some n2cs(G)g. The common divisor graph G:ˆG(G) is the graph with vertex set V(G)ˆcs(G), and two vertices n;m2cs(G) are joined if

(*) Indirizzo dell'A.: Department of Mathematical Sciences, Isfahan University of Technology, Isfahan 84156-83111, Iran.

E-mail: b.taeri@cc.iut.ac.ir

gcd(m;n)6ˆ1. The bipartite divisor graph B(G), studied in [3], is the graph with vertex set the disjoint unionr(G)[cs(G) and with edge set ffp;ng jp2r(G); n2cs(G) andpdividesng.

In [3] the authors, among other things, characterized finite groupsG such that B(G) is a path of length 5. In this paper we consider a similar problem and characterize finite groupsGsuch thatB(G) is cycle.

THEOREM1. Let G be a finite group such that B(G)is a cycle. Then GˆAS, where A is abelian, andSSL₂(q), qˆ4;8. Consequently B(G)is a cycle if andonly if B(G)is the6-cycle2 12 3 15 5 20 2 andGASL₂(4), or is the 6-cycle 2 72 3 63 7 56 2 and GASL2(8), where A is abelian.

In [6] the authors studied the groupsGsuch thatG(G) has no triangle.

In this paper we study the groups Gsuch that B(G) has no square. We prove that

THEOREM 2. Let G be a finite group such that G=Z(G), where Z(G) denotes the center of G, is simple. Then B(G)has no cycle of length4if and only if GˆAS, where A is abelian, andSSL₂(q), qˆ4;8.

Finite groups with cyclic bipartite divisor graphs.

We use the following lemma which contains some well known facts.

LEMMA1. Let G andH be groups.

(i) cs(GH)ˆ fabja2cs(G);b2cs(H)g.

(ii) If x;y2G commute andhave coprime orders, then C_G(xy)ˆ

ˆC_G(x)\C_G(y)andsojx^Gjandjy^Gjdividej(xy)^Gj.

(iii) If N is a normal subgroup of G andx2N, then jx^Nj divides jx^Gj. Also if y2G, thenj(yN)^G=Njdividesjy^Gj.

All of the cycles are 2-regular graphs and bipartite 2-regular graphs are cycles. IfB(G) is 2-regular then every noncentral conjugacy class has ex- actly 2 prime divisors. Thus we can use a result of Casolo.

LEMMA 2 (Proposition 1 of [4]). Let G be a non-solvable group with s(G)ˆ2, wheres(G)ˆmaxfjp(g^G)j:g2Gg. Then GˆAS, where A is abelian andS is isomorphic to either PSL₂(4)or PSL₂(8).

A non abelian groupGis anF-groupif for all non central elementsx;y, C_G(x)C_G(y) implies thatC_G(x)ˆC_G(y). The concept of F-group is es- sential in this paper. The structure of anF-groups is given by Rebmann.

LEMMA3 (See [17]). Let G be a non abelian group. Then G is an F-group if andonly if it is one of the following types:

(1) G has a normal abelian subgroup of prime index.

(2) G=Z(G)is a Frobenius group with Frobenius kernel L=Z(G)and Frobenius complement K=Z(G)with K andL abelian.

(3) G=Z(G)is a Frobenius group with Frobenius kernel L=Z(G)and Frobenius complement K=Z(G) with K abelian, Z(L)ˆZ(G), L=Z(G)has prime power order and L is an F-group.

(4) G=Z(G)S₄andif V=Z(G)is the Klein four-group in G=Z(G), then V is not abelian.

(5) GˆPA where P is an F-group of prime power order and A is abelian.

(6) G=Z(G)PSL₂(pⁿ) or PGL₂(pⁿ), G⁰SL₂(pⁿ), where G⁰ is the derived subgroup of G, p a prime, pⁿ>3.

(7) G=Z(G)PSL₂(9)(A₆)or PGL₂(9), andG⁰is isomorphic to the Schur cover of PSL2(9).

In what follows we find a characterization of finite groups with cyclic bipartite graphs.

PROOF OF THEOEREM 4. If G non-solvable then by Lemma 2, GASL2(q), where iˆ4;8, and A is abelian. If GASL2(4)

AA₅, then since cs(A₅)ˆ f12;15;20g, B(G) is the 6-cycle 2 12 3 15 5 20 2. If GASL₂(8), then since cs(SL₂(8))ˆ

ˆ f56;63;72g,B(G) is the 6-cycle 2 72 3 63 7 56 2.

Now we show that Gcannot be solvable. By the way of contradiction, suppose thatGis solvable andB(G) is a cycle. Thus, for all non-centralg2G, we havep(g^G)ˆ2. First note that by Corollary C of [9], if cs(G)ˆ fm;ng and gcd(m;n)6ˆ1, then eithermornis a prime power and thusB(G) cannot be a cycle of length 4. By [4, Theorem 3.7] we have jV(D(G))j 4. If jV(D(G))j ˆ2, then sinceB(G) is 2-regular,B(G) is a cycle of length 4, which is a contradiction. SojV(D(G))j 3. Also we know thatp(G=Z(G))ˆV(D(G)) (see for example [3, Corollary 7]), and so 3 jp(G=Z(G))j 4. Ifjcs(G)j ˆ2, thenB(G) is a cycle of length 4, which is impossible. Hencejcs(G)j 3.

We claim thatGis anF-group. First note that ifcˆp^aq^b2cs(G), with aandbpositive, then for all x2Gwithp2p(x^G) fp;qgwe must have jx^Gj ˆp^aq^b, sinceB(G) is 2-regular. Letx;ybe two elements ofGsuch that C_G(x)C_G(y). We have to proveC_G(x)ˆC_G(y). IfC_G(x)5C_G(y), thenjx^Gj andjy^Gjare distinct andjy^Gjdividesjx^Gj, which is impossible. HenceGis anF-group and so is one of the groups listed in Lemma 3. The groups listed in Lemma 3 (6) and (7) are non-solvable. ThusGis one of the group listed in (1)-(5).

Suppose (1) holds. ThenGhas an abelian normal subgroupNof prime indexp. We show that cs(G)ˆ fp;mg, for somem. Letxbe any non-central element ofG. Ifx2N, then sinceNC_G(x)5GandjG=Nj ˆp, we have jx^Gj ˆp. Ifx62N, thenGˆN xih and som:ˆ jx^Gj ˆ jN:CN(x)j. For any non-central elementy2G, there existsa2Nsuch thatyˆax. Therefore C_N(x)ˆC_N(y) and hencejy^Gj ˆm. Thus cs(G)ˆ fp;mg, a contradiction.

If (2) holds, thenjcs(G)j ˆ2, a contradiction. (Recall that ifGis quasi- Frobenius with abelian kernel and complement has exactly three con- jugacy class sizes, see for example [16, Lemma 4.4]).

Suppose (3) holds. Letx2KnZ(G). SinceC_G(x)ˆK we have jx^Gj ˆ

ˆ jG:Kj ˆ jL:L\Z(G)j, which is a prime power, a contradiction.

If (4) holds, thenp(G=Z(G))ˆp(S₄)ˆ f2;3g, a contradiction.

If (5) holds, then cs(G)ˆcs(P), a contradiction.

ThusGcannot be solvable. p

As mentioned above there is no group withB(G) a cycle of length 4. Let K_m;n be a complete bipartite graph with bipartite halves of sizesmandn.

SinceK_2;2is a cycle of length 4 the following question arises.

QUESTION1. Is there a group G with B(G)ˆKm;n?

Finite groups whose bipartite divisor graphs contain no cycle of length 4.

In this section we investigate the structure of groups such that B(G) contains no cycle of length 4. We say thatGsatisfies theone-prime power hypothesis, ifm;n2cs(G), then either gcd(m;n)ˆ1 or gcd(m;n) isa power of a prime. The terminology is similar to the one-prime hypothesis introduced by Lewis [15] on character degrees. One can see easily that the graphB(G) has no cycle of length 4 if and only if G satisfies the one-prime power hypothesis: SupposeB(G) has no cycle of length 4 and letm;n2cs(G). If

pqdivides gcd(m;n), thenp m q n pis a cycle of length 4 inB(G), a contradiction. Thus gcd(m;n)ˆp^a, for some primepand integera0.

Conversely if for allm;n2cs(G), gcd(m;n) is a prime power, thenGhas no cycle of length 4. In fact if p m q n pis a cycle of length 4 in B(G), thenpqdivides gcd(m;n) and so gcd(m;n) is not a prime power.

Suppose thatG=Z(G)S, whereSis a group such thatSˆS⁰. Ifxis a non-central element ofG, thenjx^Gj ˆca, wherecˆ j(xZ(G))^G=Z(G)jand a is a divisor ofjM(S)j(M(S) is the Schur multiplier ofS). To see this, let Dˆ fg2Gj[g;x]2Z(G)g, where [g;x]ˆg ¹g^x, b e the pre-image of C_G=Z(G)(xZ(G)) in G. Since C_G(x) is the kernel of the homomorphism D !G⁰\Z(G) with d7![d;x], it follows that jD=C_G(x)j divides jG⁰\Z(G)j. By [18, Theorem 9.18 (1) and (6)], jG⁰\Z(G)j divides jM(S)j and so a:ˆ jD=Cj divides jM(S)j. Now jx^Gj ˆ jG:C_G(x)j ˆ

ˆ jG=Z(G):D=Z(G)j jD=Z(G):C_G(x)=Z(G)j ˆca, as claimed.

By [10, Satz 25.7] we know that ifq6ˆ9 is odd, thenjM(PSL₂(q))j ˆ2;

if q6ˆ4 is a power of 2, then jM(PSL₂(q))j ˆ1; and jM(PSL₂(4))j ˆ2, jM(PSL2(9))j ˆ6. We can use these information in the proof of the fol- lowing lemma.

In the following lemma we also use the well known Burnside's p^a- Lemma which states that a finite group which has a conjugacy class of a prime power size is not simple.

LEMMA4. Let G be a finite non-solvable F-group. If G satisfies the one- prime power hypothesis, then G=Z(G)SL₂(4)or SL₂(8).

PROOF. Note that in Lemma 3 all groups satisfying (1)-(5) are solvable.

So a non-solvableF-group satisfying the one-power prime hypothesis sa- tisfies either (6) or (7). SupposeG=Z(G)PSL2(q) andqis odd. We want to obtain a contradiction. Recall that ifq1[mod 4] we have cs(G=Z(G))ˆ fq(q‡1);q(q 1);¹₂q(q‡1);¹₂(q 1)(q‡1)g and ifq3[mod 4] we have cs(G=Z(G))ˆ fq(q‡1);¹₂q(q‡1);¹₂(q 1)(q‡1);¹₂q(q 1)g. Suppose that we have the first case.

By above discussion, for anyx2GnZ(G) we havejx^Gj ˆ j(xZ(G))^G=Z(G)ja, wherea2 f1;2g. Note that, b y Burnside'sp^a-Lemma, there is no non-trivial conjugacy class of G=Z(G) of prime power size. Ifb₁ ˆca₁2cs(G) and b₂ˆca₂2cs(G), wherec2cs(G=Z(G)), thencdivides gcd(b₁;b₂) and so gcd(b₁;b₂) not a prime power. Thusb₁ˆb₂. Hence at most one conjugacy class size of G is a multiple of c, for every c2cs(G=Z(G)). So cs(G)ˆ fq(q‡1)a1;q(q 1)a2;¹₂q(q‡1)a3;¹₂(q 1)(q‡1)a4g, where a1, a₂, a₃, a₄2 f1;2g, and jcs(G)j 4. If b₁ˆq(q‡1)a₁2cs(G) and

b2ˆq(q 1)a22cs(G), are distinct, then 2q divides gcd(b1;b2) and so (b1;b2) not a prime power. Sob1ˆb2. Thus there exist at most one con- jugacy class size which is multiple of q(q‡1) and q(q 1). Thus cs(G)ˆ fq(q‡1)a₁;¹₂q(q‡1)a₃;¹₂(q 1)(q‡1)a₄gandjcs(G)j 3. Now considerb₁ˆq(q‡1)a₁2cs(G) andb₄ˆ¹₂(q 1)(q‡1)a₄2cs(G), then q‡1 divides gcd(b1;b2) and so (b1;b2) not a prime power (if q‡1 is a prime power, the sinceqis odd we have q‡1ˆ2^k. Since b y hypothesis qˆ4c‡1, we have 2^k 1ˆ4c‡1 and so 2^kˆ2(2c‡1), which is a con- tradiction). So b₁ˆb₄ and there exist at most one conjugacy class size which is multiple ofq(q‡1) and¹₂(q 1)(q‡1). Thus cs(G)ˆ fq(q‡1)a₁, q(q 1)a₂g and jcs(G)j 2 and G=Z(G) is solvable, which is a contra- diction. Similarly in the second case we can obtain a contradiction.

If G=Z(G)PGL2(q), where q is odd, then cs(G=Z(G))ˆ fq(q‡1), q(q 1);(q 1)(q‡1);¹₂q(q‡1);¹₂q(q 1)g. As above, at most one con- jugacy class size of G is a multiple of c, for every c2cs(G=Z(G)), and cs(G)ˆ fq(q‡1)a₁;q(q 1)a₂;(q 1)(q‡1)a₃;¹₂q(q‡1)a₄;¹₂q(q 1)a₅gand jcs(G)j 5. As above there exist at most one conjugacy class size which is multiple of q(q‡1) and q(q 1). Thus cs(G)ˆ fq(q‡1)a1, (q 1)(q‡1)a₃;¹₂q(q‡1)a₄;¹₂q(q 1)a₅g and jcs(G)j 4. Now either qˆ4k‡1 or qˆ4k‡3. Supposeqˆ4k‡1, then q‡1 is not a prime power. Since q‡1 divides gcd(b₁;b₃), where b₁ ˆq(q‡1)a₁, b₃ˆ

ˆ(q 1)(q‡1)a3 we have b1ˆb3 and cs(G)ˆ fq(q‡1)a1;¹₂q(q‡1)a4,

12q(q 1)a5g. Hence by the main theorem of [13], PGL2(q)G=Z(G)

PSL₂(2^m), a contradiction. Now suppose qˆ4k‡3. Then 2q divides gcd(b₄;b₅), where b₄ˆ¹₂q(q‡1)a₄, b₅ˆ¹₂q(q 1)a₃. Thus b₄ˆb₅ and cs(G)ˆ fq(q‡1)a₁;(q 1)(q‡1)a₃;¹₂q(q‡1)a₄g. Hence by the main theorem of [13], PGL2(q)G=Z(G)PSL2(2^m), a contradiction.

Finally suppose thatG=Z(G)PSL2(2ⁿ)ˆPGL2(2ⁿ). Then cs(G=Z(G))ˆ

ˆf(2ⁿ 1)(2ⁿ‡1);2ⁿ(2ⁿ 1);2ⁿ(2ⁿ‡1)g. Therefore cs(G)ˆf(2ⁿ 1) (2ⁿ‡1)a1, 2ⁿ(2ⁿ 1)a2;2ⁿ(2ⁿ‡1)a3g. Since 2ⁿ 1 divides gcd(b1;b2), where b1ˆ

ˆ(2ⁿ 1)(2ⁿ‡1)a1,b2ˆ2ⁿ(2ⁿ 1)a2and 2ⁿ‡1 divides gcd(b1;b3), where b3ˆ2ⁿ(2ⁿ‡1)a3, we conclude that 2ⁿ 1 and 2ⁿ‡1 are both prime power.

Thusnˆ2 ornˆ3. HenceG=Z(G)SL2(4) orSL2(8). p A generalization of Burnsidep^a-Lemma, which is due to Wielandt (see [1, Lemma 6]) states that ifxis ap-element,pa prime, ofp-power class size, thenx2O_p(G), whereO_p(G) is the largest normalp-subgroup ofG. Camina and Camina generalized this result and proved that all elements of prime power conjugacy class size are in the Fitting subgroup (see [5, Theorem 1]).

We use this result in the following lemma.

LEMMA5. Let G be a finite group satisfying the one-prime power hy- pothesis. If G=Z(G)has no solvable normal subgroup, then G is an F-group PROOF. Letx;y be two non central elements ofGsuch that C_G(x) C_G(y). We have to proveC_G(x)ˆC_G(y). IfC_G(x)5C_G(y), thenjx^Gjandjy^Gj are distinct andjy^Gjdividesjx^Gjand so gcd(jx^Gj;jy^Gj)ˆ jy^Gj. SinceGsa- tisfies he one-prime power hypothesis jy^Gj is a prime power. But since j(yZ(G))^G=Z(G)jdividesjy^Gjit follows thatj(yZ(G))^G=Z(G)jis also a prime power.

Thus, by [5, Theorem 1], the Fitting subgroup ofG=Z(G) is non-trivial, which is a contradiction. ThereforeC_G(x)ˆC_G(y) and soGis anF-group. p COROLLARY 1. Let Gbe a finite simple group. Then Gsatisfies the one-prime power hypothesis if andonly ifGSL2(4)orSL2(8).

The converse of Lemma 4 is not true. For example ifGˆSL₂(5), then G=Z(G)SL₂(4)(A₅) and cs(G)ˆ f12;20;30g. In the following lemma we characterize a finite group whose factor group over its center is the alternating group on 5 letters, in terms of its conjugacy class sizes of the group. As a corollary we find sufficient conditions for a group to satisfy the one-prime power hypothesis. It is well known that inA₅every element of order 3 or 5, is self centralizing, that isC_A5(x)ˆ hxi, for allx2A₅of order 3 or 5. Also ifxis any element of order 2 inA5, thenCA5(x) is a Sylow 2- subgroup. Thus if x2P is any element of Sylow p-subgroup P of A5, p2 f2;3;5g, thenC_G(x)ˆP.

We claim that if x is a non-central element of a group G such that jC_G(x)

Z(G)j ˆp², wherepis a prime, thenC_G(x) is abelian and thus there is no centralizer of any non-central element ofGstrictly contained inC_G(x). We use this fact in the proof of following lemma. To see the claim suppose, on the contrary that,Z(C_G(x))5C_G(x). ThenZ(C_G(x))

Z(G) 5C_G(x) Z(G) and jC_G(x):Z(C_G(x))j ˆ C_G(x)

Z(G) :Z(C_G(x)) Z(G)

ˆp:

ThusC_G(x) is abelian, contradicting toZ(C_G(x))5C_G(x). ThusZ(C_G(x))ˆ

ˆC_G(x) and soC_G(x) is abelian. Now letybe any non-central element ofG such that CG(y)CG(x). We want to show that CG(y)ˆCG(x). Let u2C_G(x). SinceC_G(x) is abelian, [u;v]ˆ1, for allv2C_G(x). In particular, since y2C_G(y)C_G(x), we have [u;y]ˆ1 and so u2C_G(y). Hence C_G(y)ˆC_G(x), as required. This completes the proof of the claim.

LEMMA 6. Let G be a finite group such that G=Z(G)A5. Then cs(G)ˆ f12;15;20g, G⁰A5 or cs(G)ˆ f12;20;30g, G⁰SL2(5).

Therefore G satisfies the one-prime power hypothesis if andonly if G⁰A₅.

PROOF. Let x be a non central element of G. Then since Z(G)5 5CG(x)5G, we have jG:Z(G)j ˆ jx^GkCG(x):Z(G)j and so 60ˆ

ˆ jx^Gj jC_G(x):Z(G)j. Since j(xZ(G))^G=Z(G)j divides jx^Gj, we have jx^Gj 2 2 f12a;15b;20cg, where a;b;c are positive integers. If jx^Gj ˆ12a, then 5ˆajC_G(x):Z(G)jand soaˆ1, alsoC_G(x) is abelian. Ifjx^Gj ˆ20c, then 3ˆcjC_G(x):Z(G)j and socˆ1, also C_G(x) is abelian. Ifjx^Gj ˆ15b, then 4ˆbjC_G(x):Z(G)jand sobˆ1 or 2, also in the casebˆ2,C_G(x) is abelian.

Thereforejx^Gj 2 f12;15b;20g, wherebˆ1 or 2. We distinguish two cases.

CASE1. Suppose that there existsx2Gsuch thatj(xZ(G))^G=Z(G)j ˆ15 withC_G(x)

Z(G) ˆC_G=Z(G)(xZ(G)). Therefore jx^Gj ˆ G

Z(G):CG(x) Z(G)

ˆ G

Z(G):C_G=Z(G)(xZ(G))

ˆ15:

We want to prove for all y2G with j(yZ(G))^G=Z(G)j ˆ15, we have C_G(y)

Z(G) ˆCG=Z(G)(yZ(G)) and sojy^Gj ˆ15.

First of all, we know that C_G=Z(G)(xZ(G)) is a Sylow 2-subgroup of G=Z(G) and, for every yZ(G)2C_G=Z(G)(xZ(G)), we haveC_G=Z(G)(xZ(G))ˆ C_G=Z(G)(yZ(G)). IfyZ(G)2C_G=Z(G)(xZ(G)), then C_G(y)

Z(G) C_G=Z(G)(yZ(G))ˆ C_G=Z(G)(xZ(G))ˆC_G(x)

Z(G). By the discussion in the paragraph preceding this Lemma, we have C_G(x)ˆC_G(y). Therefore

C_G(y)

Z(G) ˆC_G(x)

Z(G) ˆC_G=Z(G)(xZ(G))ˆC_G=Z(G)(yZ(G)):

Now suppose yZ(G)62C_G=Z(G)(xZ(G)). Then C_G=Z(G)(yZ(G)) is a Sylow 2- subgroup of G=Z(G) different from C_G=Z(G)(xZ(G)). Since A5 acts transi- tively, by conjugation, on the set of its Sylow 2-subgroups, there exists u2G such that u ¹xuZ(G)2C_G=Z(G)(yZ(G)). Hence C_G=Z(G)(yZ(G))ˆ

ˆC_G=Z(G)(u ¹xuZ(G)) and C_G(y)

Z(G) C_G=Z(G)(yZ(G))ˆC_G=Z(G)(xZ(G))^uˆ C_G(x) Z(G) _u

ˆC_G(x^u) Z(G) :

Hence, by the discussion in the paragraph preceding this Lemma, C_G(y)

Z(G) ˆC_G(x^u)

Z(G) and soC_G(y)

Z(G) ˆC_G=Z(G)(yZ(G)).

Thus we have proved that if j(xZ(G))^G=Z(G)j ˆ15 and C_G(x) Z(G) ˆ

ˆC_G=Z(G)(xZ(G)), then for all y2G with j(yZ(G))^G=Z(G)j ˆ15, we have C_G(y)

Z(G) ˆC_G=Z(G)(yZ(G)) and so jy^Gj ˆ15. Thus in this case cs(G)ˆ

ˆ f12;15;20g.

Note that ifjx^Gj ˆ12 or 20, thenC_G(x) is abelian. Also ifjx^Gj ˆ15 we saw thatC_G(x) is abelian. Hence there is no centralizer of any non-central element ofGstrictly contained inC_G(x). ThusGis anF-group. Therefore by Lemma 3, G=Z(G)PSL₂(4)A₅ and G⁰SL₂(4)A₅. Note that in this case G=Z(G)PSL2(5)A5 but we can not have G⁰SL2(5), since otherwise 30 divides a conjugacy class size of G.

CASE 2. Now suppose that C_G(x)

Z(G)5C_G=Z(G)(xZ(G)), for all for x2G with j(xZ(G))^G=Z(G)j ˆ15. Therefore

jx^Gj ˆ G

Z(G):C_G(x) Z(G)

ˆ j(xZ(G))^G=Z(G)j C_G=Z(G)(xZ(G)):C_G(x) Z(G)

and so jx^Gj ˆ30 and also C_G(x)

Z(G)

ˆ2. Thus in this case cs(G)ˆ

ˆ f12;20;30g.

Note that in the case C_G(x) is abelian, for all x2G. ThusGis an F- group and by Lemma 3,G=Z(G)PSL2(5)A5andG⁰SL2(5). p Now we investigate the structure a finite group satisfying the one- prime power hypothesis whose factor group over its center isSL₂(q).

LEMMA7. Let G be a finite group such that G=Z(G)S, where S is a simple group with trivial Schur multiplier. Then GZ(G)S. Therefore if G=Z(G)SL₂(2^m), where m3, then G satisfies the one-prime power hypothesis if andonly if GˆZ(G)SL₂(8).

PROOF. By [18, Theorem 9.18 (1) and (6)],jG⁰\Z(G)jdivides the order of the Schur multiplier of S. Therefore G⁰\Z(G)ˆ1. Now since G⁰Z(G)=Z(G) is a normal subgroup of a simple groupG=Z(G)S, it follows that GˆZ(G)G⁰Z(G)S. Now since, by [10, Satz 25.7], the Schur multiplier ofSL2(2^m),m3, is trivial, and SL2(2^m),m3, satisfies the

one-prime power hypothesis if and only ifmˆ3 (by the proof of Lemma 4)

the other assertion follows. p

Now we can prove Theorem 2.

PROOF OFTHEOREM2. Suppose that Gsatisfies the one-prime power hypothesis. Then, by Lemma 5, G is an F-groups and so, by Lemma 4, G=Z(G)SL₂(q), whereq2 f4;8g. Hence by Lemmas 6 and 7,G=Z(G)

SL₂(q)G⁰, whereq2 f4;8g. LetHˆZ(G)G⁰. ThenHˆZ(G)G⁰is a normal subgroup ofG. ThereforeG⁰H=Z(G) is a normal subgroup of G=Z(G) and soHˆG. HenceGˆAG⁰, whereAis an abelian subgroup ofG.

The converse is obvious. p

We can find some information of the socle of finite groups satisfying the one-prime power hypothesis. Recall that the socle of a finite group G is the subgroup generated by the minimal normal subgroups. It is well known that the socle of G is a direct product of minimal normal subgroups.

PROPOSITION 1. Let G be a finite group satisfying the one-prime power hypothesis. Then the socle of G is either abelian or is a direct product of a non abelian simple group and an abelian group.

PROOF. We claim that ifNˆST, whereSis a non abelian simple group, is a normal subgroup ofG, thenZ(T)6ˆ1. The results follows from this claim. To prove the claim supposeZ(T)ˆ1 and choose elementsx2S, y2T with coprime order. Then C_G(xy)ˆC_G(x)\C_G(y)C_G(x). If C_G(xy)ˆC_G(x), then TC_G(x)C_G(y) and so y2Z(T)ˆ1, a contra- diction. ThereforeC_G(xy)5C_G(x) and thusjx^Gjis a proper divisor ofj(xy)^Gj.

SinceGsatisfies the one-prime power hypothesis, gcd(jx^Gj;j(xy)^Gj)ˆ jx^Gj is a prime power. Sincejx^Sjdividesjx^Njwhich is a divisor ofjx^Gj, it follows thatjx^Sjis a prime power. Thus, by Burnsidep^a-Lemma,Sis not simple, a

contradiction. p

Now we consider finite groups satisfying the one-prime power hy- pothesis with simple socle, that is almost simple groups. First we need the following lemma.

LEMMA8. If N be a normal subgroup of G of index p, where p is a prime, then for all a2cs(N)either a2cs(G) or pa2cs(G). Thus if there exist a;b2cs(N)such that a5b,jp(gcd(a;b))j 2, andb6ˆpa, then G does not satisfy the one-prime power hypothesis.

PROOF. The first assertion is obvious. For the second assertion, sup- pose that qr divide gcd(a;b), where q and r are primes. By above, a;b2cs(G) ora;pb2cs(G) orpa;b2cs(G) orpa;pb2cs(G). In any caseqr divides two distinct conjugacy class sizes ofG. ThusGdoes not satisfy the

one-prime power hypothesis. p

In the following lemma we prove that ifG is an almost simple group whose socle isA_n, thenGdoes not satisfy the one-prime power hypothesis.

LEMMA9. The symmetric group on n letters, n5, does not satisfy the one-prime power hypothesis. In particular if A_nGAut(A_n), then G does not satisfy the one-prime power hypothesis.

PROOF. We have cs(S₅)ˆ f10;15;20;24;30gand hence S₅, do not sa- tisfy the one-prime power hypothesis. Suppose thatn6. It is obvious that

a:ˆ j(12)(34)^Snj ˆ1 2

n(n 1) 2

(n 2)(n 3) 2

andb:ˆ j(1234)^Snjˆn(n 1)(n 2)(n 3)

24 . Thusaˆ3band so gcd(a;b)ˆb.

Sincebˆ j(1234)^Anj, we have thatbis not a prime power, by Burnsidep^a- Lemma. HenceS_ndoes not satisfy the one-prime power hypothesis.

Now if n6ˆ6, then Aut(An)ˆSn and thus eitherGˆAn orGˆSn. Hence, by above,Gdoes not satisfy the one-prime power hypothesis.

The outer automorphism group of A6 is a Klein 4-group and if A₆GAut(A₆), then Gis one ofA₆,S₆,PGL₂(9),M₁₀, or Aut(A₆). In Atlas [8] notation, these are the groupsA₆,A₆:2₁,A₆:2₂,A₆:2₃, andS₆:2₂, respectively.

We haveA6PSL2(9) andAˆAut(A6)PGL2(9), wherePGL2(9)ˆ

ˆGL2(9)=Z(GL2(9)) andGL2(9) is the group of all the semi-linear trans- formation of the vector space of dimension 2 over the Galois field with q element. By [8], AˆPGL₂(9) has elements x and y of orders 2 and 5, respectively such that jC_A(x)j ˆ24 and jC_A(y)j ˆ5. Therefore jx^Aj ˆ

ˆ1440=24ˆ60 and jy^Aj ˆ1440=5ˆ288 and (jx^Aj;jy^Aj)ˆ12 is not a

prime power. Thus Aut(A6) does not satisfies the one-prime power hy- pothesis. Now, by above,S6and, by Corollary 1,A6;M10do not satisfy the one-prime power hypothesis. Finally cs(PGL₂(9))ˆ f36;45;72;80;90g and soPGL₂(9) does not satisfy the one-prime power hypothesis. p

Now we consider sporadic simple groups.

LEMMA10. If S is a sporadic simple group and SGAut(S), then G does not satisfy the one-prime power hypothesis.

PROOF. It is well known that the outer automorphism group of Sis either trivial or of order 2. IfSis one of the groupsM11,M23,M24,Co3,Co2, Co1,Fi23,Th,B,M,J1,Ly,Ru,J4, then the outer automorphism group ofS is trivial and so the result follows from Corollary 1. So suppose thatSis one of the groups M₁₂,M₂₂,J₂,Suz,HS,M^cL,He, Fi₂₂,Fi⁰₂₄,HN, O⁰N,J₃. Using Atlas, in Table 1, the groupSand the order of centralizers of two elementsx;y2S, such thataˆ jx^Sj5bˆ jy^Sjandb6ˆ2a, is given.

TABLE1.

S jSj jCS(x)j jCS(y)j

M12 2⁶3³511 11 6

M22 2⁷3²5711 11 7

J12 2⁷3³5²7 7 6

Suz 2¹³3⁷5²71113 13 11

HS 2⁹3²5³711 11 7

M^cL 2⁷3⁶5³711 11 5

He 2¹⁰3³5²7³11 17 14

Fi22 2¹⁷3⁹5²71113 14 13

Fi⁰₂₄ 2²¹3¹⁶5²7³1113172329 29 23

HN 2¹⁴3⁶5⁶71119 22 19

O⁰N 2⁹3⁴57³111931 31 19

J3 2⁷3⁵51719 19 17

Hence, by Lemma 8, G does not satisfy the one-prime power hypo-

thesis. p

Investigating the structure of a groupGsuch thatB(G) has no cycle of length 4 seems a difficult task. For instance for a finitep-groupG,G(G) is a complete graph and B(G) is a star graph, and there is no bound on the cardinality of the set of conjugacy class sizes of the group. In the case diam(G)ˆ3, the diameter ofG, (recall that, by [7], diam(G(G))3) we are able to find such bound using a result of Kazarin [14]. For non-solvable groups we make the following conjecture.

CONJECTURE1. IfGis a finite non-solvable group, thenB(G) has no cycle of length 4 if and only ifGˆAS,SSL₂(q), whereAis abelian, andSSL₂(q),qˆ4;8.

By [7] we know that diam(G(G))3. Recall also that, by [2, Theorem 1], n(G(G)), the number of the connected components of G(G) is not grater than 3. In the following we find some information on a group satisfying the one-prime power hypothesis such that diam(G(G))ˆ3.

PROPOSITION 2. Let G satisfy the one-prime power hypothesis and assumeG(G)is connectedwithdiam(G)ˆ3. ThenGis soluble andeither p(G)ˆ3,jcs(G)j ˆ6orp(G)ˆ4,jcs(G)j ˆ9.

PROOF. By [14],Gˆ(A0^jB0)C0whereA0;B0;C0are groups with pairwise coprime orders, A0 and B0 are abelian andA0B0=Z(A0B0) is a Frobenius group. Further, cs(G)ˆ fn;na;nbjn2cs(C₀)g, where aˆ

ˆ jA₀=(A₀\Z(A₀B₀))j and bˆ jB₀=(B₀\Z(A₀B₀))j. For all m2cs(C₀), withm>1, we have a patha ma m mb b. This implies thata;bare distinct prime powers and cs(C0)ˆ f1;u1;. . .;urg, whereui,iˆ1;. . .;r, are distinct prime powers other thana;b. SinceGis connected,r1. Since C0is a direct factor ofG, it satisfies the one-prime power hypothesis. Also n(G(C₀))2 and diam(G)ˆ3. Sor2. Ifrˆ1, then cs(C₀)ˆ f1;ugand so cs(G)ˆ f1;a;b;u;ua;ubg and jcs(G)j ˆ6, see Figure 1 (I). If rˆ2, then cs(C0)ˆ f1;u;vg and so cs(G)ˆ f1;a;b;u;v;ua;va;ub;vbg and

jcs(G)j ˆ9, see Figure 1 (II). p

Let GˆS₃E, whereE is an extra special group of order 5³. Then cs(G)ˆ f1;aˆ2;bˆ3;uˆ5;uaˆ10;ubˆ15g. Let H and K be non abelian groups of order 10 and 21, respectively, and put GˆHK.

Then cs(H)ˆ f1;2;5g, cs(K)ˆ f1;3;7g and cs(G)ˆ f1;aˆ3;bˆ7;uˆ2, vˆ5;uaˆ6;vaˆ15;ubˆ14;vbˆ35g. Therefore diam(G)ˆ3 and G satisfies the one-prime power hypothesis. We note that there exists a group

whose bipartite graph have diameter 3 and does not satisfy the one-prime power hypothesis. For instance, letGˆHK, whereHis a non abelian group of order 21 and K is a dihedral group of order 30. Then cs(H)ˆf1;3;7g, cs(K)ˆf1;2;15gand so cs(G)ˆ f1;2;3;6;7;14;15;45;105g and diam(G)ˆ3.

Acknowledgments. The author would like to express his gratitude to Silvio Dolfi for helpful comments. This work was done during the au- thor's sabbatical leave. He is grateful to the School of Mathematics and Statistics, University of Western Australia for their warm hospitality and facilities provided, in particular Cheryl Praeger, for numerous, in- valuable, conversations. The author thanks the Referee for valuable comments.

Fig. 1. ± GraphG(G) of a groupGsatisfying the one-prime power hypothesis with diameter3.

REFERENCES

[1] R. BAER,Group elements of prime power index. Trans. Amer. Math. Soc.,75 (1953), pp. 20-47.

[2] E. A. BERTRAM- M. HERZOG - A. MANN,On a graph relatedto conjugacy classes of groups, Bull. London Math. Soc.,22(1990), pp. 569-575.

[3] D. BUBBOLONI- S. DOLFI- M. A. IRANMANESH- C. E. PRAEGER,On bipartite divisor graphs for group conjugacy class sizes, J. Pure Appl. Algebra,213 (2009), pp. 1722-1734.

[4] C. CASOLO,Finite groups with small conjugacy classes. Manuscripta Math., 82(1994), pp. 171-189.

[5] A. R. CAMINA- R. D. CAMINA,Implications of conjugacy class size, J. Group Theory,1(1998), pp. 257-269.

[6] A. R. CAMINA- R. D. CAMINA,Coprime conjugacy class sizes, Asian-European J. Math.,2(2009), pp. 183-190.

[7] D. CHILLAG- M. HERZOG- A. MANN,On the diameter of a graph related to conjugacy classes of groups, Bull. London Math. Soc., 113, No. 2 (1993), pp. 255-262.

[8] J. H. CONWAY- R. T. CURTIS- S. P. NORTON- R. A. WILSON,ATLAS of Finite Groups, Oxford Univ. Press (Oxford, 1985).

[9] S. DOLFI- E. JABARA,The structure of finite groups of conjugate rank 2, Bull.

London Math. Soc. To appear.

[10] B. HUPPERT,Endliche Gruppen I, Springer-Verlag (Berlin, 1967).

[11] M. A. IRANMANESH - C. E. PRAEGER, Bipartite divisor graphs for integer subsets(submitted for publication).

[12] N. ITOÃ, On finite groups with given conjugate types II, Osaka J. Math.,7 (1970), pp. 231-251.

[13] N. ITOÃ,On Finite Groups with Given Conjugate Types. III, Math. Z.,117 (1970), pp. 267-271

[14] L. S. KAZARIN,On groups with isolatedconjugacy classes, Izv. Vyssh. Uchebn Zaved. Mat.,25(1981), pp. 40-45.

[15] M. L. LEWIS, Solvable groups having almost relatively prime distinct irreducible character degrees, J. Algebra,174(1995), pp. 197-216.

[16] M. FANG- P. ZHANG,Finite groups with graphs containing no triangles, J.

Algebra,264(2003), pp. 613-619.

[17] J. REBMANN,F-Gruppen, Arch. Math., Vol.XXII(1971), pp. 225-230.

[18] M. SUZUKI,Group Theory I, Springer-Verlag (New York, 1986).

Manoscritto pervenuto in redazione il 22 giugno 2009.