On the Occurrence of the Complete Graph <span class="mathjax-formula">$K_5$</span> in the Hasse Graph of a Finite Group

On the Occurrence of the Complete Graph K

in the Hasse Graph of a Finite Group.

ROLANDSCHMIDT(*)

Dedicated to Professor Guido Zappa on the occasion of his 90th birthday

1. Introduction.

In [7]we determined all finite groups with planar subgroup lattices and also those with planar Hasse graphs. Here a finite latticeLis called planar (see [4]) if it is possible to draw its Hasse diagram in the plane in such a way that no two line segments intersect; the Hasse graphLofLis its Hasse diagram considered as an undirected graph in the usual way (see §2) and for a finite groupG, the Hasse graphL(G)of its subgroup latticeL(G) is also called the Hasse graph ofG.

By Kuratowski's theorem, a finite graph is planar if and only if it does not contain a subdivision of the complete graph K₅ or the complete bipartite graphK3;3as a subgraph (see [2, p. 24]). Figure 1 shows the graphsK5and K3;3; a subdivision is obtained from a graph by subdividing some of the edges,

(*) Indirizzo dell'A.: Mathematisches Seminar, Christian-Albrechts-UniversitaÈt zu Kiel, Ludewig-Meyn-Str. 4, 24098 Kiel Germany; e-mail: [email protected]

Figure 1

that is, by replacing the edges by paths having at most their endvertices in common. We were able to show in [7]that the Hasse graph of a finite group is planar if and only if it contains no subdivision ofK3;3as a subgraph.

The aim of the present paper is to determine all finite groups whose Hasse graph contains no subdivision ofK₅as a subgraph; we shall call such a graphK₅-free, for short. Forp-groups, we obtain the following results.

THEOREMA. LetjGj ˆpⁿ where2<p2Pand5n2N. Then L(G) is K₅-free if and only if G is metacyclic of exponent at least p^{n 2}.

THEOREMB. Let jGj ˆ2ⁿ where 6n2N. Then L(G) is K5-free if and only if G is lattice-isomorphic to an abelian group of type (2ⁿ), (2^{n 1};2), or(2^{n 2};4), or G is one of the followingtwo groups

Gˆha;bja⁴ˆb²ⁿ²ˆ1;a^bˆa ¹i;

Gˆha;bja⁸ˆ1;b^{2n 3}ˆa⁴;a^bˆa ¹i:

The assumptionn5 is needed in Theorem A since for every odd primep, there are two groups of orderp⁴withK₅-free Hasse graph which are not metacyclic. Theorem B holds forn5 if one includes in the list the qua- ternion groupQ32of order 2⁵and all metacyclic groups of order at most 2⁴ (see Remarks 3.8 and 3.11).

It might be interesting to note that only rather few of theseK₅-freep- groups have planar (orK_3;3-free) Hasse graphs. In fact, we showed in [7]

that forp>2, a group of orderpⁿhas this property if and only if it is either cyclic or lattice-isomorphic to an abelian group of type (p^{n 1};p). Forpˆ2, the same groups and, in addition, only the dihedral group of order 8, the quaternion groupsQ₈andQ₁₆, and the semidihedral group of order 16 have planar Hasse graphs. This latter group is the uniquep-group with planar Hasse graph and non-planar subgroup lattice.

In [7, Lemma 3.2]we showed that the Hasse graph of a cyclic group C_k of order kis K5-free if and only if it is planar, and that C_k has this property if and only ifkis of the formpⁿ;pⁿq^m, orpⁿqrwheren;m2N and p;q;rare primes; here the groups C_pⁿ and C_pⁿ_q^m even have planar subgroup lattices whereasL(C_pqr) is non-planar. Therefore the following two theorems complete the determination of all finite groups with K5- free Hasse graph.

THEOREMC. Let G be a finite nilpotent group which is not cyclic and not of prime power order. Then L(G)is K5-free if and only if G'CpQ8

where2<p2Por G'CpCqCqwith different primes p and q.

THEOREMD. Let G be a finite group which is not nilpotent. Then L(G) is K5-free if and only if GˆPK where P/G and K is cyclic, and one of the followingholds(where p;q;r are pairwise different primes and n2N).

(a) jPj ˆp;jKj ˆqⁿandjK:CK(P)j q³. (b) jPj ˆp andjKj ˆqr.

(c) jPj ˆp², P is cyclic,jKj ˆqⁿandjK:C_K(P)j ˆq.

(d) jPj ˆp², P is elementary abelian, jKj ˆqⁿ; jK:C_K(P)j q³, jC_K(P)j q, andV(K=C_K(P))operates irreducibly on P.

(e) jPj ˆp², P is elementary abelian,jKj ˆqr and both minimal subgroups of K operate irreducibly on P.

(f) jPj ˆp³, P is nonabelian of exponent p,jKj ˆq and K operates irreducibly on P=F(P).

(g) P'Q₈; jKj ˆ3or9, G=C_K(P)'SL(2;3).

(h) jPj ˆp⁴; P is abelian of type (p²;p²); jKj ˆq and K operates irreducibly on P=F(P).

Again, by [7], only few of these groups have planar Hasse graphs, namely those in (a) of Theorem D for whichjK:C_K(P)j ˆqand those in (d) for which jKj ˆq. These groups, in fact, also have planar subgroup lattices whereas for all other groups in Theorems C and D, the Hasse graph is non-planar.

2. Notation and Preliminary Remarks.

In the whole paper,Gis a finite group andp;qare primes. We denote by L(G) the subgroup lattice ofG; soL(G)ˆ fXjXGgwith the set-theo- retical inclusion as relation and we write X\Y for the intersection and X[Yfor the join of the subgroupsX;YofG. Further we writeX<YifX is a maximal subgroup of Y.

Forn2Nand ap-groupG, we let V_nV (G)ˆ hx2Gjx^pnˆ1i

n(G)ˆ hx^pnjx2Gi and we denote by

Cnthe cyclic group of ordern

D2ⁿthe dihedral group of order 2ⁿ (n2) Q₂ⁿ the quaternion group of order 2ⁿ (n3) S₂ⁿ the semidihedral group of order 2ⁿ(n4).

Further notation is standard (see [3]and [5]).

All graphs considered are undirected graphs. In particular, we define theHasse graph Lof a finite latticeLto be the (undirected) graph with vertex setLin which the unordered pairfx;ygof elements ofLis an edge if and only if one ofx;yis a lower neighbour of the other. The degree of a vertex is the number of edges in which it is contained. For a groupG, the Hasse graph L(G) of its subgroup lattice L(G) is also called the Hasse graph ofG.

A path g (of length n) from u to v in L is a sequence of elements x₀;x₁;. . .;x_n of Lsuch thatx₀ˆu; x_nˆv and for all ithere is an edge betweenx_iandx_i‡1inL, that is,x_iis a lower or upper neighbour ofx_i‡1. We usually writegˆ(x₀;. . .;x_n) for such a path and call the vertices dif- ferent from x0 and xn the internal vertices of g. Finally, a collection fg₁;. . .;g_mgof paths is calledinternally disjointif each internal vertex of g_i(iˆ1;. . .;m) lies on nog_j (j6ˆi).

If V is a subgraph of L which is a subdivision of K₅, then there exist 5 points x₁;. . .;x₅2V and an internally disjoint set of 10 paths g_ij fromxitoxj (1i<j5) in V, and hence inL. Conversely, every subset G ˆ fx1;. . .;x5g of L with jGj ˆ5 together with an internally disjoint set of paths g_ij from x_i to x_j (1i<j5) in L yields a subgraphV ofL which is a subdivision ofK₅. We shall call such a set G a K₅-set in L and the members of G are called theK₅-points ofV.

They are uniquely determined by V since all the other vertices of V have degree 2 in V.

If we want to show, for a certain groupG, thatL(G)has such a sub- graph V, we usually only give aK₅-setGand describe the nontrivial paths g_ijbetween the members ofG; the trivial ones being edges or, for example, one of thep‡1 possible paths of length 2 between the bottom and the top of an elementary abelian section of orderp²inG.

Finally, as mentioned in the introduction, we call a graphK5-freeif it contains no subdivision ofK₅as a subgraph.

3. p-Groups and Subdivisions ofK5.

In this section we shall determine all finite p-groups whose Hasse graphs areK₅-free. We start with the following trivial, but fundamental, result.

LEMMA3.1. If G is elementary abelian of order p³, then L(G)contains a subdivision of K5.

PROOF. Let Gˆ hai hbi hci where o(a)ˆo(b)ˆo(c)ˆp and let G :ˆ f1;hbi;ha;bi;hb;ci;Gg. There are five edges between these points in L(G) and the other five paths needed are (1;habi;ha;bi), (1;hbci;hb;ci), (1;habci;hab;ci;G), (hbi;hb;aci;G), and (ha;bi;hai;ha;ci;hci;hb;ci). So G together with these paths yields a subdivision ofK₅inL(G). p REMARK 3.2. By [7, Theorem A], all other groups of order p³ have planar subgroup lattices except the nonabelian group H of exponent p (p>2). Since all subgroups of orderpdifferent fromZ(H) have degree 2 in the graph L(H), it is easy to see that L(H) isK₅-free. So the ele- mentary abelian group is the unique group of orderp³whose Hasse graph

contains a subdivision ofK5. p

Lemma 3.1 shows that ifGis ap-group withK₅-free Hasse graph, then every subgroup of G is generated by two elements. Groups with this property were studied by Blackburn [1]; most of them are metacyclic.

LEMMA 3.3. Let jGj ˆpⁿ; n5. If L(G) is K5-free, then G is meta- cyclic and has exponent at least p^{n 2}.

PROOF. We show first thatGis metacyclic. By Lemma 3.1 and [1, The- orems 4.1 and 5.1], this is true except possibly whenGis a 3-group of maximal class. (Blackburn has a weaker assumption in his Theorem 4.1, so he gets a further possibility forp>2, namely a groupGof orderp⁵in whichV(G) is nonabelian of orderp³and exponentpandG=V(G) is cyclic. However, if we takex2Gsuch thatGˆ hxiV(G) and a maximal subgroupSofGcontain- ing x, then F(S) is a cyclic normal subgroup of order p² in G and so F(S)V(G)=F(S)\V(G) is elementary abelian of orderp³.) So suppose, for a contradiction,thatGisa3-groupofmaximalclass.ThenGhasafactorgroupof order3⁵which is also of maximal class.Weshow thatthe Hasse graph of sucha group contains a subdivision ofK₅; this will be the desired contradiction.

So letG0be of order 3⁵and class 4. Then it is well-known [3, pp. 370-371]

thatG0 has normal subgroupsGiwithG0>G1>G2>G3>G4>1 such thatG1is metacyclic,G3ˆV(G1)ˆF(G1) andG2ˆ(G0)⁰is abelian. LetM be a maximal subgroup ofG₀such thatM6ˆG₁. ThenMhas maximal class [3, p. 374]and so F(M)ˆG₃ since G₀ has only one normal subgroup of order 3². Similarly, F(G₂)ˆG₄ˆZ(G₀). IfM=G₄ would contain a cyclic subgroup H=G4 of order 3², then H would be abelian and hence H\G2Z(M), a contradiction sinceMhas maximal class. ThusM=G4has exponent 3. Now we claim that G :ˆ fM;G₁;G₂;G₃;G₄g is a K5-set in

L(G0). Here we use 3 nontrivial paths, namely (M;G0;G1);(M;M1;M2;G4) with maximal subgroupsM16ˆG2 of M andM26ˆG3 of M1 (which exist sinceM=G4 has exponent 3), and (G1;SF(G1);S;V(S);1;G4) whereSis a cyclic subgroup of order 3²ofG₁not contained inG₂and not containingG₄ (which exists sinceV(G₁)ˆF(G₁)). In addition, 4 paths are edges and in the elementary abelian factors M=G₃;G₁=G₃, andG₂=G₄ there are inter- mediate subgroups not yet used. So we get a subdivision ofK5 inL(G0) and this completes the proof thatGis metacyclic.

It remains to be shown that expGp^{n 2}. So suppose, for a contra- diction, that expGp^{n 3}. SinceGis metacyclic,GˆNXwhereN/Gand N and X are cyclic. Then jN:N\Xj ˆ jG:Xj p³ and jX:N\Xj ˆ

ˆ jG:Nj p³. So G=N\X contains a subgroup isomorphic to a semi- direct product of a cyclic groupN1of orderp³by a cyclic groupX1of order p³. Forp>2, such a group has modular subgroup lattice and is therefore lattice-isomorphic to an abelian group of type (p³;p³) [5, 2.3.1 and 2.5.9].

Forpˆ2; AutC₈is elementary abelian of order 4 and henceX₁induces an automorphism of order at most 2 inN1. So to get a contradiction, we show that the Hasse graph of a semidirect product Hˆ haihbi, wherehai/H, o(a)ˆo(b)ˆp³ andb^p2C_H(a), contains a subdivision ofK5. For this let G:ˆ fH₁;. . .;H₅g where H₁ˆ ha^p2i, H₂ˆ ha^p2;b^p2i, H₃ˆ ha^p;b^p2i, H₄ˆ ha^p;b^piandH₅ˆ ha;b^pi. Clearly,H₅is abelian of type (p³;p²) and we have 4 edges fHi;Hi‡1g, 2 nondiagonal paths (H1;ha^pi;hai;ha;b^p2i;H5) and (H2;ha^p2;b^pi;H4), 3 diagonal paths (H1;ha^pb^p2i;H3); (H3;hab^pi;H5) and (H1;1;ha^p2b^p2i;ha^pb^pi;ha^p2;a^pb^pi;H4), and finally the path (H₂;hb^p2i;hb^pi;hbi;ha^p2;bi;ha^p;bi;H;H₅). SoL(H)contains a subdivision of K₅. But L(G) has a subgraph isomorphic to such anL(H), a contra-

diction. Thus expGp^{n 2}. p

Forp>2, the converse of Lemma 3.3 also holds. To prove this, we need some preliminaries which will also be used in the casepˆ2.

DEFINITION3.4. LetGbe a group.

(a) We denote byL5(G) the set of all subgroupsHofGfor which there exists aK5-setG inL(G) such thatH2G.

(b) For every subsetLofL(G) containingL₅(G) we define the graph L^in the following way: the set of vertices ofL^isLand a two element subset fH;KgofLis an edge if and only if there exists a path (X₀;. . .;X_r) inL(G) such thatX0ˆH,XrˆK andXi2L(G)n Lfor alliˆ1;. . .;r 1.

Then we have the following trivial result.

LEMMA3.5. Let L5(G) L L(G). IfGis a K5-set in L(G), then it is also a K5-set inL.^

PROOF. Let Gˆ fH1;. . .;H5g. SinceL5(G) L, clearly G L. Fur- thermore there exist internally disjoint paths g_ij from H_i to H_j in L(G) (1i<j5). If we remove in these paths all the elements X2L(G)n L, we get internally disjoint paths^g_ijfromH_itoH_jin the graph L. Thus^ Gis aK5-set inL.^ p LEMMA3.6. Let p2P; 2m2Nand Gˆ hai hbiwhere o(a)ˆp^m and o(b)ˆp². Then L(G)is K₅-free.

PROOF. Let A:ˆ hai;B:ˆ hbi and L:ˆ fAiBjj0im;0j2g whereA_i:ˆ ha^{pm i}iis the subgroup of orderpⁱofAandB_j:ˆ hb^p2jiis the subgroup of orderp^jofB. We want to apply Lemma 3.5 with this latticeLand therefore have to show thatL₅(G) L.

For this letH2L₅(G) and suppose first thatG=His cyclic. Since every element ofL5(G) clearly has degree at least 4 in the graphL(G), it follows thatHis not cyclic. Suppose, for a contradiction, thatHhas type (p^k;p) for some k1. Ifkˆ1, then HˆV(G) andG=Hwould not be cyclic. Thus k2 and Hhas exactly one noncyclic andp cyclic maximal subgroups.

There are 4 internally disjoint paths fromHto the otherK₅-points of aK₅- set inL(G). SinceG=His cyclic, at least two of these paths have to use cyclic maximal subgroups ofH, at most one of these can go on toF(H). So at least one path has to stop at a maximal cyclic subgroupXofHor to go on to a subgroupY 6ˆHcoveringX. In the first case,X2L₅(G) and again there exists Y 6ˆH covering X. Clearly, Y is cyclic; otherwise V(G) Y\HˆX, a contradiction. Hence XˆF(Y)F(G) (see [3, p. 269]) and soHˆXV(G)F(G); but this contradicts the fact thatG=His cyclic.

ThusHcannot have type (p^k;p) and sinceGis abelian of type (p^m;p²), the type ofHmust be (p^k;p²) for somek2. ThereforeHcontains the unique subgroupV₂(G) of type (p²;p²) ofG. We have shown:

(1) IfH2L5(G) andG=His cyclic, thenV2(G)H.

It follows thatHˆAiBfor somei2 and soH2 L. SinceL(G) is self-dual [5, p. 454], (1) also implies that a cyclic subgroupK inL5(G) is contained in V

2(G) and hence KˆA_i2 L (where im 2). Finally, if neither H nor G=H is cyclic, then V(G)HF(G) and hence HˆA_iB₁2 L(where 1im 1). So we have shown that

(2) L5(G) L.

We now determineL. By Definition 3.4, every edge in^ L(G)between two elements of L is also an edge inL. Furthermore, every elementary^ abelian sectionA_i‡1B_j‡1=A_iB_j (0i<m; 0j<2) yields an edge betweenA_iB_jandA_i‡1B_j‡1. Finally, there is an edge betweenA_iand A_i‡2B(0im 2) since there is a diagonal path in the factor group A_i‡2B=A_i'C_p2C_p2 given by

ha^{pm i}i; ha^{pm i1}b^pi; ha^{pm i 2}bi; ha^{pm i2}b;b^pi; ha^{pm i 2}i hbi : We claim that there are no more edges inL:^

(3) The edges fAiBj;Ai‡1Bjg(i<m), fAiBj;AiBj‡1g (j<2); fA_iB_j;A_i‡1B_j‡1g (i<m; j<2), and fA_i;A_i‡2Bg (im 2) are all the edges of the graphL.^

For this, letH;Kbe inLwithH6ˆKand let (H;X1;. . .;Xr;K) be a path inL(G)such thatX_i2L(G)n Lfor alli(so thatfH;Kgis an edge inL).^ Clearly, we may assume thatr1 and we suppose first thatHˆA_k for some k. Then km 1; and if kˆm 1, it follows that rˆ1 and KˆA_mB₁, so that fH;Kg is one of the edges in (3). Therefore let km 2. Then we prove by induction onithat foriˆ1;. . .;r,

(a) H<X_i<A_k‡2Bˆ:L,

(b) A\XiˆHˆHB\Xi ifXiis cyclic, and (g) Xi<L ifXiis not cyclic.

This is clear foriˆ1. So suppose it holds for somei<rand consider first the case thatX_iis cyclic. IfX_i‡1<X_i, then by (a),HX_i‡1and since Xi‡162 L, we have thatH<Xi‡1. SinceL(Xi) is a chain, (b) holds forXi‡1. IfX_i<X_i‡1andX_i‡1 is cyclic, then again X_i‡1 satisfies (b) sinceX_idoes;

furthermore, X_i‡1A_k‡2B since o(b)ˆp². Finally, if X_i<X_i‡1 and X_i‡1 is not cyclic, then X_i‡1ˆX_iV(G)L. Furthermore, since Xi‡16ˆAk‡1B1, we have jXi:Hj ˆp² and so Xi‡1<L. Now suppose that Xi is not cyclic. Then V(G)Xi and since Xi62 L, we have V2(G)6X_iandX_i6F(G); in particular,G=X_iis cyclic. So ifX_i<X_i‡1, it follows that X_i‡1X_iV₂(G)L. By (g), X_i<L and so X_i‡1ˆL, a con- tradiction. Thus X_i‡1<X_i<L. Since A_k‡2B₁6ˆX_i6ˆA_k‡1B, every cyclic maximal subgroup ofX_isatisfies (b) and the only noncyclic maximal subgroup ofXiisF(L)ˆAk‡1B12 L. So we have shown that (a) - (g) hold. IfXr is cyclic and jXr:Hj ˆp², then (b) shows that neither F(Xr) norXrV(G) belong toL. ThusjXr:Hj ˆpandKˆXrV(G)ˆA_k‡1B₁. IfX_r is not cyclic, then by (g),X_r is one of thep 1 maximal subgroups

different from Ak‡2B1 and Ak‡1B of L. Then G=Xr is cyclic and it follows thatK ˆL orKˆF(L)ˆAk‡1B1. In all cases, fH;Kg is one of the edges in (3).

Since L(G) is self-dual, the above argument also covers the case that HˆA_kBfor somek. It remains to consider the case thatHˆA_kB₁for somek. Here, ifkˆ0 orkˆm, thenrˆ1 andKˆA₁BorK ˆA_{m 1}, respectively. So suppose that 1km 1. Again using the duality of L(G), we may assume thatX1is a cyclic maximal subgroup ofA_kB1. Then F(X₁)ˆA_{k 1}and so eitherKˆA_{k 1} or (A_{k 1};X₁;. . .;X_r;K) is one of the paths considered above. In the latter case, K is one of the groups A_k, A_{k 1}B₁,A_k‡1B, as we have shown; sofH;Kgis one of the edges in (3).

We finally show that the graph (4) L^ is planar.

For this we just map the elementA_iB_j ofLto the point (j;i) inR². Then we connect pairs of points of the form f(r;s);(r;s‡1)g, f(r;s);(r‡1;s)gandf(r;s);(r‡1;s‡1)gin the obvious way by straight line segments. Finally we have to connect the points (0;s) with (2;s‡2) and we can do this using the straight line segment from (0;s) to ( 2s 2; s 2) together with the lower half of the circle with center ( s;0) and radius (s‡2) 

p2

. Since these circles do not intersect, this is a planar representation of the graphL; a similar one is shown in Figure 2 for^ mˆ4.

Figure 2

It is obvious that Lemma 3.6 follows from (4) and Lemma 3.5. For, if L(G) would contain a subdivision of K5, then so would L; but a planar^

graph contains no subdivision ofK5. p

Now we prove Theorem A; we also determine the occurring groups.

THEOREM3.7. LetjGj ˆpⁿwhere2<p2Pand5n2N. Then the followingproperties are equivalent.

(a) L(G)is K₅-free.

(b) G is metacyclic andexpGp^{n 2}.

(c) G is either abelian of type(pⁿ);(p^{n 1};p), or(p^{n 2};p²), or G is one of the followingnonabelian groups:

(c4) Gˆ ha;bja^pn1 ˆb^pˆ1; a^bˆa^1‡pn2i, (c5) Gˆ ha;bja^pn2 ˆb^p2 ˆ1; a^bˆa^{1‡pn 3}i, (c6) Gˆ ha;bja^pn2 ˆb^p2 ˆ1; a^bˆa^{1‡pn 4}i, (c7) Gˆ ha;bja^p2ˆb^pn2 ˆ1; a^bˆa^1‡pi, (c8) Gˆ ha;bja^p3ˆ1; b^pn3ˆa^p2; a^bˆa^1‡pi.

PROOF. By Lemma 3.3, (a) implies (b), and that (c) implies (b) is ob- vious. So suppose that (b) holds. Then again by Iwasawa's and Baer's the- orems [5, 2.3.1 and 2.5.9], since p>2; L(G) is modular and therefore is isomorphic to the subgroup lattice of an abelianp-group which clearly also has exponent at leastp^{n 2}. Thus (a) follows from Lemma 3.6 and [7, The- orem A]. Furthermore, Gis one of the groups in (c) if expGp^{n 1}. If expGˆp^{n 2}andGhas a cyclic normal subgrouphaiof orderp^{n 2}, then by [5, 2.3.11],haihas a complementhbiinGandGis abelian or the group in (c5) or (c6). Finally, suppose that expGˆp^{n 2} and G has no cyclic normal subgroup of orderp^{n 2}. By [5, 2.3.18]there are an abelian normal subgroup A, an element b of order p^{n 2} in G and s2N such that GˆAhbi and a^bˆa^1‡ps for all a2A. Then jA:A\ hbij ˆ jG:hbij ˆp² and since G⁰6 hbi, it follows thatsˆ1 andA=A\ hbiis cyclic. Soxˆx^bˆx^1‡pfor hxi ˆA\ hbiand hencejA\ hbij p. It follows easily thatGis the group in

(c7) or (c8). p

Forp>2, Theorem 3.7 gives the groups of orderpⁿwithK₅-free Hasse graphs ifn5. Of course, it is not difficult to determine also the groups of orderp⁴with this property. We give the result without proof.

REMARK3.8. LetjGj ˆp⁴; p>2. ThenL(G)isK5-free if and only if Gis metacyclic or one of the following groups:

(a) Gˆ hx;y;zjx^pˆy^pˆz^p2 ˆ1ˆ[y;z];y^xˆyz^p;z^xˆzyi, (b) Gˆ hx;y;zjx^pˆy^pˆz^p2 ˆ1ˆ[y;z];y^xˆyz^sp;z^xˆzyi wherep5 andsis a quadratic nonresidue modulop,

(c) Gˆ hx;y;zjy³ˆz⁹ˆ1ˆ[y;z];x³ˆz³;y^xˆyz ³;z^xˆzyi.

We turn to the casejGj ˆ2ⁿ. Here, for simplicity, we shall prove our main result only for n6. But for this, we first have to study certain groups of order 2⁵.

LEMMA3.9. If G is a dihedral or semidihedral group of order2⁵or if Gˆ ha;bja⁸ˆb⁴ˆ1;a^bˆa ¹i, then L(G)contains a subdivision of K₅. PROOF. Suppose first thatG'D32orG'S32. Then in both cases,G has exactly three maximal subgroupsA;D;M such thatAˆ haiis cyclic, D'D₁₆ and M is quaternion or dihedral; furthermore Z(G)ˆ ha⁸iand G=Z(G)'D₁₆. LetG:ˆ fZ(G);ha⁴i;U;V;DgwhereUandV are the two dihedral subgroups of order 8 ofD. SinceU\V ˆ ha⁴i, we see that there are 5 edges inL(G) between members ofG. Further obvious paths are (ha⁴i;ha²i;D) and (D;G;M;M1;M2;Z(G)) with noncyclic M1<M and ha⁴i 6ˆM2<M1. SinceU=Z(G) andV=Z(G) are elementary abelian of or- der 4, there are trivial paths (Z(G);U₁;U) and (Z(G);V₁;V) with U₁6ˆ ha⁴i 6ˆV₁and there are further noncyclic maximal subgroupsU₂ofU and V2 of V. So we finally get a path (U;U2;X;1;Y;V2;V) with jXj ˆ jYj ˆ2 andX6ˆZ(G)6ˆY. All these paths are internally disjoint and soGis aK5-set inL(G).

Now let Gˆ ha;bja⁸ˆb⁴ˆ1;a^bˆa ¹i. Then Z(G)ˆ ha⁴;b²i and G=hb²i 'D₁₆. FurthermoreGhas three maximal subgroupsMˆ hai hb²i andM1;M2 for whichMi=hb²i 'D8 (iˆ1;2). ClearlyF(G)ˆM1\M2 ˆ

ˆ ha²;b²iand we letG:ˆ fM1;M2;F(G);Z(G);hb²ig. This time there are 4 edges between members ofG, two trivial paths fromM_itoZ(G) and a third noncyclic maximal subgroupH_iinM_iwhich yields a path fromM_itohb²ifor iˆ1;2. The final two paths are (M₁;G;M₂) and (hb²i;1;ha⁴i;ha²i;F(G)). So

Gis aK5-set inL(G). p

LEMMA3.10. Let6n2Nand suppose that G is a nonabelian group of order2ⁿsuch that L(G)is K₅-free. Then G is one of the followinggroups.

(a) Gˆ ha;bja²ⁿ¹ˆb²ˆ1;a^bˆa^1‡2n2i (b) Gˆ ha;bja²ⁿ²ˆb⁴ˆ1;a^bˆa^1‡2n3i (c) Gˆ ha;bja²ⁿ²ˆb⁴ˆ1;a^bˆa^1‡2n4i (d) Gˆ ha;bja⁴ ˆb²ⁿ²ˆ1;a^bˆa ¹i (e) Gˆ ha;bja⁸ ˆ1;b^{2n 3}ˆa⁴;a^bˆa ¹i

PROOF. By Lemma 3.3,Gis metacyclic and has exponent at least 2^{n 2}. If expGˆ2^{n 1}, then eitherGis the group with modular subgroup lattice in (a) orGis dihedral, quaternion, or semidihedral [3, p. 90]. In the latter three cases,G=Z(G)'D₂ⁿ¹which is impossible by Lemma 3.9 sincen6. Thus Gsatisfies (a).

So let expGˆ2^{n 2}. SinceGis metacyclic, there existsN/Gsuch that Nˆ haiis cyclic andGˆNhxifor somex2G. Consider first the case that no such N has order 2^{n 2} and let HˆNhx²i. Since a^x2ˆa^r with r1(mod 4);L(H) is modular [5, 2.3.4]and since His generated by two elements of order at most 2^{n 3}, it follows from [5, 2.3.5]that expH2^{n 3}. Hence there existsb2GnHsuch thato(b)ˆ2^{n 2}. Then GˆNhbi and jN:N\ hbij ˆ jG:hbij ˆ4. Clearly, a^bˆa^t for some t2N and if t1(mod 4), then [a;b]2 hbiandhbi/G which would contradict our as- sumption in this case. So t61(mod 4) and it follows that jN\ hbij 2.

Thus either jNj ˆ4 andGis the group in (d) orjNj ˆ8; b²ⁿ³ ˆa⁴ and a^bˆa ¹ ora^bˆa³. In the latter case, (ab²ⁿ⁴)^bˆa³b^{2n 4}ˆ(ab²ⁿ⁴) ¹ and we may replaceabyab²ⁿ⁴ to obtain the group in (e).

It is left to consider the case that GˆNhxi where Nˆ hai/G and o(a)ˆ2^{n 2}. Againa^xˆa^t for somet2Nand ift1(mod 4), then by [5, 2.3.4], L(G) is modular. Since expG2⁴,G is not hamiltonian; then [5, 2.3.11]implies that N has a complement hbiin G. Thus Gis one of the groups in (b) or (c).

So lett61(mod 4). Then againjN\ hxij 2 and we show that this case cannot occur. Ifxinduces an automorphism of order 2 inN, thena^xˆa ¹ ora^xˆa ^1‡2n3and soG=hx²iis a dihedral or semidihedral group of order 2^{n 1} or 2^{n 2}. This contradicts Lemma 3.9 except when nˆ6 and N\ hxi 6ˆ1. But in this case, G=ha⁸i is the third group of order 2⁵ in Lemma 3.9, again a contradiction. Thus x induces an automorphism of order 4 in N. Since AutN has only two cyclic subgroups of order 4 [3, p. 84]andt61(mod 4), it follows that a^bˆa ^{1‡2n 4} forbˆxorbˆx ¹. Then (ab)⁴ˆa²ⁿ³b⁴; so ifN\ hbi 6ˆ1, we have (ab)⁴ˆ1 and can replaceb byab. Thus we may assume thatN\ hbi ˆ1. ThenG=ha^{2n 3};b²iis semi- dihedral of order 2^{n 2} and Lemma 3.9 implies that nˆ6. So Gˆ ha;bja¹⁶ˆb⁴ˆ1;a^bˆa³i and G has the following maximal sub- groups: H:ˆ ha;b²i has modular subgroup lattice, S:ˆ ha²;bi and T:ˆ ha²;abi. Put W:ˆV₂(H)ˆ ha⁴;b²i and Z:ˆZ(G)ˆ ha⁸i and let

G:ˆ fZ;W;F(G);S;Tg. There are 3 edges (connectingF(G) toW;S;T)

and 4 obvious paths (S;G;T);(Z;Zhb²i;W);(Z;ha⁴i;ha²i;F(G)) and (Z;1;hb²i;hbi;ha⁸;bi;ha⁴;bi;S). Since G=W 'D8, there are trivial paths (S;S₁;W) and (T;T₁;W) with T₁6ˆF(G)6ˆS₁6ˆ ha⁴;bi. Finally, (ab)²ˆ

ˆa¹²b²and so ifXˆ h(ab)²i, thenT=X'D8andZ<X. Thus there exists a noncyclic subgroup T2=X of order 4 with T2 6ˆT1 and a path (T;T2;T3;X;Z) so that all the paths considered are internally disjoint. So L(G)contains a subdivision ofK₅, a contradiction. p PROOF OFTHEOREM B. Let jGj ˆ2ⁿ where 6n2N. Again by [5, 2,3.1 and 2.5.9], the groups in (a) - (c) of Lemma 3.10 have modular sub- group lattice and are therefore lattice-isomorphic to the abelian groups mentioned in Theorem B. So ifL(G)isK₅-free, then by Lemmas 3.3 and 3.10,Gis one of the groups given in this theorem. Conversely, ifGis lat- tice-isomorphic to an abelian group of type (2ⁿ);(2^{n 1};2), or (2^{n 2};4), then by Lemma 3.6 and [7, Theorem A],GisK5-free.

So, finally, let Gbe one of the other two groups given in Theorem B.

Then in both cases, b²2Z(G); let hb²i ˆ:Z and H:ˆ ha;b²i. Then His abelian of exponent 2^{n 3}and hence there exists a complementhcitoZinH;

soHˆZ hciwitho(c)ˆ4.

Clearly,G=Z'D8. So everyx2GnHsatisfiesx²2Zandhaihxi ˆG since H is the unique maximal subgroup of G containing hai. Then Hˆ hai(H\ hxi)ˆ haihx²iand sincex²2Z, it follows thathx²i ˆZ. Since every subgroup ofGnot contained inHcontains an elementx2GnH, it follows that L(G) consists of the subgroups ofH, four cyclic subgroups X1;. . .;X4 containing Zas a maximal subgroup, two maximal subgroups M1andM2different fromH, andG.

Now it is easy to see that L5(G)L(H). For, these cyclic sub- groups X_i have degree 2 inL(G), G has degree 3, and both M_i have degree 4; however, also M_i cannot be a member of a K₅-set since two of the four paths from Mi to the other members would have to start with a cyclic maximal subgroup of Mi and then use F(Mi), which is impossible.

Now suppose, for a contradiction, thatG is aK₅-set inL(G). Then by Lemma 3.5, it is also a K₅-set inL^ whereL ˆL(H). Clearly,L^ contains L(H) and a further edge can only connect two of the three groups Z;Z hc²i;Hsince no other subgroup of His covered by an element of L(G)nL(H). Since there are already edges fromZ hc²itoZand toHin L(H), the only additional edge isfZ;Hg. LetZbe a cyclic 2-group con- tainingZas a maximal subgroup and considerHˆZ hcias a subgroup of index 2 in G:ˆZ hci. In the subdivision ofK₅ given byG in L, we^ replace the edgefZ;Hg- if it appears in one of the paths - by the path (Z;Z;Z hc²i;G;H). Then we get a subdivision of K5 in L(G); but this

contradicts Lemma 3.6. p

As in the casep>2 it is not difficult to determine the 2-groups of order less than 2⁶ withK5-free Hasse graph. Again we give the result without proof.

REMARK3.11. LetjGj ˆ2ⁿ;n5. ThenL(G)isK₅-free if and only if one of the following holds:

(a) n4 andGis metacyclic,

(b) Gis one of the groups in Theorem B fornˆ5, (c) G'Q₃₂.

4. Proof of Theorem C.

It remains to determine the nonprimary finite groups with K₅-free Hasse graphs. For this, we first look at direct products of lattices.

LEMMA4.1. Let LˆL1L2with finite lattices Li. Then Lcontains a subdivision of K5if one of the followingholds.

(a) L₁ contains L(C₂C₂)as a sublattice andjL₂j 3.

(b) L1 contains L(C4C2)as a sublattice andjL2j 2.

PROOF. (a) We may assume that L1ˆL(H)ˆ f1;M1;M2;M3;Hg wherejHj ˆ4 andjM_ij ˆ2 foriˆ1;2;3 and thatL₂is a chainA<B<C of length 2. Then

G:ˆ fHA;1B;M1B;M2B;HBg

is aK5-set inLsince there are 5 trivial paths between members ofGin the interval [HB=1B], furthermore 4 obvious paths fromHAtoKB viaKA, and finally the path (M₁B;M₁C;HC;M₂C;M₂B).

(b) Here we may assume that L₁ˆL(H) where Hˆ hai hbi with o(a)ˆ4,o(b)ˆ2 and thatL2ˆ f0;Igis a chain of length 1. Then

G :ˆ f10;ha²i 0;ha²;bi 0;ha²i I;ha²;bi Ig

is aK5-set inL since there are 5 edges between members ofG and we have the further paths (10;ha²bi0;ha²;bi0);(10;1I;ha²iI), (10;hbi0;hbiI, ha²;biI), (ha²i0;habi0;habiI;HI;ha²;biI),

and (ha²;bi0;H0,hai0;haiI;ha²iI). p

There is a similar result for the Hasse graph of a semidirect product of two groups. This graph contains a subdivision ofK5 if a certain smaller

configuration occurs in the Hasse graph of the complement. We first give this configuration a name.

DEFINITION4.2. A (3,6)-gonin a graph is a set of 3 distinct pointsR;S;T together with an internally disjoint set of 6 paths between them, one be- tweenRandS, two betweenSandT, and three betweenRandT.

Since at most one path between two points can be an edge, there are intermediate points in one of the paths betweenSandTand in two of the paths betweenRandT(see Figure 3).

LEMMA 4.3. Let GˆNK where 16ˆN/G and N\Kˆ1. If L(K) contains a (3,6)-gon, then L(G)contains a subdivision of K5.

PROOF. Let R;S;T be the points of a (3,6)-gon in L(K), let e;d₁;d₂;r₁;r₂;r₃be the appropriate paths fromRtoS,StoT, andRtoT with interior points U;V;W of d₂;r₂;r₃, respectively. For every path gˆ(X0;. . .;Xr), we letg ¹:ˆ(Xr;. . .;X0) be the path with the same end- points in opposite direction. The isomorphism theorem for groups implies that if all theX_iare subgroups ofK, then^g:ˆ(NX0;. . .;NXr) is a path in L(G)and that the setf^e;^d₁;^d₂;^r₁;^r₂;^r₃gis internally disjoint. Finally, for every subgroup X of K, we let g(X)ˆ(X₀;X₁;. . .;X_r) be a fixed path in L(G) from X to NX such that Xi<Xi‡1 for all i. By Dedekind's law, X_iˆ(N\X_i)X and hence X_i\KˆX for all i. This shows that for X6ˆYK, the pathsg(X) andg(Y) are disjoint.

Figure 3

We now let G:ˆ fR;S;T;NR;NTg. There are obvious paths e;d1;r₁ betweenR;S;T,g(R) betweenRandNR,g(T) betweenTandNT, and^r₁ between NRand NT. Furthermore,g(S) and^e ¹ yield a path from Sto NR(viaNS), and we get a path fromStoNTusingd₂fromStoU, then g(U), and finally^d₂ from NU to NT. Similarly we get paths from R to NTviar₂,g(V);^r₂and fromTtoNRviar₃¹;g(W);^r₃¹. ThusGis aK₅-set

in L(G). p

We shall need the following result on groups without (3,6)-gons.

PROPOSITION4.4. LetKbe a group such thatL(K)contains no (3,6)- gon. ThenKhas one of the followingproperties (wherepandqare primes and2n2N).

(a) K is cyclic.

(b) K is abelian of type(2ⁿ;2)or lattice-isomorphic to this group.

(c) K is elementary abelian of order p²or nonabelian of order pq.

(d) K is isomorphic to Q8; C4C4, or CpC2C2 where p>2.

PROOF. LetKbe a minimal counterexample and suppose first thatKis a 2-group. The trivial subgroupTand the two noncyclic subgroupsR;Sof order 4 clearly yield a (3,6)-gon inL(D₈). SoD₈cannot be involved inKand [5, 2.3.3]

shows thatL(K) is modular. SinceL(D8)is a subgraph ofL(C2C2C2), K is generated by two elements and so, again by [5, 2.3.1 and 2.5.9], either K'Q8orL(K)'L(H) whereHis abelian of type (2ⁿ) or (2ⁿ;2^m). However, ifC₈C₄'H₀H, thenRˆV₂(H₀);TˆV(H₀);Sˆ V

2(H₀) would yield a (3,6)-gon inL(H), a contradiction. It follows thatK 'H'C₄C₄or that mˆ1. In every case,Kis one of the groups in (a) - (d), a contradiction. SoKis not a 2-group.

Now suppose, for a contradiction, that there are subgroupsA/PK such that P=Ais elementary abelian of orderp²;p>2, or nonabelian of orderpq;p>q;pandqprimes. Suppose first thatPˆK. Then sinceKis a counterexample, A6ˆ1. LetA<S<K. IfSwould have a maximal sub- group B6ˆA, thenRˆK,TˆA andSwould yield a (3,6)-gon inL(K) with 5 paths inside [K=A]and the sixth connectingStoTviaBandT\B.

This contradiction shows that all subgroups strictly betweenAandK are cyclic of prime power order; sinceA6ˆ1, it follows thatKis ap-group all of whose maximal subgroups are cyclic. Hence K'Q₈ (see [3, p. 311]), a contradiction sincep>2.

ThusP<Kand the minimality ofKimplies thatAˆ1 and thatPis a maximal subgroup of K. Let x2KnP . If P^xˆP and x normalizes a

minimal subgroupMofP, thenRˆ1,TˆPtogether withSˆP\ hxiif P\ hxi 6ˆ1 andSˆMifP\ hxi ˆ1 yield a (3,6)-gon inL(K)with 5 paths inside [P=1], the sixth path being (S;hxi;K;T) or (S;Shxi;K;T), respec- tively. This is a contradiction; so ifP^xˆP, thenxoperates irreducibly on P. In particular,hxi \Pˆ1; nowRˆP; Tˆ1 and SˆK yield a (3,6)- gon, again a contradiction. It follows that P^x6ˆP. If P\P^x6ˆ1, again Rˆ1; TˆP and SˆP\P^x would yield a (3,6)-gon with sixth path (S;P^x;K;T). So, finally, P\P^xˆ1 for all x2KnP; but then again RˆP; Tˆ1 andSˆKyield a (3,6)-gon. This is the desired contradiction which shows that for p>2, neither C_pC_p nor a nonabelian group of orderpq(p>q2P) are involved inK.

It follows that Sylow p-subgroups are cyclic; moreover, Burnside's theorem [3, p. 419]implies that K has a normal p-complement. The in- tersection of all these normalp-complements (2<p2P) is a normal Sylow 2-subgroup HofK with cyclic factor group. The minimality ofK implies thatHis one of the groups in (a) - (d) andKˆHCwith cyclic groupCof odd order. IfCwould operate nontrivially onH, then it would also operate nontrivially on H=F(H) (see [3, p. 275]). So jH=F(H)j ˆ4 and it would follow that the alternating groupA4would be involved inK. ButL(A4)has a (3,6)-gon given by RˆA₄; Tˆ1 and the subgroupS of order 4. This contradiction shows thatKˆHC.

SinceKis not a 2-group,C6ˆ1; sinceKis a counterexample,His not cyclic. So jH:F(H)j ˆ4 and if F(H)6ˆ1, then RˆH; TˆF(H) and SˆTCwould yield a (3,6)-gon inL(K), a contradiction. ThusjHj ˆ4.

Since K is a counterexample, C contains a subgroup D'C_pC_q or E'C_p2, p and q primes. However, in L(HD) we get a (3,6)-gon RˆH; SˆDandTˆ1 with obvious paths. InL(HE) we may take RˆE; SˆV(E) andTˆHS; here, ifA;B;Care the three subgroups of order 2 of H, we have the paths (S;1;A;H;T);(S;AS;T) and (R;AR;HR;T);(R;BR;BS;T);(R;CR;CS;T) to get a

(3,6)-gon. This is a final contradiction. p

Since one of the three points of a (3,6)-gon has degree at least 5 in the graph, it is easy to see that the Hasse graphs of the groups in (b) - (d) of Proposition 4.4 contain no (3,6)-gon and that the cyclic groups with this property are those with planar subgroup lattice and those whose order is the product of 4 pairwise different primes. We shall not need this.

An immediate consequence of 4.1 - 4.4 is the following result on co- prime direct products which we shall use quite often. First of all, it com-

pletes the determination of nilpotent groups withK5-free Hasse graphs, that is, it implies Theorem C.

LEMMA4.5. Let GˆAB where A6ˆ16ˆB and(jAj;jBj)ˆ1. If L(G) is K₅-free, then either G is cyclic or GˆPK wherejPj ˆp and K is isomorphic to Q₈or elementary abelian of order q²or nonabelian of order qr with pairwise different primes p;q;r.

PROOF. Suppose thatGis not cyclic. Then one of the two factorsAand B, sayB, is not cyclic. By [6, Theorem 1.1],L(B) containsL(C₂C₂) as a sublattice. By [5, 1.6.4],L(G)'L(A)L(B) and so Lemma 4.1, (a) shows that jL(A)j 2. ThusjAj ˆp for some prime pand now Lemma 4.1, (b) yields thatL(B) does not containL(C4C2) as a sublattice. By Lemma 4.3, L(B) contains no (3,6)-gon and so Proposition 4.4 implies thatGis one of the groups in Lemma 4.5 orB'C_qC₂C₂ withq6ˆ26ˆp. But in this case,G'(C_pC_q)(C₂C₂) which would contradict Lemma 4.1,(a). p PROOF OFTHEOREMC. IfL(G)isK5-free, then by Lemma 4.5,Gis one of the groups given in the theorem. Conversely, ifGˆPQwithjPj ˆp is one of these groups, thenL(G) contains only four points of degree at least 4, namelyZ(Q);Q;PZ(Q);GifQ'Q₈and 1;Q;P;GifQ'C_qC_q.

Hence there is noK5-set inL(G). p

5. Proof of Theorem D.

LEMMA5.1. All the groups in (a) - (h) of Theorem D have K₅-free Hasse graphs.

PROOF. In most cases this is rather obvious; in some cases it even follows already from the fact that there do not exist 5 subgroups ofGwith degree at least 4 inL(G). This, for example, holds ifGsatisfies (a). For, thenC_K(P) and the subgroups properly containingPC_K(P) have degree p‡1 orp‡2, whereas all other subgroups have degree at most 3; since jK:CK(P)j q³, there are at most 4 subgroups of degree at least 4. And ifG satisfies (b), then Ghas precisely 4 subgroups of degree at least 4. If G satisfies (d), then in addition to C_K(P) and the subgroups properly con- taining PC_K(P), also 1;P, and PC_K(P) have degree at least 4. But every path between one of the groups 1;P, orCK(P) and one of the groups containingPCK(P) has to useCK(P) orPCK(P). Since we would need

at least 4 paths of this type for aK5-set, it follows that L(G) isK5-free.

Similarly, ifGsatisfies (e), then the only subgroups of degree at least 4 are 1;P;PQ;PR;GwhereKˆQRwithjQj ˆqandjRj ˆr; however, there is no path fromPtoGwhich does not usePQ;PR, or 1. Finally, ifGsatisfies (f), thenq6 jp 1 since K is irreducible onP=F(P). Hence K centralizes F(P) and it follows that 1;F(P);P;Gare the only subgroups of degree at least 4 inL(G).

Now suppose thatGsatisfies (c). Then subgroups of orderqⁿhave degree 2 and subgroups of orderpqⁿofGhave degreep‡2 inL(G); however, all but 2 of the paths from such a subgroup to the other members of aK₅-set would have to start with a cyclic subgroup of orderqⁿand then useC_K(P), which is impossible. So if we let H:ˆPCK(P), it follows that L5(G) L(H)[ fGg ˆ:L. SinceC_K(P) hxifor every x2GnH, the edges of L^ are the edges of L(H) together with fH;C_K(P)g;fH;Gg;fC_K(P);Gg and fV(P)C_K(P);Gg. Thus L^ is planar and by Lemma 3.5, L(G) is K₅-free.

Now let G be the group in (g) with jKj ˆ9. Then there are only 6 subgroups of degree at least 4, namely F(P);P;CK(P);F(P)CK(P), PC_K(P), andG. SoL5(G) LwhereLis the set of these 6 subgroups. It is easy to see thatP;C_K(P) andGhave degree 3 inL. By Lemma 3.5,^ L(G) isK₅-free. The other group in (g) is isomorphic toG=C_K(P).

Finally, suppose thatGsatisfies (h). Then every maximal subgroupM ofPhas degreep‡2; however, all but 2 of the paths fromMto a member of a K5-set would have to start with a cyclic maximal subgroup ofMand then use F(M). This is impossible. SinceK is also irreducible onF(P)ˆ

ˆV(P), a dual argument shows that alsoF(M) cannot be a member of aK₅- set. Furthermore,F(P)Khas degreep²‡2, but again all but 2 paths would start with cyclic subgroups of orderqand contain 1, which is impossible.

This leaves only 1;F(P);P;Gas possible members ofL5(G). So there is no

K₅-set inL(G). p

We now prove in a couple of steps that every nonnilpotent finite group withK₅-free Hasse graph is one of the groups in (a) - (h) of Theorem D.

First we study groups with a normal Sylowp-subgroup. For this we need two simple results on small groups.

LEMMA 5.2. Let p;q2P such that p6ˆq and suppose that GˆPQ where P is an elementary abelian normal subgroup of order p²,jQj ˆq and [P;Q]6ˆ1. If L(G)is K5-free, then Q operates irreducibly on P.

PROOF. If not, then by Maschke's theorem, PˆP1P2 where Pi/G(iˆ1;2) and P1Q, say, is nonabelian of order pq. Then G:ˆ f1;P2;P;P1Q;Gg is aK5-set inL(G); here 7 paths are trivial in the sense of §2 and since p>q, there are further paths (P;P₁;P₁Q), (P₂;P₂Q;Q;P₁Q) and (1;Q^x;P₂Q^x;G) for somex2P. This contradicts our

assumption. p

LEMMA5.3. Let GˆPK where P/G; jPj ˆp or p², p does not divide jKjand C_K(P)ˆ1. If L(G)is K₅-free, thenjL(K)j 4.

PROOF. Suppose, for a contradiction, thatjL(K)j 5. IfPis cyclic, we may assume thatjPj ˆp. ThenKis cyclic, sojKj>8 and hencejPj>10. If Pis not cyclic, then by Lemma 5.2, every nontrivial subgroup ofKoperates irreducibly onP. HencejKjis odd andjL(K)j 5 implies thatjPj>10. In both cases,NG(K)ˆK; so there are at least 10 complements toPinGand any two of them intersect trivially (see [5, 4.1.1]).

By assumption, there are pairwise different subgroups H_i satisfying P<H_iG(iˆ1;. . .;4); let H₅ˆ1. We choose 10 pairwise different complements K_ij (1i<j5) to P in G and paths d_ij in L(K_ij) from Hi\Kij toHj\Kij via the join of the two groups. Then the set of paths (Hi;dij;Hj) for 1i<j4 and (Hi;di5) for 1i4 is internally dis- joint. SofH₁;. . .;H5gis aK5-set inL(G), a contradiction. p LEMMA5.4. Let GˆPK where P is a normal p-subgroup, p a prime, K a p⁰-group and [P;K]6ˆ1. If L(G) is K5-free, then there are primes q;r such that either

(a) K is a cyclic q-group andjK:CK(P)j q³, or (b) K is cyclic of order qr.

PROOF. LetGbe a minimal counterexample. Since then alsoG=F(P) is a counterexample [3, p. 275],F(P)ˆ1 and Lemma 3.1 implies that

(1) jPj ˆporPis elementary abelian of orderp².

Let K₀:ˆC_K(P). By Lemma 5.3, jL(K=K₀)j 4. Since the subgroup lattice of every noncyclic group contains L(C2C2) as a sublattice [6, Theorem 1.1], it follows thatK=K0is cyclic and that there are primesq;r such that

(2) K=K0is cyclic of orderqrorq^kwhere 1k3.

Finally, by Lemma 4.3,

(3) L(K) contains no (3,6)-gon.

Now suppose first thatK=K₀ is cyclic of orderqr; q6ˆr. SinceGis a counterexample, K₀6ˆ1. By Proposition 4.4, K is cyclic; otherwise K'C_qC₂C₂ and PH with C₂C₂'HK would be a counter- example of order less thanjGj. LetQandRbe the subgroups betweenK0

andKsuch thatjQ:K0j ˆqandjR:K0j ˆr. The minimality ofGimplies thatQandRsatisfy (a) or (b) of the lemma; sinceK₀6ˆ1, it follows thatjK₀j is a prime andjK₀j 6ˆr, say. Lemma 4.3, applied withNˆK₀, yields that L(PR) contains no (3,6)-gon; by Proposition 4.4, jPj ˆp. We claim that G :ˆ f1;K0;PQ;PR;Gg is aK5-set in L(G). Here only 3 of the requir- ed paths are trivial, but, in addition, there are 3 obvious paths (1;P;PK₀;PQ); (K₀;Q;PQ) and (K₀;R;PR). Since p5, there exist a;b;c;d2Psuch thatK;K^a;K^b;K^c;K^dare pairwise different. So if we let S be the subgroup of order r in K, we have 4 further paths (1;S^a;PS;PR); (1;S^b;K0S^b;K^b;G); (K0;Q^c;K^c;G) and (PQ;PQ\K^d;K^d, PR\K^d;PR). Since any two different conjugates ofKintersect inK0, the set of these paths is internally disjoint. SoG is aK5-set, a contradiction.

It follows thatK=K₀is cyclic of orderq^kwhere 1k3. We claim that kˆ1 and

(4) K is elementary abelian of order q² or nonabelian of order qt withq<t2P.

For this, suppose first thatKis aq-group. Ifk2, then every maximal subgroup ofKwould operate nontrivially onPand hence would be cyclic by the minimality of G. So K would be isomorphic toCqCq or Q8 [3, p. 311], which would be impossible since k2. Thus kˆ1 and hence F(K)K0. SoF(K)/Gand the minimality ofGimplies thatF(K)ˆ1. By Lemma 3.1, (4) holds. - Now suppose thatKis not aq-group. IfKˆST with S'C₂C₂ andjTj 2P, then the minimality ofGwould imply that tˆq; soSˆK0 andGˆPTS, contradicting Lemma 4.5. Thus Propo- sition 4.4 implies that (4) holds or thatKis cyclic. In the latter case, letQbe the Sylowq-subgroup ofK. Then Lemma 4.3, applied with aq-complement N in K₀, yields that L(PQ) contains no (3,6)-gon; by Proposition 4.4, jPQj ˆpq. It follows thatkˆ1 andGˆPQK₀. Now Lemma 4.5 implies that jK₀jis a prime; but this is impossible since G is a counterexample.

Thus (4) holds in all cases.

LetQKsuch thatjQj ˆqandQ6ˆK0. By Lemma 4.3, applied with NˆK₀; L(PQ) contains no (3,6)-gon and Proposition 4.4 implies