6-BFC groups

6-BFC Groups.

CLIFFDAVID(*) - JAMESWIEGOLD(**)

To Professor Guido Zappa on his 90th birthday

1. Introduction and preliminaries.

A group is said to be BFC if its conjugacy classes (of elements) are boundedly finite, andn-BFC if the largest conjugacy classes have order n.

B.H. Neumann proved in [4] that a group is BFC if and only if its derived group is finite; in [8], the second author showed that the derived group of ann-BFC group is of order bounded in terms of n. He formulated there the following conjecture, in whichl…n†stands for the number of prime factors ofn, multiplicities included.

CONJECTURE. For every n-BFC group G, the order of G⁰ is at most n^{12…1‡l…n††}.

There are nilpotent groups of class 2 and arbitrarily largenwhere this bound is achieved. Further, it is proved in [8] that the conjecture is true whennis prime and whennˆ4, in which caseG⁰is of order 4 or 8. There is a wide literature attacking this problem; the best bound achieved so far is that of Segal and Shalev [5], namely n^{12…13‡log2…n††}. Vaughan-Lee [6] estab- lished the conjecture for nilpotent groups. The smallest value of n for which the conjecture is not known to be true is 6, and the aim of this note is to rectify this by proving the following result.

THEOREM. Let G be a6-BFCgroup. Then G⁰is either C₆or Q₈. Throughout, notation is standard unless otherwise stated. For example, we writenˆb…G†for ann-BFC-groupG. It is easy to see [8] that the proof

(*) Indirizzo dell'A.: 68, Ty'n-y-Twr, Baglan, West Glamorgan, Wales, UK.

(**) Indirizzo dell'A.: School of Mathematics, Cardiff University, Senghennydd Road, Cardiff CF24 4AGWales, UK.

of the theorem reduces to one for finite groups, and henceforward we shall consider finite groups only.

We state some obvious facts here.

1.1. For any groups A and B,b…AB† ˆb…A†b…B†.

1.2. If N is normal subgroup of a group G and g an element of G, then the number of conjugates of gN in G=N divides the number of conjugates of g in G.

A useful tool in the proof of the Theorem is this. Letabe an element of a group Gwith exactlytconjugatesaˆa₁;a₂;. . .;a_t:Then the mapm…a†ofG into the symmetric groupS_tdefined by

gm…a† ˆ a1 a2 . . . at

a^g₁ a^g₂ . . . a^g_t

!

is a homomorphism ofGonto a transitive subgroup ofS_t. 1.3. The kernel ofm…a†isCore_GC…a†.

Note that both alternatives for G⁰ mentioned in the Theorem occur.

There are in fact four 6-BFC group of order 24, the smallest possible order.

Three of them have derived group cyclic of order 6, collected together as Examples 1.4; see Coxeter and Moser [1].

EXAMPLE1.4. The dihedral group of order24is6-BFCand its derived group is C6. The same is true of the dicyclic group of order 24and the group with the following presentation:

ha;b:a⁴ˆb⁶ˆ …ab†²ˆ …a ¹b†²ˆ1i:

EXAMPLE1.5. The binary tetrahedral group, with presentation ha;b;c:a²ˆb²ˆ ‰a;bŠ;c³ˆ1;a^c ˆb;b^c ˆa ¹b ¹i;

is6-BFCand has derived group Q₈:

Finally in this section, we show that the problem reduces to one in f2;3g-groups. Specifically, we have the following result.

LEMMA1.6. (1)Every6-BFCgroup is soluble.

(2)If G is a6-BFCgroup, then GˆAB, where A is af2;3g-group and Bis an abelianf2;3g⁰-group.

PROOF. The proof of (1) is almost immediate. If there is an insoluble 6- BFC group, some composition factorHof it is a nonabelian simple group withb…H† 6. By the remarks preceding 1.3, there is an isomorphism ofH onto a simple subgroup ofS₆;these areA₅andA₆, and they have conjugacy classes of order more than 6.

Now letGbe any 6-BFC group. All Sylowp-subgroups ofGwithp>5 are central. Otherwise, there is an element g of p-power order not com- muting with some elementaofG, and the conjugatesa^giofawith 0i<p are all distinct, too many for a 6-BFC group.

Less obvious but still easy is that the Sylow 5-subgroup is central. We shall show first that every elementxof 5-power order inGcommutes with all elements having 1, 2, 3, 4, or 6 conjugates. The first four are proved in a manner like that in the previous paragraph. Now suppose that y has 6 conjugates. Then, by 1.3 above,G=CoreC…y†is a transitive group of degree 6. Ifyfails to commute withx, then the elementxCoreC…y†is of order 5, and thusG=CoreC…y†is doubly transitive and therefore primitive; but it is soluble and thus imprimitive because 6 is not a prime-power. Thusxdoes indeed commute with all elements having 1, 2, 3, 4 or 6 conjugates. Ifxis not central, thenGmust be generated by elements having exactly 5 con- jugates, namely the elements outside the subgroup generated by elements having 1, 2, 3, 4 or 6 conjugates. Finally, suppose thatuis an element of 3- power order not commuting with an element v with 5 conjugates. Then G=CoreC…v†is a transitive group of degree 5 having the non-trivial ele- mentuCoreG…v†of order 3; thus it is eitherS5orA5and that is impossible becauseb…G† ˆ6. So every elementuof 3-power order does commute with every element with 5 conjugates; since these generateG, we have that the Sylow 3-subgroup P is central. This is the final contradiction, as then GˆPRwithRa 3⁰-group and so by 1.1,Gcannot have an element with 6 conjugates. Thus the elements of 5-power order are central, which is all that is needed to complete the proof of part (2) of the lemma. All Sylowp- subgroups withp>3 are central and the direct decomposition stated in (2) is obvious.

ThusG⁰ˆA⁰andb…G† ˆb…A†, so we have reduced the problem to one inf2;3g-groups.

2. Proof of the Theorem.

The proof goes by induction on the group order. The theorem holds for the smallest 6-BFC groups, namely those in Examples 1.4 and 1.5. So we

assume that Gis of order more than 24 and that the theorem holds for smaller groups. We divide the proof into several sections. Recall that we may assume thatGis af2;3g-group.

2.1. Suppose thatGis nilpotent. ThenGˆAB, whereAis a 2-group andBa 3-group. ThenAmust be 2-BFC andBmust be 3-BFC by 1.1, so thatA⁰is of order 2 andB⁰of order 3, andG⁰ˆA⁰B⁰is cyclic of order 6, as required.

2.2. Suppose thatGhasA4as a homomorphic image, and letNbe such that G=NA4. We shall show that NˆZ, the centre of G, and that G⁰Q₈. The conjugacy classes of A₄ are: the identity class, the three elements of order 2 in A⁰, and two classes consisting of four elements of order 3 outsideA⁰. The possible conjugacy class sizes ofGare 1, 2, 3, 4, 6;

by 1.2, elements with 1, 2, 3 or 6 conjugates map toA⁰₄modN. The set of elements ofGmapping toA⁰₄ isG⁰N, and the elements outside G⁰N have four conjugates each and map to elements with 4 conjugates each outside A⁰₄. TakeainG G⁰N. ThenjG:C…a†j ˆ4 andjG=N:C…aN†j ˆ4, which means thatC…a†containsN. ButGis generated by elements likea, so that N is central. As G=N is A4, this means that NˆZ, as claimed, and G=ZA4. Next, G⁰=…G⁰\Z† A⁰₄C2C2. Further, G⁰\Z is iso- morphic to a subgroup of the Schur multiplier ofA₄, that is, ofC₂. Indeed it must be of order 2, elseG⁰is of order 4, too small for a 6-BFC group. Thus G⁰is of order 8. LetHbe a stem-group [3] in the isoclinism class ofG. Then Z…H† H⁰,H⁰G⁰,G=Z…G† H=Z…H†by definition, and it is very easy to see thatb…G† ˆb…H†. AsHis not nilpotent, this means thatZis of order 2 andHof order 24. It is now an easy matter to check that the only group with the required properties is that in Example 1.5, and soG⁰ isQ₈, as claimed.

Thus we may now assume from now on:

2.3.G is not nilpotent and does not have A₄as a homomorphic image.

LEMMA2.3.1. G is not generated by the set of all elements with1, 2,or4 conjugates.

PROOF. Suppose the contrary, namely that G is so generated. The elements with 1 conjugate are central. If a has 2 conjugates, thenG⁰C…a†

asjG:C…a†j ˆ2. If a has 4 conjugates, thenG=CoreC…a†is a transitive group on 4 symbols that is at most 6-BFC. SinceA4is not a candidate, the only possibilities areD₈,C₄,C₂C₂. ThusG=CoreC…a†is nilpotent of class

at most 2 and so g₃…G† C…a†in all cases considered. But theng₃…G†is central andGis nilpotent, which is not allowed. This proves the lemma.

ThusGis generated by the set of all elements with 3 or 6 conjugates, and we prove:

LEMMA 2.3.2. The Sylow subgroups of G=Z are elementary abelian, and the derived group is a 3-group.

PROOF. As before, we may assusme thatZG⁰; that is, we replaceGif necessary by the stem-group of its isoclinism class. LetSbe the set of all elements with 3 or 6 conjugates. ThenZˆ T

g2SC…g† ˆ T

g2SCoreC…g†, so that G=Zis a subgroup of the direct product Dr

g2GG=CoreC…g†; so all we have to do is to establish that eachG=CoreC…g†has elementary abelian Sylow sub- groups and that the derived group is a 3-group.

When g has 3 conjugates, the result is clear since the choices for G=CoreC…g† are C3 and S3. When g has 6 conjugates, G=CoreC…g†is a soluble transitive group of degree 6 which is at most 6-BFC, and the only such groups are (abstractly)A₄,S₃,S₃C₂,C₂wrC₃,C₃wrC₂,C₆. But C₂wrC₃ maps toA₄, so in our case G=CoreC…g† must be one ofC₆,S₃, S3C2,C3wrC2and so it has Sylow subgroups of the required type.

To sum up:Gis af2;3g-group, not nilpotent, does not map toA4, the Sylow subgroups are elementary abelian, and…G=Z†⁰is a 3-group.

CASE1.Z6ˆ1. ThenG=Zis smaller thanGand is not nilpotent. Fur- ther, it not 2-BFC nor 4-BFC since the derived groups of such groups are 2-groups [8]. Further, it is not 6-BFC either, as if it were the induction hypothesis would give that …G=Z†⁰ is C6 or Q8, neither of which are 3- groups. Thus G=Z is 3-BFC and by [8] again, …G=Z†⁰ is of order 3. In particular,G⁰is abelian.

Suppose first that Zcontains an elementx of order 2. IfZˆ hxi, we have G⁰C6 as G⁰=ZC3, and all is well. If hxi 6ˆZ, then G⁰=hxiis of order more than 3 and so G=hxiis 6-BFC; it cannot be 3-BFC since its derived group is of order more than 3, and it cannot be 2-BFC nor 4-BFC as G⁰=hxiis not a 2-group. The induction hypothesis now applies to give thatG⁰hxiisC₆, since it cannot beQ₈asGis metabelian. ThusG⁰is of order at most 12. IfG⁰as order less that 12, it must have order 6 and so it isC6

sinceS3is not a derived group. Thus we may assume thatG⁰is of order 12, and we can writeG⁰ˆAZ, whereAis of order 3 and the centreZofGis of order 4. Then…G=A†⁰is of order 4, soG=Ais 4-BFC; it is also nilpotent of

class 2 since its derived group is ZA=A. The nilpotent residual of G is contained inA, and indeed it must beAsinceAis of order 3 andGis not nilpotent. By [2],Gsplits overA, sayGˆAU, whereA\Uˆ1. Note that UG=A, soUis 4-BFC andUis nilpotent of class 2. The Sylow 3-sub- groupXofUis abelian and central inU, being a direct factor; sinceAis normal and of order 3,XcentralizesA. ThusXis in the centre ofGand is therefore trivial since the centre is of order 4. SoUis a 2-group. By Lemma 2.3 of [7],Uis generated by elements with four conjugates; sinceAdoes not centralizeU, it must fail to centralize an elementuinUwith fourU- conjugates. Letabe a generator ofA. Since‰a;uŠis not 1, it must beasince a^u ˆa ¹. Thusudoes not commute with any element ofGof the formav, with v in U; that is, the G-centralizer of u is inUand thus it is the U- centralizer of u. This is a contradiction: jG:C_G…u†j ˆ jG:C_U…u†j ˆ

ˆ3jU:C_Uj ˆ12, which givesu12 conjugates, impossible asGis 6-BFC.

So still in Case 1, we may assume thatZis a 3 -group and thus thatG⁰is a 3-group sinceG⁰=Zis of order 3. We shall show thatZmust be of order 3.

If not, there is an element yof order 3 such that hyi 6ˆZ; as above, the factor-groupG=hyimust be 6-BFC and by induction this is a contradiction sinceG⁰is a 3-group. ThusZis of order 3 andG⁰of order 9.

IfG⁰is cyclic, sayG⁰ˆ hti, thenZˆ ht³i. For everyginG,t^Gˆt^mfor some integermand sot³ˆ …t³†^gˆt^3m, which means that 9j3…m 1†and 3j…m 1†, saymˆ3r‡1 and then‰t;gŠ ˆt ¹t^gˆt^3m2Z. Thus‰G⁰;GŠis central, a contradiction asGis not nilpotent. ThusG⁰is not cyclic. There are two possibilities for the nilpotent residual ofG. It is eitherG⁰ or a non-central subgroup of order 3 in G⁰. We deal with the two cases se- parately.

Suppose first that G⁰ is the nilpotent residual ofG. By [2] again, G splits overG⁰, sayGˆG⁰UwhereG⁰\Uˆ1 and thusUis abelian. We shall show thatGhas order at most 54 in these circumstances. The 6 non- central elements inG⁰can split into G-conjugacy classes only of the fol- lowing sizes: 6; 3, 3; 4, 2; 2, 2, 2. Suppose that G⁰ contains an elementx with just two conjugates. ThenjG:C…x†j ˆ2 and thusjUC…x†:C…x†jis at most 2, that 2, that is, jU:U\C…x†j is at most 2. But G⁰ˆ hZ;xi so C…G⁰† ˆC…x† and thus U\C…G⁰† has index at most 2 in U. But U is abelian andGis generated byUandG⁰, so thatU\C…G⁰†is central. But U\Zˆ1, soUhas order at most 2 andGhas order at most 18. Such a group cannot be 6-BFC since it has centre of order 3: all centralizers are bigger than the centre. If there is a conjugacy class of size 3, a similar argument shows thatGhas order 27, impossible as groups of that order cannot be 6-BFC. Thus we may assume that the 6 non-central elements in

G⁰form a conjugacy class, and an argument like the one above shows that Uhas order at most 6, and the only non-trivial case is where it has order 6 and soGhas order 9:6ˆ54. A rather fussy argument now completes the proof. Ifuis an element ofUof order 2, then its centralizer has index a 3- power and therefore index 3; soucentralizes a subgroupXof order 9. If the Sylow 3-subgroup P is abelian, this means that X is central, im- possible as Z has order 3. Otherwise P is one of the two non-abelian groups of order 27, and it is readily proved that it does not have an au- tomorphism of order 2 such that the splitting extension ofP by it pro- duces a group with the required properties.

Thus we may assume that the nilpotent residualVhas order 3, and thus G⁰ˆZV. Again by [2],GˆVU for some subgroupUwithV\Uˆ1.

ThenG⁰ˆV⁰U⁰‰V;UŠ ˆVU⁰ andU⁰ is of order 3, so Uis 3-BFC. Some elementuof Uwith 3 conjugates inUmust fail to commute with a gen- erator v of V. It follows as above that C_G…u† ˆC_U…u† and so jG:C_G…u†j ˆ9, false asGis 6-BFC. This completes Case 1.

CASE2.Zˆ1.

By Lemma 2.3.2,Ghas elementary abelian Sylow subgroups andG⁰is a 3-group. Let Pbe the Sylow 3-subgroup. ThenPˆG⁰Lfor some sub- groupL. By Maschke's theorem,Lcan be chosen to be normal inG. But thenLˆ1 sinceLis a normal subgroup missingG⁰and therefore central, and we have G⁰ˆP. Next, C…G⁰†is nilpotent and therefore of the form G⁰X, whereXis a 2-group characteristic inG⁰and therefore normal inG.

Since it missesG⁰, it too is trivial and soC…G⁰† ˆG⁰.

We claim that G⁰ contains a normal subgroup of order 3. LetM be a maximal subgroup ofGcontainingG⁰. ThenMis normal and thus of index 2. It is smaller thanGand therefore, by the induction hypothesis, it is not 6-BFC since its derived group is a 3-group. It is not 4-BFC nor 2-BFC, because such groups have 2-group derived groups. If it is 1-BFC, that is, abelian, thenMisG⁰because its 2-part is normal and therefore trivial. So whenMis abelian,GisG⁰hai, whereais of order 2. The centralizer ofahas 3-power index and so is of index 3; asG⁰is evidently of order more than 3 (at least 6 since Gis 6-BFC), this means thata centralizes a non-trivial subgroupYofG⁰. ButYis central since it centralizesaandG⁰, and this is a contradiction. ThusMmust be 3-BFC, its derived groupM⁰is the normal subgroup ofGof order 3 that we claimed exists. Note thatG=M⁰is smaller than G, and the by now familiar argument shows that it is 3-BFC, which means thatG⁰=M⁰is of order 3 andG⁰of order 9.

Further,M⁰is a direct factor ofG⁰, and so by Maschke again, there is a G-normal subgroup A such that G⁰ˆM⁰A. Further, G=C…G⁰† is an elementary 2-group; as a subgroup of Aut…C3C3†, it has order at most 4 andGhas order at 18 or 36 sinceC…G⁰† ˆG⁰. WhenGhas order 18, we have GˆG⁰haifor some elementaof order 2; as in the previous case,Ghas non- trivial centre and this is a contradiction. Suppose finally thatG⁰has order 36. We have that G⁰ is the direct product of two G-normal subgroups hai;hbi, of order 3. A Sylow 2-subgroup ofGis a four-grouphc;di. As above, c has centralizer of index 3 and must centralize a subgroup of order 3, generating the centre ofha;b;ciand therefore normal inG. Without loss, we may assume thatccentralizesa; sinceais not central, (conjugation by) dmust inverta. Sincecis not central, it must invertb. Ifdinvertsb, thend inverts everything inG⁰and so has centralizer of order 4, meaning that it has 9 conjugates, which is impossible. Thusdcentralizesband invertsa.

But thencdinvertsaandband so has too many conjugates. This completes the Case 2 and the theorem is proved.

The next lowest value of n for which the conjecture is not known is nˆ8. To confirm it in this case would be a much longer undertaking than that in this short note.

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Manoscritto pervenuto in redazione il 3 gennaio 2006.