Some results on the partitions of groups

Some Results on the Partitions of Groups

M. FARROKHID. G. (*)

ABSTRACT - We investigate some properties of partitions of groups. Some in- formation on the structure of nilpotent groups as well the embedding of groups with nontrivial partitions into a larger one are discussed. Moreover, we consider the structure of groups admitting a nontrivial partition with1,2or3subgroups (considering the whole group) with nontrivial partition and the relation between the number of subgroups with nontrivial partition and the number of primes dividing the order of the group. Finally we analyze the relation between the number of components of a nontrivial partition of a group and its order.

Introduction.

LetGbe a nontrivial group. A collectionPof nontrivial subgroups ofG is said to be apartitionofGif every nontrivial element ofGbelongs to a unique subgroup inP. IfjPj ˆ1, thenPis said to be the trivial partition.

The subgroups inP are the component of the partition.

The study of groups partitions was first considered by Miller in 1906 and have been continued by many author's such as Young, Hughes, Thompson, Khukhro, Isaacs, Herzer, Schulz, Pannone, Kontorovich, Wall etc. Baer [2,3], Kegel [10] and Suzuki [16] in 1960-1961 obtained the clas- sification of those finite groups, other thanp-groups, admitting a nontrivial partition. We refer the reader to [21] for a history of partitions.

THEOREM (The Classification Theorem). Let G be a finite group ad- mitting a nontrivial partition. Then G is isomorphic with exactly one of the following groups

(1) S₄;

(2) a p-group with H_p(G)6ˆG;

(*) Indirizzo dell'A.: Department of Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran.

E-mail: m.farrokhi.d.g@gmail.com

(3) a group of Hughes-Thompson type;

(4) a Frobenius groups;

(5) PSL(2;pⁿ)with pⁿ4;

(6) PGL(2;pⁿ)with pⁿ5and p odd;

(7) Sz(2^2n‡1),

where p is a prime and n is a natural number.

This paper is organized in four different sections investigating several problems concerning partitions of groups. In section 1 we obtain some results on the structure of nilpotent groups with partition, in particular we prove that ap-group of nilpotent class less thanp admitting a non- trivial partition has exponentp. In section 2, we use a group admitting a nontrivial partition with a component containing its derived subgroup to generate an infinite sequence of groups with the same property con- taining the first group as subgroup. This result can be applied to show that all finite groups admitting a nontrivial partition can be embedded in larger groups with nontrivial partitions. In section 3, we classify all finite groups with a nontrivial partition, which admit 1, 2 or 3 subgroups (considering the whole group) with a nontrivial partition. Also we give an upper bound for the number of subgroups admitting a nontrivial partition of a group with a nontrivial partition in terms of the order of group. Fi- nally, in section 4, we use the classification theorem of finite groups ad- mitting a nontrivial partition to show that if P is a partition of a finite groupG, then the greatest common divisor between the order ofGand the number jPj 1 is greater than one. Finally we show that a finite groupGwith nontrivial partition has a partition Psuch thatjPj 1 di- videsjGj.

1. Nilpotent groups with a nontrivial partition.

We start with some elementary lemmas, the first of which is well known (see for instance [2]).

LEMMA 1.1. Let G be a group and let P be a partition of G. If x;y2Gn f1gwith xyˆyx and either

(1) jxj 6ˆ jyj, or

(2) jxj ˆ jyjis a composite number, then x;y belong to the same component of G.

In order to prove Theorem 1.3 we need the following lemma of which we give a proof (see also [11])

LEMMA1.2. Let G be a group which admits a nontrivial partition. Then Z(G)is isomorphic with one of the following groups:

(1) trivial group;

(2) an elementary abelian p-group; or (3) a torsion-free group,

where p a is prime number.

PROOF. LetPbe a nontrivial partition ofGand supposeZ(G)6ˆ1. Let x;y2Z(G)n f1g and suppose that X and Y are components of P with x;y2X[Y. Letg2Gn(X[Y). Sincexgˆgxandygˆgywe have, by Lemma 1.1, that jxj ˆ jgj ˆ jyj is a prime or jxj ˆ jgj ˆ jyj is infinite.

Therefore Z(G) is either an elementary abelianp-group or a torsion-free

group. p

THEOREM1.3. If G is a nilpotent group admitting a nontrivial parti- tion, then either G is a p-group for some prime p, or G is a torsion-free group.

PROOF. Since G is nilpotent Z(G)6ˆ1 and so, by Lemma 1.2, either there exists a primepsuch thatZ(G) is an elementary abelianp-group or Z(G) is a torsion-free abelian group. IfZ(G) is torsion-free, then by a well- known result of Mal'cev [15, 5.2.19]Gis torsion-free too. Also ifZ(G) is an elementary abelianp-group, then by [15, 5.2.22]Gis ap-group. p Concerning thep-groups with small nilpotent classes we have the fol- lowing structural result, which will be used in sections 3 and 4. We refer the interested reader to [13, 20] for related topics. Recall that ifGis a group andpis a prime, then the Hughes subgroupH_p(G) is the subgroup ofG generated by all the elements of order different fromp. As in [12, p. 183] it can be proved that ap-group of nilpotent classnhas a nontrivial partition if and only ifHp(G)6ˆG.

THEOREM 1.4. Let G be a p-group of nilpotent class less than p ad- mitting a nontrivial partition. Then G is a p-group of exponent p.

PROOF. Let Gbe a minimal counterexample with respect to the nil- potent classnwhich is less thanp. Observe that 16ˆZ(G)6ˆGbecauseGis

not abelian by Lemma 1.2. Moreover we have thatHp(G)6ˆGby a previous observation andZ(G)Hp(G) by Lemma 1.1 and Lemma 1.2. Note also thatHp(G=Z(G))Hp(G)=Z(G). SoHp(G=Z(G))6ˆG=Z(G) andG=Z(G) has a nontrivial partition. Now the class ofG=Z(G) isn 1. Then, by minimality ofG, we have thatG=Z(G) has exponentpandx^p2Z(G), for everyx2G.

MoreoverHˆ hx;yiis finite for allx;y2G(see [15, 5.2.18]). Sincen5pwe have that His regular. So for any couplea;b2Hwe have 1ˆ[a^p;b]ˆ [a;b]^p(see [6, Satz 10.6]). Also, by the regularity ofHwe have thatV1(H) is the set of all the element of order p. Then H⁰ has exponent p. Since x^py^pˆ(xy)^pc(see [6, p. 322]) wherec2

1(H⁰), we have thatcˆ1 and so x^py^pˆ(xy)^p. Now since Gˆ hGnH_p(G)i the group G has exponent p,

which is a contradiction. p

2. Embedding of groups with a nontrivial partition into a larger one.

It is straightforward to show, by using Lemma 1.1, that a nontrivial direct product of finite groups has a nontrivial partition if and only if there exists a prime p such that each of the groups in the product, except probably one group, is ap-group of exponentpand the subgroup gener- ated by all elements of order6ˆpis proper. But the situation is not too clear for subgroups of direct products. Here we consider groupsGadmitting a nontrivial partition with a component containingG⁰to show that all direct products ofGhave subgroups containingGproperly as an isomorphic copy with the same property asG.

DEFINITION. LetXbe a subgroup ofGandnm. Then

G_m;n(G;X):ˆ f(g₁;. . .;g_m)2G^m:(g₁ g_n;. . .;g_{m n‡1} g_m)2X^{m n‡1}g LEMMA2.1. If XG and2nm, thenG_m;n(G;X)is a subgroup of G^mif and only if G⁰X.

PROOF. First assume thatGm;n(G;X) is a subgroup ofG^m. Ifx;y2G, then by the definition

(x ¹;x;1;z‚‚‚‚}|‚‚‚‚{ⁿ . . .^{2 times};1

;x ¹;x;1;z‚‚‚‚}|‚‚‚‚{ⁿ . . .^{2 times};1

;. . .)2Gm;n(G;X) and

(y ¹;y;1;z‚‚‚‚}|‚‚‚‚{ⁿ . . .^{2 times};1

;y ¹;y;1;z‚‚‚‚}|‚‚‚‚{ⁿ . . .^{2 times};1

;. . .)2G_m;n(G;X):

Hence

(x ¹y ¹;xy;1;z‚‚‚‚}|‚‚‚‚{ⁿ . . .^{2 times};1

;. . .)ˆ(x ¹;x;1;z‚‚‚‚}|‚‚‚‚{ⁿ . . .^{2 times};1

;. . .)(y ¹;y;1;z‚‚‚‚}|‚‚‚‚{ⁿ . . .^{2 times};1

;. . .) belongs toGm;n(G;X), whence

[x;y]ˆ(x ¹y ¹)(xy)1 12X:

HenceG⁰X.

Now suppose that G⁰X and (x1;. . .;xm);(y1;. . .;ym)2Gm;n(G;X).

Thenx_i‡1 x_i‡n;y_i‡1 y_i‡n 2Xso that

x_i‡1 x_i‡ny_i‡n¹ y_i‡1¹ 2X;

for eachiˆ0;1;. . .;m n. Now sinceG⁰Xwe conclude that (x_i‡1y_i‡1¹) (x_i‡ny_i‡n¹ )2X

that is

(x1;. . .;xm)(y1;. . .;ym) ¹ˆ(x1y₁¹;. . .;xmy_m¹)2Gm;n(G;X):

ThereforeGm;n(G;X) is a subgroup ofG^m. p Utilizing the above result we can prove the main result of this section.

THEOREM2.2. Let G be a group admitting a nontrivial partitionP. If P has a component X containing G⁰, then alsoGm;2(G;X)admits a non- trivial partition with X^mas a component, in whichG_m;2(G;X)⁰X^m.

PROOF. By the definition X^mG_m;2(G;X) and (g;g ¹;. . .)2 Gm;2(G;X). LetV be the set of all m-tuples (X1;. . .;Xm)2P^m such that X1;. . .;Xm6ˆXand

G_m;2(G;X)\(X₁ X_m)6ˆ1:

Then

G_m;2(G;X)ˆX^m[ [

(X1;...;Xm)2V

(G_m;2(G;X)\(X₁ X_m));

for if XiˆX for some i and (x1;. . .;xm)2Gm;2(G;X)\(X1 Xm), then

x₁x₂;. . .;x_{i 1}x_i;x_ix_i‡1;. . .;x_{m 1}x_m2X

and we have in turn

x_{i 1};x_i‡12X xi 2;xi‡22X

...;

hence we deduce that G_m;2(G;X)\(X₁ X_m)X^m. Now assume that

(x1;. . .;xm)2(Gm;2(G;X)\(X1 Xm))\(Gm;2(G;X)\(Y1 Ym));

in which (X₁;. . .;X_m);(Y₁;. . .;Y_m)2Vare distinctm-tuples. ThenX_i6ˆY_i for someiandxi2Xi\Yiˆ1. By the same reason which has been con- sidered before it follows that (x1;. . .;xm)2X^m that is (x1;. . .;xm)ˆ (1;. . .;1). Hence

P_mˆ fX^mg [ fG_m;2(G;X)\(X₁ X_m):(X₁;. . .;X_m)2Vg is a nontrivial partition of G_m;2(G;X). Moreover, G_m;2(G;X)⁰X^m as

G⁰X. The proof is complete. p

LEMMA 2.3. If G is a finite group and X is a subgroup of G, then jGm;2(G;X)j ˆ jXj^{m 1}jGj for each natural number m. In particular, for every finite group G admitting a nontrivial partition with a component X containing G⁰and each natural number m, there exists a group of order jGkXj^{m 1}, which admits a nontrivial partition.

PROOF. Let (g₁;. . .;g_m)2G_m;2(G;X). Then there exists x_i2X such that gigi‡1ˆxi, for each iˆ1;. . .;m 1. Hence every element of Gm;2(G;X) is of the form

(g₁;g₁¹x₁;g₂¹x₂;. . .;g_m¹₁x_{m 1});

where g₁2G, x_i2X and g_i‡1ˆg_i¹x_i, for each iˆ1;. . .;m 1. On the other hand, we observe that different choices of m-tuples

(g1;x1;. . .;xm 1)2GX^{m 1} produces different elements of Gm;2(G;X).

Therefore Gm;2(G;X) has exactly jGjjXj^{m 1} elements, as required. p REMARK. An easy observation shows that if we replace G_m;2(G;X) in Theorem 2.2 by

G1(G;X)ˆ f(g1;g2;. . .):gi2G;gigi‡12X;8i2Ng or G₂(G;X)ˆ f(. . .;g ₁;g₀;g₁;. . .):g_i2G;g_ig_i‡12X;8i2Zg;

then Theorem 2.2 still remains true and we can obtain infinite groups with a nontrivial partition.

PROPOSITION 2.4. Let m; n2, G be a finite group admitting a nontrivial partition with a component X containing G⁰ and let Xⁿ be the corresponding component ofG_n;2(G;X)containingG_n;2(G;X)⁰. Then

Gmn;2(G;X)Gm;2(Gn;2(G;X); Xⁿ):

PROOF. If ((g1;1;. . .;g1;n);. . .;(gm;1;. . .;gm;n))2Gm;2(Gn;2(G;X);Xⁿ), then we know from the definition thatg_i;jg_i‡1;jandg_i;jg_i;j‡12X, wheni‡1 in the former andj‡1 in the latter does not exceedmandn, respectively.

LetTbe anmngrid and letfbe a space-filling curve ofT, which is a bijective map fromAˆ f1;. . .;mngtoTwith the property that the image of every two consecutive members ofAare adjacent blanks ofT. It is evi- dent thatfinduces a map fromGm;2(Gn;2(G;X);X^m) toGmn;2(G;X), which sends an element ((g_1;1;. . .;g_1;n);. . .;(g_m;1;. . .;g_m;n)) to (g_f1;. . .;g_fmn).

Clearly, this map is an injective homomorphism. What remains is to show that this is indeed an epimorphism. For this let (g⁰₁;. . .;g⁰_mn)2G_mn;2(G;X) and putg⁰_iforf(i)-blank ofT. Now ifg⁰_i;g⁰_jare placed in two adjacent blanks ofT, then the path defined byfconnectingg⁰_iandg⁰_jhas odd length. From this and the fact that the product of each two consecutive elements on the path belongs to X, we conclude that g⁰_ig⁰_j belongs to X. Hence ((g_1;1;. . .;g_1;n);. . .;(g_m;1;. . .;g_m;n)), where g_i;j is the element located at (i;j)-blank, belongs toGm;2(Gn;2(G;X);X^m) and will be sent to (g⁰₁;. . .;g⁰_mn)

by the map. p

REMARK. According to the above statements, ifGis a group admitting a nontrivial partition with a componentXcontainingG⁰, then the groups

G2;2(G;X);G3;2(G;X);. . .

are the only groups which can be obtained in this way up to isomorphism.

Now we attempt to obtain the structure of all finite groupsGadmitting a nontrivial partition with a component containing G⁰. Recall that ifGis a finite group, thenGis a group ofHughes-Thompson type, ifGis not a p-group and that G6ˆH_p(G) for some prime divisor p of jGj.

PROPOSITION2.5. A finite groupGhas a nontrivial partition with a component containing G⁰if and only if Gis isomorphic with one of the following groups:

(1) a p-group with Hp(G)6ˆG;

(2) a group of Hughes-Thompson type; or (3) a Frobenius group with cyclic complement.

PROOF. It is easy to see that all of the groups in (1), (2) and (3) have the desired property. Thus we assume that Gis a finite group,P is a nontrivial partition of G and X is a component of P such that G⁰X. Let Y 6ˆX be a component of P and let x;y2Yn f1g such that x;y are conjugate in G. Then yˆx^g for some g2G and so x ¹yˆ[x;g]2X\Yˆ1 that is xˆy. Now if g2N_G(Y) and x2Y, then x^g2Y so that x^gˆx and consequently g2C_G(Y). Thus NG(Y)ˆCG(Y). IfYˆNG(Y) for someY2Pn fXg, thenY ˆCG(Y) is abelian and so Y\Y^gˆ1, for all g2GnY, since Y has no conjugate elements. Hence G is a Frobenius group with complement Y. By [15, 10.5.6], the Sylow p-subgroups of Y are cyclic, which implies that Y is cyclic. Now suppose that G is not a Frobenius group. Then for each componentY6ˆXofP, we haveCG(Y)ˆNG(Y)Y and so by Lemma 1.1, it follows that every component of P other than X is an elemen- tary abelian p-group for some prime p. Let Y;Z6ˆX be distinct components of P and let y2Y and z2Z be of orders p and q, re- spectively. Then jyG⁰j ˆp and jzG⁰j ˆq and if p6ˆq, then jyzG⁰j ˆpq so that pqkyzj. Hence yz2GnX is of composite order, which is im- possible. Therefore there exists a prime p such that each element of GnX has orderp so that Hp(G)X6ˆG. Consequently G is either a p-group with a nontrivial partition or a group of Hughes-Thompson

type, the desired conclusion. p

EXAMPLE. Letp;q be primes such thatqjp 1. Then there exists a non-cyclic groupGof orderpqsuch that its nontrivial subgroups form a nontrivial partition for G with a component XˆH_q(G) of order p con- tainingG⁰. LetMˆGZ^m_q. ThenH_q(M)6ˆMand soMhas a nontrivial partition. Moreover jHq(M)j ˆpq^m and Hq(M)M⁰. Hence, by Lemma 2.3, we can construct a group of orderp^n‡1q^mn‡m‡1, which admits a non- trivial partition for eachn0.

Utilizing the above results we obtain the following embedding theorem.

THEOREM2.6. Let G be a finite group admitting a nontrivial partition.

Then G can be embedded in a group of larger size with a nontrivial par- tition.

PROOF. By a theorem of Dickson [6, p. 214] and a theorem of Suzuki [17], each of the groupsPSL(2;pⁿ),PGL(2;pⁿ) andSz(2^2n‡1), wherepis a prime and nis a natural number can be embedded into a finite group of larger size with a nontrivial partition. Also, by Theorem 2.2 and Proposition 2.5, everyp-group with a nontrivial partition and every group of Hughes- Thompson type can be embedded in a finite group with larger size admit- ting a nontrivial partition. Now ifGis a Frobenius group with kernelNand complementH, thenGcan be embedded into a larger Frobenius group with kernel NN and complement H, where H acts on each component of

NNasHacts onN. p

3. Number of subgroups with a nontrivial partition.

IfGis an arbitrary finite group containing a subgroup with a nontrivial partition, then Ghas minimal subgroups with the same property. In this section, we attempt to describe the structure of these subgroups and to interrelate the number of subgroups with a nontrivial partition with the order of groupG, whenGadmits a nontrivial partition.

We first show the existence of finite groups with arbitrary number of subgroups admitting a nontrivial partition.

LEMMA3.1. For each natural number n, there exists a finite group G with exactly n subgroups admitting a nontrivial partition.

PROOF. Letqbe a prime number and letnbe a natural number. Using a famous theorem of Dirichlet on arithmetic progressions [1, Theorem 7.9], there exists a prime number psuch that qⁿjp 1. Now sinceU(Zp), the group of units ofZp under multiplication is cyclic, there exists a natural numberrsuch thatris of orderqⁿmodulop. Let

Gˆ hx;y:x^pˆy^qnˆ1;x^yˆx^ri:

We can see thatGis a Frobenius group with kernelK ˆ hxiand comple- ment Hˆ hyi and the subgroups KH_i, in which H_iˆ hy^qii, for

iˆ0;1;. . .;n 1 are distinguished as the subgroups ofG, which admit a

nontrivial partition, as required. p

Now we determine the structure of all finite groups admitting a non- trivial partition, which have one, two or three subgroups with a nontrivial partition, respectively. The first of which is the most important because it

gives a criterion for a group to have only trivial partition. We use this criterion in section 4 to show that the complements of a Frobenius group have only trivial partition.

PROPOSITION 3.2. Let G be a finite group admitting a nontrivial partition. Then the maximal subgroups ofGhave no nontrivial partitions if and only if Gis a non-cyclic group of order p² or pq, where p; qare distinct primes.

PROOF. Let G be a finite group admitting a nontrivial partition.

Clearly, if jGj ˆp² orpq, where p;q are distinct primes, then all of the maximal subgroups ofGhave only trivial partitions. Now assume thatGhas no maximal subgroups with a nontrivial partition and letPbe a nontrivial partition of G. IfM is a maximal subgroup ofG, then MX for some componentXofPand soMˆX, otherwisePinduces a nontrivial partition onM, which is a contradiction. On the other hand, ifXis a component ofP, thenXMfor some maximal subgroupMofG. So in conjunction with the previous result we deduce thatXˆM. HenceP is exactly the set of all maximal subgroups ofG. Suppose thatXˆN_G(X) for all componentsX. If X2P, thenG6ˆ [_g2GX^gand consequently there exists a componentY of P, which is not conjugate toX. Hence

jGj [

g2G

X^g[[

g2G

Y^g

ˆ[G:X](jXj 1)‡[G:Y](jYj 1)‡1

>jGj;

which is a contradiction. Thus there is a component X of P such that XN_G(X) so thatX3GsinceXis a maximal subgroup ofG. Ifg2GnXis of prime orderp, thenXhgi ˆGthat isjGj ˆpjXjand consequently every componentY 6ˆX of P has order p. If Xis a p-group of orderp^m, then mˆ1 sinceGis ap-group with maximal subgroups of orderp. Now sup- pose thatXis not ap-group andjXjhas a prime divisorq6ˆp. IfpkXj, then Sylow p-subgroups of G are of order at least p², which contradicts the maximality of components ofPdifferent fromX. LetQbe a Sylowq-sub- group ofX. ThenGˆN_G(Q)X. Asp does not divide the order ofX, we should havepkNG(Q)jso thatNG(Q)ˆG, forjNG(Q)jis divisible bypqbut no maximal subgroups ofGhas this property. HenceQ3Gand soGˆQY for all componentsY6ˆX, asQYdoes not lie in a maximal subgroup. Then it

followsXˆQ, i.e.,Xis aq-group. IfZ6ˆ1 is a characteristic subgroup ofX, thenZ3Gand similarlyGˆZYfor eachY2Pn fXg. So it follows that XˆZ. In particular,F(X) the Frattini subgroup ofXis trivial andXis an elementary abelianq-group. SinceXhas no nontrivial partitions we obtain jXj ˆqso thatjGj ˆpqand the proof is complete. p We observe that if the maximal subgroups of a finite groupGhave no nontrivial partition, then by Proposition 3.2, G has no subgroup with nontrivial partition. So we can deduce the following

COROLLARY3.3. A finite groupGwith nontrivial partition has only one subgroup with nontrivial partition if and only if Gis non-cyclic of orderp² orpq, wherepandqare different primes.

IfGis a group andPis a partition ofG, thenPis said to be amaximal partition, if each component ofP has only trivial partition. In fact, our maximal partitions are just the primitive partitions of Young [19]. But since the primitive partitions are the maximal elements in the set of all partitions ordered by, whereP P⁰if and only if every component ofP isP⁰-admissible, in what follows we shall use the name maximal partition instead of primitive partition. Clearly, maximal partitions have the max- imal cardinal among all partitions of a group. Note that, Young [19] and also Zorn's lemma guarantee the existence of maximal partitions for any group and the definition assures the uniqueness of maximal partition.

PROPOSITION 3.4. Let G be a finite group admitting a nontrivial partition. ThenGhas exactly two subgroups with a nontrivial partition if and only ifGis a Frobenius group with kernelKand complementH, in whichH; K satisfy the following properties:

(1) K is non-cyclic of order p², which is the normalizer of its nontrivial subgroups and H is cyclic of order q; or

(2) K is cyclic of order p and H is cyclic of order q², where p;q are distinct primes.

PROOF. LetGbe a group admitting a nontrivial partition and letMbe the unique proper subgroup ofG, which admits a nontrivial partition. Also letPandP⁰be the maximal partitions ofGandM, respectively and letAbe the set of all components ofPintersectingMnontrivially. By Proposition 3.2,Mis a maximal subgroup ofGand sinceM3^cGwe havejGj ˆrjMjfor

some primer. IfMXfor some componentXofP, thenMˆX, which is impossible asXhas only trivial partition. Then it follows thatP\Mis a nontrivial partition ofMand by Proposition 3.2, it is exactlyP⁰. In parti- cular, there exist primesp;qand componentsX;Y₁;. . .;Y_p2Psuch that

Mˆ(M\X)[[^p

iˆ1

(M\Y_i);

where jM\Xj ˆp,jM\Y₁j ˆ ˆ jM\Y_pj ˆqand M\X3M. First suppose thatpˆqˆr. ThenGis ap-group of orderp³of nilpotent class at most 2. Ifpis odd, then by Theorem 1.4,Gis of exponentpso that every maximal subgroup ofGadmits a nontrivial partition, which is impossible.

Also ifpˆ2, then asGcannot be of exponent 2 we should have GD8ˆ hx;y:x⁴ˆy²ˆ1;x^yˆx ¹i:

ButD8has two different subgroups

hx²;yi Z₂Z₂; hx²;yxi Z₂Z₂;

which have nontrivial partitions. Hence G is not p-group. Now we in- vestigating the possibilities thatpˆq orp6ˆq. First suppose thatpˆq that isjMj ˆp². We observe thatM is a completely reducibleF_p hxi- module. IfMis not irreducible then there is a submoduleNof orderpon whichhxiacts nontrivially. In this caseGhas at least two proper subgroups which have nontrivial partition, namelyMandNhxi. So the action ofhxion Mmust be irreducible. In particularGis a Frobenius group andrdivides p² 1 but it does not dividep 1.

Finally, assume thatp6ˆq. In this case M\Xis a characteristic sub- group ofMand hence normal inG. IfM\XX, thenX3Gand sincejGj is the product of three primes, we haveGˆXZfor each componentZ6ˆX ofP. In particular,GˆXY₁, which implies thatjZj ˆ jY₁j ˆqis of prime order for all componentsZ6ˆXofP. Clearly,Xis cyclic and ifjXj ˆpqor pr, wherer6ˆp;q, thenXhas a characteristic subgrouphxiof orderqorr, which is normal inGand hencehxiY1is a proper subgroup with a nontrivial partition of orderq²orqr, which is different fromM, a contradiction. Thus jXj ˆp²and sincejPj ˆp²‡1>p‡1ˆ jP⁰jthere exists a componentZ ofP such that (M\X)Zis a proper subgroup with a nontrivial partition different fromM, which is impossible. Therefore we must haveM\XˆX that isXM. Ifr6ˆq, then there exists an elementx2GnXof orderr such thatXhxiis a proper subgroup with a nontrivial partition different fromM. Hence every component ofP other thanXis a cyclicq-subgroup

and moreoverjGj ˆpq². If jPj>jP⁰jand Z2PnP⁰, then XV1(Z) is a proper subgroup with a nontrivial partition other than M, which is im- possible. ThusjPj ˆ jP⁰j. Now sincejZj q² for each component Z6ˆX we get

pq²ˆ jGj ˆ jXj ‡(jY1j ‡ ‡ jYpj) pp‡pq² pˆpq²; which is possible only ifjY1j ˆ ˆ jYpj ˆq². ThereforeGis a Frobenius group with cyclic kernelXof orderpand cyclic complementsY_iof orderq². Conversely, each of the groups in parts (1) and (2) satisfies the hypothesis

of the theorem. p

PROPOSITION 3.5. Let G be a finite group admitting a nontrivial partition. ThenGhas exactly three subgroups with a nontrivial partition if and only if eitherGD₈orGis a Frobenius group with kernelKand complementH, in whichH; Ksatisfy the following properties:

(1) K is non-cyclic of order p², H is cyclic of order q² and K is the normalizer of its nontrivial subgroups;

(2) K is cyclic of order p and H is cyclic of order q³. In this case, Gˆ hx;y:x^pˆy^q3ˆ1;x^yˆxⁱ;i^q26^p 1;i^{q3 p} 1i;or (3) K is cyclic of order p and H is cyclic of order qr. In this case,

Gˆ hx;y:x^pˆy^qrˆ1;x^yˆxⁱ;i^q6^p 1;i^r6^p 1;i^qrp 1i;

where p;q;r are distinct primes and15i5p.

PROOF. Let Gbe a finite group admitting a nontrivial partition and let M;N be the proper subgroups ofG, which admit a nontrivial parti- tion. Also let P, P⁰ and P⁰⁰ be the maximal partitions of G, M and N, respectively. SinceM;Nare the only proper subgroups ofGadmitting a nontrivial partition, eitherM;N are normal inGorM;N are conjugate.

First suppose thatM;N are conjugate and thatNˆM^g for someg2G.

Now sinceM;Nhave only two conjugates [G:N_G(M)]ˆ[G:N_G(N)]ˆ2 so that N_G(M);N_G(N)3Gand consequently

N_G(N)ˆN_G(M^g)ˆN_G(M)^gˆN_G(M):

This implies that M;NN_G(M)G, which is impossible by invoking Proposition 3.2. HenceM;Nare normal subgroups ofG. Now we consider two possibilities:

(i) One ofM;Nis contained in the other. Without loss of generality we may assume that MN so that N is a maximal subgroup of G and [G:N]ˆrfor some primer. Also, by Proposition 3.4,Nis of orderpq²and Mis of orderpqorq², for some distinct primesp;q. SinceNis a maximal subgroup ofGadmitting a nontrivial partition it does not lie in a component of P. Also we should have jPj ˆ jP\Nj, otherwise we can choose a componentX2Psuch thatN\Xˆ1. Then it follows thatMXis a proper subgroup with a nontrivial partition other thanM;N, which is impossible.

Suppose that MX for some component X of P. Thus X is a normal subgroup of G and its order is a product of three primes. Moreover, GˆXY for each component Y2Pn fXg. Hence there exists a primes such thatjYj ˆsfor each componentY2Pn fXgandjPj ˆ jXj ‡1. On the other hand,jP\Nj ˆ jMj ‡1 and sinceN\XˆM, each component ofP\N other thanMis of ordersandjNj ˆ jMjs. Then it follows that jPj>jP\Nj, which is impossible as it has been observed before. More- over, as forNwe haveM6ˆXfor eachX2P. Now invoking Proposition 3.4, we obtain

P\MˆP⁰andP\NˆP⁰⁰:

If r6ˆp;q, then G has an element x of order r and Mhxi is a proper subgroup with a nontrivial partition different from M;N, which is im- possible. Thusrˆporq. In this case, we have two possibilities by Pro- position 3.2. First assume thatjMj ˆpq. IfX⁰2P⁰ is the Sylow p-sub- group ofMas in Proposition 3.4, thenX⁰is a characteristic subgroup ofM and consequently X⁰3G. Also X⁰ˆX⁰⁰2P⁰⁰ and for each component Y⁰⁰2P⁰⁰n fX⁰⁰g, the subgroups M\Y⁰⁰ and Y⁰⁰ are of orders q and q², respectively. LetXbe a component ofPcontainingX⁰. IfpkZj, for some componentZ2Pn fXg, thenGhas a proper subgroup with a nontrivial partition of orderp², which is impossible. Hence the components of P other thanXareq-subgroups. AlsoX3GforX⁰3G, from whichXY⁰is a proper subgroup with a nontrivial partition different from N for each Y⁰2P⁰n fX⁰g. ThusXY⁰ˆM and we deduce thatXˆX⁰is of orderp.

Since N is a Frobenius group with complements Y⁰⁰6ˆX⁰⁰, the compo- nents ofP⁰⁰other thanX⁰⁰are conjugate pairwise, which induces the same property for components of P different from X. Consequently jYj ˆq³ for each Y2Pn fXg. If the component Y2Pn fXg is not cyclic and y2Y₁6ˆY\N is nontrivial, then y is of order q or q² and Mhyi is a proper subgroup with a nontrivial partition different fromM;N, which is impossible. Hence the components ofP other thanXare cyclic. There- foreGis a Frobenius group with kernelXof orderp and complements

Y 2Pn fXgof orderq³and we have

Gˆ hx;y:x^pˆy^q3ˆ1;x^yˆxⁱ;i^q2 6^p 1;i^q3p 1i;

for some 15i5p. In the second case, we havejMj ˆq², the components of P⁰are of orderqandjP⁰j ˆq‡1. From thisP⁰P⁰⁰, the components in P⁰⁰nP⁰are of orderpandjP⁰⁰j ˆq²‡q‡1. IfX⁰⁰2P⁰⁰nP⁰andqkXjfor some componentXofPcontainingX⁰⁰, thenXhas an elementxof orderq andMhxiis a proper subgroup with a nontrivial partition of orderq³, which is impossible. Moreover, since the components of P⁰⁰nP⁰ are pairwise conjugate, the same property is valid for the components ofP, which in- tersectMtrivially. Hence these components arep-subgroups of the same order. If there is a componentX2Psuch thatX\M6ˆ1 andq²kXj, then Xhas a subgroupHof orderq²possessingM\X, which implies thatHMis a proper subgroup with a nontrivial partition of orderq³, a contradiction. On the other hand, if there is a component X2P such thatX\M6ˆ1 and pkXj, thenXhas an elementxof orderp, which does not belong toNand thusMhxiis a proper subgroup with a nontrivial partition different from M;N, which is impossible. HenceP⁰Pand the components inPnP⁰are cyclic of orderp². ThereforeGis a Frobenius group with non-cyclic kernel Mof orderq²and cyclic complementsY 2PnP⁰of orderp². Moreover,M is the normalizer of its nontrivial subgroups andp²dividesq² 1 but it does not divideq 1.

(ii) None of the M;N is contained in the other. In this case, both of M;N are maximal subgroups of G. Thus the components ofP are cyclic subgroups of orders, which are the products of at most two primes. Also M\N6ˆ1 is a subgroup of order a primep. LetM\Nˆ hxiandXbe the component ofPcontainingx. Thenhxi3Gand consequentlyX3G. IfGis a p-group, then jGj ˆp³ andGhas nilpotent class at most 2. Gis not of exponent p, otherwise every nontrivial element ofGlies in a proper sub- group with a nontrivial partition so thatGˆM[N, which is impossible.

Hence, by Theorem 1.4, GD₈. Now suppose that G is not a p-group.

Then at least one between M;N is non-abelian. Assume that M is non- abelian of orderpq. IfMY for some component Y2P, then MˆY, which is impossible asYis cyclic. The same statement is also true forN. If hxi X, then for each componentY 2Pn fXgwe haveGˆXY, forXis a maximal subgroup of G. Thus every component of P other than X has orderqasMhas elements of orderqoutsideX. Now sincehxi3M;N, we havejP⁰j ˆ jP⁰⁰j ˆp‡1 andjPj 2p‡1. ButjPjcannot exceed 2p‡1, otherwise M\YˆN\Yˆ1 for some componentY ofP andhxiY is a

proper subgroup with a nontrivial partition different fromM;N, which is a contradiction. Now letjXj ˆpr. Then

pqrˆ jGj ˆ jXj ‡(jPj 1)(q 1)ˆpr‡(jPj 1)(q 1)

and we have thatjPj ˆpr‡1. Thus we should haverˆ2. Ifpis odd, then X has a characteristic subgroup hyiof order 2. Hence hyi3G andhyiY is a proper subgroup with a nontrivial partition for each component Y2Pn fXg, which is impossible because jMj ˆ jNj ˆpq6ˆ2qˆ jhyiYj.

Thuspˆ2 and soq is odd, asGis not a 2-group. In this term, Ghas a unique element x of order 2, which implies that x2Z(G) and this is in contradiction with Lemma 1.1. Hence M\Nˆ hxi ˆX. Clearly, the components of P⁰ other than X are conjugate by powers of x. Now if Y;ZW for some component W2P, Y2P⁰ and Z2P⁰⁰ such that Y;Z6ˆX, thenY^xi;Z^xi W^xi foriˆ0;1;. . .;p 1. In particular, we de- duce thatNis also non-abelian, because the conjugates ofZby powers ofx should be different. IfjZj ˆr, thenjWj ˆqrand we havejPj ˆp‡1. In this case,Gis a Frobenius group with cyclic kernelXof orderpand cyclic complementWof orderqr. We note that,q6ˆrotherwiseYˆZ, and so we would haveMˆN. Hence

Gˆ hx;y:x^pˆy^qrˆ1;x^yˆxⁱ;i^q6^p 1;i^r6^p 1;i^qrp 1i;

for some 15i5p. Now assume that no components of P⁰ and P⁰⁰ other thanXlie in a component ofP simultaneously. ThenjPj ˆ2p‡1 and we can separate the components ofPother thanXinto two classesY1;. . .;Yp

and Z1;. . .;Zp in such a way that M\Y_i6ˆ1 and N\Z_i6ˆ1 for

iˆ1;. . .;p. Also we have jY₁j ˆ ˆ jY_pj and jZ₁j ˆ ˆ jZ_pj. If

M\Y_iY_i, then jY_ij is a product of two primes, and we would have GˆX[Y₁[ [Y_p, a contradiction. Hence M\Y_iˆY_i and Y_iM.

SimilarlyZ1;. . .;ZpN. LetjZ1j ˆ ˆ jZpj ˆr. Then pqrˆ jGj

ˆ jXj ‡(jY1j 1)‡ ‡(jYpj 1)‡(jZ1j 1)‡ ‡(jZpj 1)

ˆp‡p(q 1)‡p(r 1) 5pqr;

which is a contradiction. The proof is complete. p As it is shown in Propositions 3.2, 3.4 and 3.5, the number of subgroups with a nontrivial partition depends on the number of prime divisors of the order ofG. In the sequel, we are investigating this relation more precisely.

Recall that for each natural number nˆp^a₁¹ p^a_m^m, where n>1 and

p1;. . .;pmare distinct primes, the functionV(n) determines the number of

prime divisors ofnconsidering repetitive primes, i.e., V(n)ˆa₁‡ ‡a_m:

THEOREM3.6. Let G be a finite group admitting a nontrivial partition and let n be the number of subgroups of G with a nontrivial partition. Then V(jGj)n‡1 with equality if and only if the subgroups of G with a nontrivial partition form a chain.

PROOF. For an arbitrary group G let n(G) denotes the number of subgroups ofGadmitting a nontrivial partition. Now letGbe a finite group, which admits a nontrivial partition. Ifn(G)ˆ1, then by Proposition 3.2, V(jGj)ˆ2n(G)‡1. Assume that the result holds for all groups withn- value less than that of G. According to Proposition 3.2, G possesses a maximal subgroup M with a nontrivial partition. Hence n(M)n(G) 1 and consequentlyV(jMj)n(M)‡1. IfM3G, then [G:M] is a prime so that V(jGj)ˆV(jMj)‡1n(G)‡1. Now suppose that M3@@ G, then N_G(M)ˆM. Since G, the conjugates of M and proper subgroups ofM admitting a nontrivial partition are all distinct, we obtain

n(G)1‡[G:M]‡n(M) 1ˆ[G:M]‡n(M):

Let [G:M]ˆp^a₁¹ p^a_nⁿ be the canonical decomposition of [G:M]

into primes. Then [G:M]2^a1‡‡anˆ2^V([G:M]) that is V([G:M]) log₂([G:M]). Therefore

V(jGj)ˆV(jMj)‡V([G:M])

n(G) [G:M]‡1‡log₂([G:M]) 5n(G)‡1:

Also the last inequality implies that for a finite group G with a nontrivial partition V(jGj)ˆn(G)‡1 if and only if the subgroups of G admitting a nontrivial partition form a characteristic chain

H₁3^cH₂3^c 3^cH_n(G)ˆG;

where H_i is a maximal subgroup of H_i‡1. p

We close this section by determining the structure of those finite groups Gadmitting a nontrivial partition such that the subgroups ofGwith a non- trivial partition form a chain. To do this we first need the following lemma.

Recall that a partitionPof a groupGis normal ifP^gˆPfor eachg2G.

LEMMA3.7. Let G be a finite group admitting a nontrivial partition.

Then G has exactly one maximal subgroup with a nontrivial partition if and only if G is a Frobenius group with elementary abelian kernel K and cyclic complement H of prime power order. Moreover, G has no nontrivial normal subgroups properly contained in K.

PROOF. LetPbe a nontrivial normal partition ofGand letMbe the unique maximal subgroup ofG, which admits a nontrivial partition. Then M3Gand [G:M]ˆqis a prime. If no component of P lies in M, then jXj ˆ jX\Mjqfor each componentXofP. Moreover,P\Mform a par- tition forMand we have

qjMj ˆ jGj

ˆX

X2P

jXj (jPj 1)

ˆX

X2P

qjX\Mj (jPj 1)

ˆq(jMj ‡(jPj 1)) (jPj 1):

Then it follows thatqˆ1, a contradiction. LetAbe the set of all compo- nents ofP, which are contained inM. IfX2Pn A, thenX6Mand we conclude that X is a maximal subgroup of G as M is the only maximal subgroup ofGadmitting a nontrivial partition. Now we consider two pos- sibilities:

(1) NG(X)ˆX, for some component X2Pn A. In this term G is a Frobenius group with complementXand since

[

g2G

(XnX\M)^g

ˆ[G:X]jXnX\Mj ˆjGj

jXj jXj jXj q

ˆ jGj jMj

the components ofPn Aare pairwise conjugate. This meansKˆ [_Y2AYis the Frobenius kernel of G, which is nilpotent by [15, 10.5.6]. If K has a nontrivial characteristic subgroupH, thenH3Gand for eachY2Pn A, HY is contained in a maximal subgroup ofG with a nontrivial partition different fromM, a contradiction. ThusKis an elementary abelianp-group, which is the normalizer of its nontrivial subgroups. Letx2GnMbe of the least possible order. Thenjxj ˆq^tfor somet1. SinceKhxi 6Mwe have GˆKhxi, otherwiseGhas a maximal subgroup with a nontrivial partition containing Khxi different from M, which is impossible. Therefore the components ofPn A, as the Frobenius complements ofGare cyclic of order

q^t. Conversely, every group with the aforementioned properties has a un- ique maximal subgroup, which admits a nontrivial partition.

(2) X3G for every component X2Pn A. Let X2Pn A and let x2GnX. Then GˆXhxi as X is a maximal subgroup of G and hence there is a prime psuch that jxj ˆpfor eachx2GnX. IfjPj>jAj ‡1, then there exists a componentY 2Pn A different fromX and primer such thatjxj ˆrfor eachx2GnY. But sinceGnX[Y is nonempty we deduce thatpˆr. ThereforeGis ap-group of exponentpand we can see that G has maximal subgroups admitting a nontrivial partition different from M. Therefore jPj ˆ jAj ‡1 and consequently GˆM[X, where

Pn A ˆ fXg, which is impossible. p

THEOREM3.8. Let G be a finite group admitting a nontrivial partition.

Then the subgroups of G, which admit a nontrivial partition form a chain if and only if G is a Frobenius group with kernel K and complement H satisfying the following conditions:

(1) K is cyclic of order p and H is cyclic of order qⁿ. In this case Gˆ hx;y:x^pˆy^qnˆ1;x^yˆxⁱ;i^qn16^p 1;i^qnp 1i; or

(2) K is non-cyclic of order p²and H is cyclic of order qⁿ. In this case Gˆ hx;y;z:x^pˆy^pˆz^qnˆ1;xyˆyx;x^zˆy;y^zˆxⁱy^ji;

where

Aˆ 0 i 1 j

2GL(2;p); A^{qn 1}6ˆI; A^qnˆI; det(A^m kI)6ˆ0;

in which mˆ1;2;. . .;qⁿ 1, kˆ1;2;. . .;p 1 and p;q are distinct primes.

PROOF. By Lemma 3.7 and the fact that the subgroups of G, which admit a nontrivial partition form a chain,Gis a Frobenius group with ele- mentary abelian kernelKof orderporp²and cyclic complementHof order qⁿfor some distinct primesp;qand natural numbern. For, ifjKj p³, then it has subgroups with nontrivial partition which are not members of a chain.

Also in the case where jKj ˆp²,K is the normalizer of its subgroups of orderpinG. Considering the order ofKwe have two possibilities:

(i) Kis cyclic of orderp. LetKˆ hxiandHˆ hyi. SinceK3G,x^yˆxⁱ for some i and consequently x^yt ˆx^it for each t2N. Then it follows xˆx^yqn ˆx^iqn that is i^qnp 1. Also x^{yqn 1} ˆx^{iqn 1}. Then it follows that

i^{qn 1}6^p 1, because xy^qn1 6ˆy^qn1x. On the other hand, every group with these properties is a Frobenius group with cyclic kernel of order p and cyclic complement of order qⁿ satisfying the hypothesis of theorem.

Therefore

Gˆ hx;y:x^pˆy^qnˆ1;x^yˆxⁱ;i^qn1 6^p 1;i^qnp 1i:

(ii) K is an elementary abelian p-group of order p². Let Hˆ hzi, x2Kn f1gandyˆx^z, theny62 hxiand henceKˆ hx;yi. Lety^zˆxⁱy^j, where i;jare assumed to be inZp and let Abe the matrix, which is in- troduced in (ii). Then

(x^uy^v)^zmˆx^umy^vm and u_m v_m

ˆA^m u v ;

for all u;v2Z_p and integers m. Now since K is the normalizer of its nontrivial subgroups, for eachu;v2Z_psuch thatk6ˆ0, (u;v)6ˆ(0;0) and 1m5qⁿ

(x^uy^v)^zm6ˆ(x^uy^v)^k; which is equivalent to

A^m u

v 6ˆk uv ;

i.e., det(A^m kI)6ˆ0. On the other hand, A^qn16ˆIand A^qnˆI, for the same is true forz. Conversely, we can verify that every group with these properties is a Frobenius group with elementary abelian kernel of orderp² and cyclic complement of orderqⁿsatisfying the hypothesis of the theorem.

Therefore

Gˆ hx;y;z:x^pˆy^pˆz^qnˆ1;xyˆyx;x^zˆy;y^zˆxⁱy^ji;

where A^qn16ˆI, A^qnˆI and A^m has no eigenvalues in Z_p for each

mˆ1;2;. . .;qⁿ 1. p

REMARK. Note that the structure of groups in Proposition 3.4(1) and Proposition 3.5(1) can be obtained from Theorem 3.8(2) by putting nˆ1 and 2, respectively.

The case of infinite group is more difficult. In the spirit of Proposition 3.2 we have the following result, but a precise classification of those infinite groups admitting a nontrivial partition without proper subgroups admit- ting a nontrivial partition remains open.

THEOREM3.9. Let G be an infinite group, which admits a nontrivial partition. If G has no proper subgroups admitting a nontrivial partition, then either G is a simple group or G⁰ˆHp(G),[G:G⁰]ˆp and G⁰ˆG⁰⁰for some prime p.

PROOF. LetP be a nontrivial partition ofG. IfX2Pis not maximal, then there exist a subgroupHofGcontainingXproperly. But thenP\H is a nontrivial partition forH, which is impossible by hypothesis. ThusGhas a unique nontrivial partition with maximal subgroups as its components.

Suppose thatGis not simple. ThenGhas a nontrivial normal subgroupN, which lies in a componentXofP. Ifx2GnX, thenNhxihas a nontrivial partition so thatGˆNhxiandNˆX. Hence there exists a primepsuch that jxj ˆp for each x2GnX, from which NˆHp(G). As N has no characteristic subgroupsNˆN⁰. Moreover, sinceG⁰N, we should have

NˆG⁰, as required. p

4. Number of components of a partition.

Schulz [16] had shown by using the classification theorem of finite groups admitting a nontrivial partition that a finite group with a linear partition is either an elementary abelianp-group or a Frobenius group. In this section, we shall invoke a similar procedure to prove the following result concerning the number of components of a nontrivial partition and the order of group itself.

THEOREM4.1. Let G be a finite group admitting a nontrivial parti- tionP.

(1) If G6PSL(2;2ⁿ), thengcd(jPj 1;jGj)6ˆ1and if GPSL(2;2ⁿ), then there exists a nontrivial partition P⁰ such that gcd(jP⁰j 1;jGj)ˆ1;

(2) IfP is normal, thengcd(jPj 1;jGj)6ˆ1; and

(3) There exists a nontrivial partitionP of G such thatjPj 1 di- videsjGj.

PROOF. Using the classification theorem of finite groups admitting a nontrivial partition it is enough to deal with each case separately. So, letG be a finite group with a nontrivial partition. Then

(i) GˆS₄. LetPybe the maximal partition ofG. Then it is easy to see that the components ofPyare cyclic subgroups ofG. LetPbe a Sylow 2-

subgroup ofG. ThenPD8andPcontains exactly one cyclic subgroup of order 4. IfPwerePy-admissible, then considering an elementx2GnP withjhxij ˆ4 we would haveP\ hxi ˆ1 andjPhxij ˆ32, a contradiction.

Since there is no partition of G containing two different components of order 6, the length of a nontrivial partition ofGis either 10 or 13, which satisfy gcd(10 1;jGj)6ˆ1 and 13 1kGj, respectively.

(ii) Gis ap-group. LetP be a nontrivial partition ofGand letd(H) be the number of cyclic subgroups of orderpfor an arbitrary subgroupHof G. Then, by [6, Satz I.7.2],d(H)^p 1 and so

jPj ˆX

X2P

1^p X

X2P

d(X)ˆd(G)^p 1

that is gcd(jPj 1;jGj)6ˆ1. Also if M is a maximal subgroup of G con- taining H_p(G), then M together with the cyclic subgroups of order p generated by elements ofGnMform a nontrivial partitionP ofGand we have pjMj ˆ jGj ˆ jMj ‡(jPj 1)(p 1). HencejPj 1ˆ jMj and sojPj 1 dividesjGj.

(iii) Gis a group of Hughes-Thompson type. Letpbe a prime such that H_p(G)G, then [G:H_p(G)]ˆp, forGis not ap-group. Also, by [9],H_p(G) is a nilpotent group. Ifpj⁴⁴jH_p(G)j, thenGis a Frobenius group and we do this in part (iv). Hence we can assume thatH_p(G) is not a group of prime power order, whence by Theorem 1.3,Hp(G) has only the trivial partition.

ThusHp(G) together with the cyclic subgroups of order pgenerated by elements of GnH_p(G) form the only nontrivial partitionP of Gand we have pjH_p(G)j ˆ jGj ˆ jH_p(G)j ‡(jPj 1)(p 1). Then it follows that jPj 1ˆ jH_p(G)jand sojPj 1 dividesjGj.

(iv) Let G be a Frobenius group and suppose that K andH are re- spectively the Frobenius kernel and a complement ofG. By [15, 10.5.5] and Corollary 3.3,Hhas no nontrivial partitions. In addition, by [15, 10.5.6],K is nilpotent and ifKhas a nontrivial partition, then by Theorem 1.3,Kis a p-group. Clearly, K together with the conjugates of Hform a nontrivial partitionPforGand we havejPj ˆ jKj ‡1 so thatjPj 1 dividesjGj.

Also ifK has no nontrivial partitions, thenP is the only nontrivial par- tition ofG. Now assume that K is ap-group with a nontrivial partition.

Thus the maximal partition ofK together with the conjugates ofHmake the maximal partition Py of G. If H is a proper subgroup of a P-ad- missible subgroupMofG, thenM is also a Frobenius group with kernel KMK. Thereby every nontrivial partitionP ofGhas the form

P ˆ fK1;. . .;Km;M1;. . .;Mn;H^g1;. . .;H^gkg;

whereKiare subgroups ofKandMiare Frobenius subgroups ofGwith kernelKMias subgroups ofKand complementsHMias conjugates ofH. As M_i contains exactly jK_Mij conjugates ofH andK1;. . .;Km;K_M1;. . .;K_Mn form a partition forK, we get

jPj ˆm‡n‡ jKj jKM1j jKMnj ^p m‡n

^p d(K1)‡ ‡d(Km)‡d(K_M1)‡ ‡d(K_Mn) ^p d(K)

^p 1:

Therefore gcd(jPj 1;jGj)6ˆ1.

(v) GˆPSL(2;pⁿ) withpⁿ4. By [6, Satz II.8.5], there are subgroups H,KandLofGof orderspⁿ, (pⁿ 1)=dand (pⁿ‡1)=d, respectively, where dˆgcd(pⁿ 1;2) such that the conjugates ofH,KandLmakes a nontrivial partition forG. Moreover,His an elementary abelianp-group,KandLare cyclic groups, andH,KandLhavepⁿ‡1,pⁿ(pⁿ‡1)=dandpⁿ(pⁿ 1)=d conjugates, respectively. LetPybe the maximal partition ofGand letHbe a Py-admissible proper subgroup ofG, which is not a component ofPy. W e shall use [6, Hauptsatz II.8.27] of Dickson to determine the structure ofH. If HA₄, then asKandLare cyclic Hall subgroups ofGandA₄has no ele- ments of order 4, we have that 4 dividesjHj ˆpⁿ. Then it follows thatpˆ2 anddˆ1. SinceHisPy-admissible we havepⁿ1ˆ3. Then it follows that nˆ2 andGˆPSL(2;4)A5. IfHA5, then similarly we can show that pˆ2 and dˆ1. SinceA₅ has no elements of order 15 andHis Py-ad- missible, we should have pⁿ 1ˆ3 and pⁿ‡1ˆ5. Then we have GA5H, a contradiction. IfHS4, then as the Sylow 2-subgroups ofH are not cyclic andK,Lare cyclic Hall subgroups we should have 8jpⁿ. But then the Sylow 2-subgroups ofHshould be elementary abelianp-groups, which is impossible. IfHis a dihedral group of order 2mwithmj(pⁿ1)=d, then mˆ(pⁿ1)=d as m6ˆ1. Also if p is odd, then (pⁿ1)=dˆ2.

Therefore we havepⁿˆ5 andGPSL(2;5)PSL(2;4)A₅. IfHis the semidirect product of an elementary abelianp-group of orderp^mwith a cyclic subgroup of order k such that k divides both p^m 1 and pⁿ 1, then kj(pⁿ 1)=dand hencekˆ(pⁿ 1)=d. On the other hand,kjp^gcd(m;n) 1 so thatpⁿ 1jd(p^gcd(m;n) 1). Sinced2 we have gcd(m;n)ˆn, i.e.,mˆn.

If HPSL(2;p^m), in which mjn, then jPSL(2;p^m)j ˆp^m(p^2m 1)

d and

p^m 1jpⁿ 1. Ifp^m 16ˆ1, then we should havep^m 1ˆpⁿ 1. Then we

have mˆn, a contradiction. Thus p^mˆ2 and so HPSL(2;2)S3. Therefore pⁿ1ˆ3 and so nˆ2 and GˆPSL(2;4). Finally, if HPGL(2;p^m) then we must have that 2mdividesn. On the other hand,H contains cyclic subgroups of orderp^m 1 and p^m‡1. So, if HwerePy- admissible then we would have pⁿ 1

gcd(2;p 1)p^m‡1. This is possible if and only ifnˆ2m,pˆ3 orpˆ2 andmˆ1. In the casepˆ3 we have that GˆPSL(2;9) and HPGL(2;3)S₄. In such a case H is not Py-ad- missible, because all the involutions ofHcannot be covered by the 3 cyclic subgroups of H of order 4. Therefore we must have pˆ2 and so GˆPSL(2;4) andHPGL(2;2)PSL(2;2)S3and this case has been considered before.

Now assume thatpis odd andG6PSL(2;4). LetPbe a nontrivial par- tition ofGand letXbe a component ofP, which is neither a conjugate ofKand Lnor a Sylowp-subgroup. ThenXis a semidirect product of ap-subgroupH_X oforderpⁿandacyclicsubgroupoforder(pⁿ 1)=2, whichisaconjugateofK.

HenceXis a Frobenius group with kernelHX. Suppose thatX1;. . .;Xmare such components andH_XiandK_Xi;1;. . .;K_Xi;pⁿare their Frobenius kernel and complements, respectively. Letkbe the number ofp-components ofP. Then

jPj ˆk‡m‡([G:N_G(K)] mpⁿ)‡[G:N_G(L)]

ˆk‡m‡ pⁿ(pⁿ‡1)

2 mpⁿ

‡pⁿ(pⁿ 1) 2 ^p k‡m:

Butk‡mis the number ofp-components ofP⁰, whereP⁰is obtained from Pby omittingXifromPand putting the partition ofXiinstead. As Sylowp- subgroups ofGare pairwise disjoint we can classifyp-components ofP⁰in such a way that the union over components of each class is a Sylowp-sub- group ofG. Now ifAis the set of allp-components ofP⁰, then

jPj ^p X

X2A

d(X) ^p X

g2G

d(H^g) ^p [G:N_G(H)]d(H) ^p pⁿ‡1

^p 1:

Therefore gcd(jPj 1;jGj)6ˆ1. Now let GˆPSL(2;2ⁿ), where n2.

ThenGhas a dihedral subgroupMof order 2(2ⁿ 1), which is aPy-ad- missible subgroup, wherePyis the maximal partition ofG. Moreover, since [G:N_G(H)]ˆ2ⁿ‡1, there exist Sylow 2-subgroupsH₁andH₂ofGsuch that H₁\Mˆ H₂\Mˆ1. Since H₁ and H₂ are elementary abelian 2- groups, if x₁;y₁2 H₁ and x₂;y₂2 H₂ are distinct elements such that hx1i \ hy1i ˆ hx2i \ hy2i ˆ1, then hx1;y1i ˆ hx1i [ hy1i [ hx1y1i and hx2;y2i ˆ hx2i [ hy2i [ hx2y2i. Thus we can construct a partitionP⁰ from Pyby omitting the components ofM,hx₁;y₁iandhx₂;y₂iand replacingM, hx₁;y₁iandhx₂;y₂iinstead. Therefore

jP⁰j ˆ jPj (2ⁿ 1) 2 2

ˆ[G:N_G(H)](2ⁿ 1)‡[G:N_G(K)]‡[G:N_G(L)] 2ⁿ 3

ˆ2²ⁿ 1‡2ⁿ(2ⁿ‡1)

2 ‡2ⁿ(2ⁿ 1)

2 2ⁿ 3

ˆ2^2n‡1 2ⁿ 4

and we can see that gcd(jP⁰j 1;jGj)ˆ1. Now letPbe a normal partition of G. IfHis both a dihedral group of order 2(2ⁿ1) and a component ofP, then the conjugates ofHare disjoint as the components ofP. Then it follows that Ghas at least 2ⁿ(2²ⁿ 1)=2 involutions whileGhas only 2²ⁿ 1 involutions and this is possible only ifnˆ1. Also ifHis both a semidirect product of a Sylow 2-subgroup with a cyclic subgroup of order 2ⁿ 1 and a component of P, then K has at least [G:N_G(H)]jHj ˆ2ⁿ(2ⁿ‡1) different conjugates while the number of conjugates ofKis [G:N_G(K)]ˆ2ⁿ(2ⁿ‡1)=2, which is impossible. Therefore every component of P is either ap-subgroup or a conjugate ofKorL. Now ifAis the set of allp-components ofP, then

jPj ˆ jAj ‡[G:N_G(K)]‡[G:N_G(L)]

² X

X2A

d(X)‡2ⁿ(2ⁿ‡1)

2 ‡2ⁿ(2ⁿ 1) 2 ² X

g2G

d(H^g) ² (2ⁿ‡1)d(H) ² 1:

So it follows that gcd(jPj 1;jGj)6ˆ1. Finally, assume thatGˆPSL(2;pⁿ)