### A graph related to the join of subgroups of a finite group

HADIAHMADI(*) - BIJANTAERI(**)

ABSTRACT- For a finite groupGdifferent from a cyclic group of prime power order, we introduce an undirected simple graphD(G) whose vertices are the proper subgroups ofGwhich are not contained in the Frattini subgroup ofGand two verticesHandKare joined by an edge if and only ifG hH;Ki. In this paper we studyD(G) and show that it is connected and determine the clique and chromatic number ofD(G) and obtain bounds for its diameter and girth. We classify finite groups with complete graphs and also classify finite groups with domination number 1. Also we show that if the independence number of the graphD(G) is at most 7, thenGis solvable.

MATHEMATICSSUBJECTCLASSIFICATION(2010). 20D99; 05C25, 05C83.

KEYWORDS. Graph on groups, subgroup graph, join of subgroups.

1. Introduction

There are different ways to associate to a group a certain graph. In this context, it is interesting to ask for the relation between the structure of the group, given in group theoretical terms, and the structure of the graph, given in the language of graph theory.

As a pioneer, BosaÂk [5], in 1964, defined the graph of subsemigroups of a semigroup. Inspired by his work, CsaÂkaÂny and PollaÂk [6] in 1969, studied the intersection graph of subgroups of a finite group. Zelinka continued the work on the intersection graph of subgroups of a finite abelian group [12]. Also Shen [11] in 2009, classified finite groups with

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

E-mail: hadiahmadi@math.iut.ac.ir

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

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

disconnected intersection graph of subgroups and solved a problem posed by CsaÂkaÂny and PollaÂk [6]. More work on graphs associated to subgroups of a group or submodules of a module or ideals of a ring can be found in [9, 4, 1, 7].

Motivated by previous studies on the graph of algebraic structures, for
any finite groupGdifferent from a cyclic group of prime power order, we
define an undirected simple graph D(G) whose vertices are the proper
subgroups ofGwhich are not contained in the Frattini subgroup ofGand
two vertices H_{1} andH_{2} are joined by an edge ifhH_{1};H_{2}i G. Our main
goal is to study the connection between the algebraic properties of a group
and the graph theoretic properties of the graph associated to it.

Throughout this paper all groups are finite different from a cyclic pÿgroup. For a groupG,p(G) denotes the set of all prime divisors ofjGj.

Forp2p(G), the set of all Sylowpÿsubgroups ofGis denoted by Syl_{p}(G)
andn_{p}: jSyl_{p}(G)j.M(G) denotes the set of all maximal subgroups ofG
andF(G) is the Frattini subgroup ofGwhich is defined as the intersection
of all maximal subgroups ofG.

LetDbe a graph with vertex setV(D). The order ofDis the number of
its vertices, that isjV(D)j. The degree of a vertexvdenoted by deg(v) is the
number of edges incident tov. For distinct adjacent verticesv_{1}andv_{2}we
write v_{1}v_{2}. If D is connected we denote by d(v_{1};v_{2}) the length of a
shortest path between v1 and v2. The diameter of D is defined as
d(D):maxfd(v1;v2)jv1;v22V(D)g. The girth of D, denoted by g(D), is
the length of a shortest cycle and a graph with no cycle has infinite girth. A
tree is a connected graph which does not contain a cycle.

A star is a tree consisting of one vertex adjacent to all others. A complete graph is a graph in which every pair of distinct vertices are adjacent. A complete graph withnvertices is denoted byKn. By a clique in a graphDwe mean a complete subgraph ofDand the order of a largest clique ofDis called the clique number ofDand is denoted byv(D). For a graphD, letx(D) denote the chromatic number ofD, that is the minimum number of colors which can be assigned to the vertices of Dsuch that every two adjacent vertices have different colors. Let SV(D), denote byND[S] the set of vertices inDwhich are inSor adjacent to a vertex inS.

IfN_{D}[S]V(D), thenSis said to be a dominating set of vertices inD. The
domination number of a graphD, denoted byg(D) is the minimum size of a
dominating set of the vertices inD. A subsetXof the vertices ofDis called
an independent set if the induced subgraph on X has no edges. The
maximum size of an independent set in a graph D(G) is called the in-
dependence number ofDand is denoted bya(D). All other notations and

definitions are standard; in the scope of group theory and graph theory;

see for instance [10, 3].

In this paper, we study some graph properties of the graph D(G). We see thatD(G) is always connected with diameter at most 3 and if it contains a cycle, its girth is always 3 or 4. Also we classify finite groups with com- plete graphs and also groups whose graphs have domination number 1. We show that the clique and chromatic number of this graph is equal to the number of maximal subgroups ofG. Some results about the independence number ofD(G) are established. Regularity ofD(G), in some partial cases, is dealt with.

2. Connectivity, clique, chromatic and domination number

It is easy to see that a groupGhas a unique maximal subgroup if and only ifGis a cyclicpÿgroup. We emphasize that in this workGis a finite group different from a cyclicpÿgroup wherepis a prime number.

DEFINITION 2.1. Let G be a group and F(G) denotes its Frattini subgroup. We associate a graph D(G) to G whose vertex set is fH<GjH<ÿj F(G)g, consists of proper subgroups H which are not con- tained in F(G), with two vertices H1 and H2 adjacent if and only if G hH1;H2i.

REMARK2.2. LetGbe a group andHG. We haveHF(G) if and
only if hH;Ki^{<}_{ÿ}_{4}Gfor each 16K^{<}_{ÿ}_{4}G. Because F(G) is exactly the set of
non-generator elements of the group. Thus to avoid isolated vertices we
consider proper subgroups which are not contained inF(G).

In this section we firstly find the order of this graph and determine its clique and chromatic number. We classify groups with complete graphs. In the sequel we show that this graph is connected and classify groups whose graphs have domination number 1.

PROPOSITION 2.3. D(G) is an (n1;n2;. . .;nt)-regular graph for some
ni1,1it andjV(D(G))j P^{t}

i1jG:NG(Hi)j.

PROOF. The vertex setV(D(G)) of the graph, isGÿinvariant under the conjugacy action ofG. Let [H1]G;[H2]G;. . .;[Ht]Gbe theGÿorbits of this

action. HenceV(D(G)) [^{t}_{i1}[Hi]Gand so
jV(D(G))j X^{t}

i1

j[H_{i}]_{G}j X^{t}

i1

jG:N_{G}(H_{i})j:

Also for eachi2 f1;2;. . .;tgand eachg2G, we have deg(H_{i})deg(H^{g}_{i}).

Because there exists a bijectionu:N_{D}[H_{i}]ÿ!N_{D}[H^{g}_{i}] such thatu(H^{x}_{j})H^{xg}_{j} ,
where ND[H_{i}] fH^{x}_{j} 2V(D(G))jH^{x}_{j} H_{i}g forx2G;1jt. Therefore
jND[Hi]j jND[H^{g}_{i}]jandD(G) is (n1;n2;. . .;nt)ÿregular. p
PROPOSITION 2.4. For any group G we have x(D(G))v(D(G))
jM(G)j.

PROOF. Firstly note thatv(D(G))x(D(G)). Let the vertexM2 M(G) be colored with a colorC. Now we can color all subgroups contained inM(if any exist!) withCand since all vertices out ofM(G) are contained in ele- ments ofM(G), we have a coloring forD(G). Hencex(D(G)) jM(G)j.

Now note thatM(G) is a clique forD(G) and sojM(G)j v(D(G)). This

finishes the proof. p

THEOREM2.5. D(G)is complete if and only if G is one of the following types:

(1)GPP, where P is a group of prime order.

(2)GP^{j}Q, where P and Q are groups of prime order.

(3)GQ_{8}, the quaternion group of order 8.

PROOF. IfGis one of the types (1), (2), (3), then clearly each vertex of the graph is a maximal subgroup ofGand soD(G) is complete.

Conversely, letD(G) be complete. We havejM(G)j 2. FixM2 M(G)
and x2MnF(G). Thus hxi 2V(D(G)). Now hhxi;Mi M, that is hxi is
not adjacent to M and by completeness of D(G) we have M hxi. Let
jxj p^{n}_{1}^{1} p^{n}_{k}^{k}, where thep_{i}'s are distinct prime numbers,k1,n_{i}1 and
1ik.Thenthereexistelementsxi2Gsuchthatjxij p^{n}_{i}^{i};1ikand
M hxi hx1i hx2i hxki. Obviously there exists j2 f1;2;. . .;kg
such thathx_{j}i<ÿj F(G). Again by completeness ofD(G) we haveM hx_{j}iand so
Mis a cyclicpÿgroup, wherepp_{j}. Now two cases hold:

Case 1. F(G)1. If jp(G)j 1, then G is an elementary abelian pÿgroup, that is,G G

F(G)Z_{p}^{n}, for somen2. Ifn3, then clearly

D(G) is not complete. Hencen2 and soGZpZp and thus Gis of
type (1). Now letjp(G)j 2. Since all maximal subgroups ofGare cyclic, all
Sylow subgroups of Gare also cyclic and by [10, Theorem 10.1.10], Gis
solvable and splits over G^{0}, the derived subgroup ofG. So there exists a
subgroup K ofGsuch thatGG^{0}^{j}K andjG^{0}j andjKjare prime num-
bers. HenceGhas type (2).

Case 2. F(G)61. Let M1;M22 M(G). There exist prime numbers
p1;p2 and positive integersn1;n2 such thatjM1j p^{n}_{1}^{1},jM2j p^{n}_{2}^{2}. Since
F(G)61,p_{1}p_{2}. This means that all maximal subgroups ofGare cyclic
pÿgroup, for some primep. HencejGj p^{n}, for somen1 and all max-
imal subgroups ofGare cyclic. Now by [10, Theorem 5.3.4] which classifies
finitepÿgroups with a cyclic maximal subgroup, up to isomorphism,Ghas
one of the following types:

(i) Z_{p}^{n}
(ii) Z_{p}^{nÿ1}Zp

(iii) hx;ajx^{p}1a^{p}^{nÿ1};a^{x}a^{1p}^{nÿ2}i;n3
(iv) The dihedral groupD_{2}^{n};n3, of order 2^{n}

(v) Generalized quaternion groupQ_{2}^{n} hx;yjx^{2}^{nÿ1}1;y^{2}x^{2}^{nÿ2},
x^{y}x^{ÿ1}i;n3

(vi) Semidihedral group SD2^{n} hx;ajx^{2}1a^{2}^{nÿ1};a^{x}a^{2}^{nÿ2}^{ÿ1}i;

n3.

Recall that all groups in the paper are different from a cyclic group of prime power order, so the case (i) cannot happen.

Suppose that (ii) holds. If n3, then G has a non-cyclic maximal subgroup and son2. ThusGhas type (1).

Now suppose that (iii) holds. Then clearly hxi and hx;a^{p}^{nÿ2}i are two
distinct vertices of D(G) which are not adjacent, contradicting the com-
pleteness ofD(G) so this case does not happen.

Similarly the cases (iv) and (vi) cannot happen, as we can find two distinct vertices which are not adjacent.

Finally suppose that (v) holds. Ifn3, thenGQ_{8}andD(G)K_{3}so
D(G) is complete. Whenn4, it suffices to consider two distinct vertices
hyi andhy;x^{2}^{nÿ3}i of D(G) which are not adjacent again contradicting the
completeness ofD(G). Hence in this case the only possibility isQ8. p

LEMMA2.6. For a group G we have the following properties:

(1)D(G)is a connected graph with d(D(G))3.

(2)IfD(G)contains a cycle, then3g(D(G))4.

(3) For an arbitrary normal subgroup N of G with NF(G); we have a(D(G))a

DG

N .

PROOF. (1) Let H1;H2 be arbitrary vertices of V(D(G)). Since
H_{i}<ÿj F(G), i1;2, there exist M_{i}2 M(G) such that H_{i}M_{i} (may be
M_{1}M_{2}). ThereforehH_{1};M_{1}i G hH_{2};M_{2}iandH_{1}M_{1}M_{2}H_{2},
which means thatD(G) is connected andd(D(G))3.

(2) Let M_{1};M_{2}2 M(G) and take z2Gn(M_{1}[M_{2}). If hzi^{<}_{ÿ}_{4}G, then
we have a triangle in D(G) with vertices fM1;hzi;M2g, hence
g(D(G))3. Thus let G hzi and jzj p1n1 p_{k}^{n}^{k}, where the p_{i}'s are
distinct prime numbers k2 and n_{i}1 for 1ik. Hence
GZ_{p}_{1}^{n}1...pknk. If k3, then jM(G)j 3 and clearly g(D(G))3. Now
letk2. Ifn_{1}1 or n_{2}1, then D(G) is a star graph and so does not
contain a cycle; a contradiction. Thus n12 and n22, from which it
follows easily that g(D(G))4.

(3) This is straightforward. p

PROPOSITION2.7. (1)Ifa(D(G))7, then G is a solvable group.

(2)If G is a nilpotent group such thatD(G)is non-complete, then
a(D(G))maxfjG:N_{G}(H)j jH2V(D(G))g 1:

PROOF. (1) Let G : G

F(G). By (3) of Lemma 2.6, we have a(D(G)) a(D(G)). Hencea(D(G)) 7. Thus each maximal subgroupM ofG contains at most 7 proper non-trivial subgroup. Therefore ifjp(M)j 2 andM is non- nilpotent, by an inspection we can show that each Sylow subgroup ofM is cyclic. Hence M is supersolvable, thus G is solvable (see [10, Theorems 10.1.10 and 10.3.4]). SinceF(G) is solvable,Gis solvable.

(2) Suppose, if possible, that there exists a maximal subgroup ofGof
prime orderp. Then by nilpotency ofG,jGj pq, whereqis a prime. Since
Gis non-cyclic of prime power order, we haveGZ_{p}Z_{q}, which implies
thatD(G) is complete, a contradiction. Therefore the order of every max-
imal subgroup ofGis not prime.

Now let H2V(D(G)). There exists a maximal subgroup M of Gsuch
that HM. Since G is nilpotent, every maximal subgroup M of G is
normal. Therefore H^{g}M for each g2G. Hence [H]_{G}[ fMg is an in-
dependent set ofD(G). SinceHis arbitrary, the result follows. p

REMARK2.8. The converse of Proposition 2.7 (1), does not hold, as it is
shown by the elementary abelian groupZ^{m}_{p}, wherem3,p7.

Recall that, by [10, Theorem 9.2.1], every finite solvable groupGhas a Sylow basis, namely a set of mutually permutable Sylow subgroups, one for each prime dividing the group order.

THEOREM 2.9. g(D(G))1 if and only if G is solvable with a Sylow
basisfP1;P2gsuch that P1is a cyclic maximal subgroup of G andF(G)is a
maximal subgroup of P_{1}, or G is a pÿgroup of order p^{n}, where n is a pos-
itive integer and G is isomorphic to one of the groups:

(1) ZpZ_{p}^{nÿ1}, p is a prime number

(2) hx;ajx^{p} 1 a^{p}^{nÿ1};a^{x} a^{1p}^{nÿ2}i; p is a prime number and
n3

(3) The dihedral group D_{2}^{n};n3, of order2^{n}
(4) Generalized quaternion group Q2^{n};n3

(5) Semidihedral group SD2^{n} hx;ajx^{2} 1a^{2}^{nÿ1};a^{x}a^{2}^{nÿ2}^{ÿ1}i;

n3.

PROOF. First let G be a solvable group with Sylow basis fP_{1};P_{2}g,
whereP1is a maximal subgroup ofGandF(G) is a maximal subgroup ofP1.
We show that fP1g is a dominating set for D(G). Suppose that H2
V(D(G))n fP_{1}g. IfH<P_{1}, thenHF(G)P_{1}and maximality ofF(G) inP_{1}
implies that HF(G)P_{1}. Since GP_{1}P_{2}HF(G)P_{2}HP_{2}, and thus
HP_{1} a contradiction. Therefore H<ÿj P_{1} and so hH;P_{1}i G, that is
HP1.

Now letGbe one of the types (1);(2);(3);(4);(5), thenG : G

F(G)Z^{m}_{p},
m2. By [10, Theorem 5.3.4] G has a cyclic maximal subgroupA. Let
A : A

F(G), which is a cyclic maximal subgroup ofG, so A haifor some

a2G andjAj p^{mÿ1}. SinceG is an elementary abelianpÿgroup,m2.

ThusjF(G)j p^{nÿ2}. IfK2V(D(G))n fAg, thenK<ÿj A, becauseAis cyclic.

ThereforehK;Ai Gand soKA. Thus ifGhas one of the types (1)-(5), theng(D(G))1.

Conversely suppose that g(D(G))1 and D fAg is a dominating
set forD(G). We claim thatAis a cyclic maximal subgroup ofG. If not,
then there exists a maximal subgroup Mof Gsuch that A^{<}ÿ4MandAis
not adjacent to M, which is a contradiction. Hence A is a maximal
subgroup of G. Now take a2AnF(G). Then the vertex hai is not ad-

jacent to A and so A haiand the claim is proved. Clearly F(G) is a
maximal subgroup of A. Therefore G is a solvable group and A is of
primary index (see [10, Exercise 10.5.7 and 5.4.3-(iii)]). Suppose that
jG:Aj p^{n}, wherep is prime andn1 andP is a Sylowpÿsubgroup
of G. If G is a pÿgroup, then by [10, Theorem 5.3.4] G is one of the
types (1), (2), (3), (4) and (5). Therefore suppose that G is not a
pÿgroup. Since G is a solvable group, it posseses a Sylow basis
fP1;P2;. . .;P_{l};Pg, l1.

Without loss of generality we can assume thatP_{i}A;1il, (see
[8]). If P_{1}^{<}_{ÿ}_{4}A, then P_{1} is not adjacent with A, a contradiction. Thus
AP_{1}, that isfA;Pg is a Sylow basis ofG as desired. This completes

the proof. p

3. Bipartiteness and regularity

In this section we firstly investigateD(G), when it is a cycle or when it is bipartite. Then we classify groups whose graphs are rÿregular for r2 f3;4g.

PROPOSITION3.1. (1)D(G)is a cycle if and only if GQ_{8}orZ_{p}2
1p^{2}_{2} or
Z2Z2, where p1;p2are distinct prime numbers.

(2)D(G)is bipartite if and only if GZ_{p}^{n}1

1 p^{n}_{2}^{2} where n_{i}1;i2 f1;2g.

PROOF. (1) IfGis one of the groupsQ_{8} orZ_{2}Z_{2}, thenD(G)K_{3}.
Also if GZ_{p}2

1p^{2}_{2}, then D(G)C_{4} where C_{n} denotes the cycle of length
n;n1.

Conversely let D(G) be a cycle. By Lemma 2.6, D(G) is connected
and d(D(G))3. Hence D(G)2 fC3;C4;C5;C6;C7g. Suppose, if possible,
that D(G)C_{5} and V(D(G)) fH_{1};H_{2};H_{3};H_{4};H_{5}g and D(G) is:

H_{1}H_{2}H_{3}H_{4}H_{5}H_{1}. It is easy to see that jM(G)j 2, say
M(G) fH1;H2g. Thus H4F(G) so H42=V(D(G)), a contradiction.

Similarly D(G)2 fC6;C7g, is impossible.

ThereforeD(G)2 fC3;C4g. First letD(G)C3 soD(G) is complete and
by Theorem 2.5,Gis isomorphic to one of the following:Z_{p}Z_{p}orZ_{p}^{j}Z_{q}
orQ_{8}. ClearlyV(D(G)) M(G). IfGZ_{p}Z_{p}, thenD(G)K_{p1}and so
p2. HenceGZ_{2}Z_{2}. Now suppose, if possible, thatGZ_{p}^{j}Z_{q}. We
can assume thatp6qand so we haveV(D(G))Syl_{p}(G)[Syl_{q}(G). Thus
jV(D(G))j npnq1p>123 a contradiction. Therefore this
case is impossible.

Now let D(G)C4. Then jM(G)j 2 and G hgi is a cyclic group.

Suppose that jgj p^{n}_{1}^{1}p^{n}_{2}^{2} p^{n}_{k}^{k}, where the pi's are distinct primes and
n_{i}1;1ik. Since jM(G)j 2 we conclude that k2 and
GZ_{p}^{n}1

1p^{n}_{2}^{2}. LetV(D(G)) fH1;H2;H3;H4gsuch thatM(G) fH1;H2g
andD(G) is the cycle:H1H2H3H4H1. Ifn13 orn23, then
D(G) contains K_{1;3} as a subgraph, a contradiction. Therefore (n_{1};n_{2})2
f(1;1);(1;2);(2;1);(2;2)g. If (n_{1};n_{2})2 f(1;1);(1;2);(2;1)g, thenjV(D(G))j 2
f2;3g. contradictingjV(D(G))j 4. Thus (n_{1};n_{2})(2;2), that isGZ_{p}2

1p^{2}_{2}

and the proof is complete.

(2) If D(G) is bipartite, it contains no odd cycle and this implies that jM(G)j 2. Hence Gis cyclic and the proof of (1) gives the result. The

converse is obvious. p

PROPOSITION3.2. (1)D(G)is3ÿregular if and only if G has one of the
typesZ_{p}3

1p^{3}_{2},Z_{3}Z_{3}or S_{3}, where p_{1} and p_{2}are distinct primes and S_{3} is
the symmetric group on three symbols.

(2)D(G)is4ÿregular if and only if GZ_{p}^{4}

1p^{4}_{2}.
PROOF. (1) IfGZ_{p}^{3}

1p^{3}_{2} , then one can see thatD(G) has exactly six
vertices and G has exactly two maximal subgroups with no common
neighbors. Thus D(G) is 3ÿregular. In fact D(G)K_{3;3}. If GS_{3} or
GZ_{3}Z_{3}, thenD(G)K_{4}.

Conversely let D(G) be 3ÿregular. Thus jM(G)j 4. First let
jM(G)j 2. ThereforeGis cyclic with two distinct prime divisorsp1;p2.
Note that the maximal subgroups have no common neighbors. So
jV(D(G))j 6 and thus GZ_{p}^{3}

1p^{3}_{2}. Now suppose that jM(G)j 4. Then
D(G)K4and by Theorem 2.5, GP^{j}QorGPP, whereP andQ
are groups of prime order. IfGPP, thenD(G)Kp1 and sop3,

Fig. 1.

that isGZ3Z3. IfGP^{j}Q, thenV(D(G))Syl_{p}(G)[Syl_{q}(G) and
obviously we havejPj 3 andjQj 2. ThusGS3.

Finally letjM(G)j 3. By [2, Lemma 1]Gis a cyclic group of order
p^{n}q^{m}r^{k}, where p;q;r are distinct primes and n;m;k1 or G is a
2ÿgroup. In both cases one can easily see that jV(D(G))j 6 and any
two of the maximal subgroups of G have no common neighbors as
shown in Figure 1. Now note that F(G)Mi\Mj, for each Mi;Mj2
M(G);1i;j3. So if H is a non-maximal vertex of D(G), then
HF(G), which is a contradiction. Thus this case does not hold and the
proof is complete.

(2) LetGZ_{p}4

1p^{4}_{2}. ThenjV(D(G))j 8 andjM(G)j 2. SoD(G) has six
non-maximal vertex and clearly it is 4ÿregular. In factD(G)K4;4.

Conversely ifD(G) is 4ÿregular, thenjM(G)j 5. First letjM(G)j 5.

SoD(G)K_{5}and by Theorem 2.5, we haveGPPorGP^{j}Qwhere
PandQare groups of prime orders. By an easy inspection we can see that
neither case can happen. We claim that ifjM(G)j 4, thenD(G) cannot be
4ÿregular and this case also does not hold. In fact ifD(G) is 4ÿregular, then
Ghas a non-maximal subgroupHwhich contained in exactly one maximal
vertex of D(G), as F(G)M_{i}\M_{j}, for eachM_{i};M_{j}2 M(G);1i;j4.

Since deg(H)4,D(G) has another non-maximal vertex different fromH, sayK. SimilarlyKis contained in exactly one maximal vertex ofD(G). This shows that there exists a maximal vertex ofD(G) whose degree is greater than 4, which is a contradiction. Thus the claim is proved.

Now letjM(G)j 3. Similar to the proof of (1), we havejV(D(G))j 6 and so any two maximal subgroups of Ghave a common neighbour and every maximal subgroup of G contains exactly one vertex of the graph which is shown in Figure 2.

Fig. 2.

Again by [2, Lemma 1]Gis a cyclic group of orderp^{n}q^{m}r^{k}, wherep;q;r
are distinct primes and n;m;k1 or G is a 2ÿgroup such that

F(G)G Z_{2}Z_{2}. If G is of the first type, then obviously each maximal
subgroup ofGcontains more than one vertex ofD(G), which is a contra-
diction. If G is of the second type, then each maximal subgroup of Gis
cyclic. To see this suppose thatHis a non-maximal vertex of the graph.

Then there is a maximal subgroup M of G such that H^{<}ÿ4M. Now take
x2Mn(H[F(G)). Then hxi 2V(D(G)) and hxi M. But exactly one
vertex of D(G) is contained in M. Hence M hxi. Therefore F(G) is a
maximal subgroup ofHand alsoHis maximal inMand so we have:

jG:F(G)j jG:MjjM:HjjH:F(G)j 2^{3};
which is a contradiction.

Finally letjM(G)j 2. ThenGis cyclic with two distinct prime divisors
p1;p2. Since D(G) is 4ÿregular, every maximal subgroup of G contains
exactly three non-maximal vertex of the graph. Also maximal subgroups of
Ghave no common neighbor. ThereforeGZ_{p}^{4}

1p^{4}_{2}andD(G)K_{4;4}, which
completes the proof.

Acknowledgments. The authors wish to thank the referee for his/her valuable remarks and suggestions.

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Manoscritto pervenuto in redazione il 26 Maggio 2013.