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


Academic year: 2022

Partager "A graph related to the join of subgroups of a finite group"


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


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 ifGˆ 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.


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 H1 andH2 are joined by an edge ifhH1;H2i ˆG. 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 Sylp(G) andnp:ˆ jSylp(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 verticesv1andv2we write v1v2. If D is connected we denote by d(v1;v2) the length of a shortest path between v1 and v2. The diameter of D is defined as d(D):ˆmaxfd(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.

IfND[S]ˆV(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 Gˆ hH1;H2i.

REMARK2.2. LetGbe a group andHG. We haveHF(G) if and only if hH;Ki<ÿ4Gfor each 16ˆK<ÿ4G. 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 ˆPt


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))ˆ [tiˆ1[Hi]Gand so jV(D(G))j ˆXt


j[Hi]Gj ˆXt



Also for eachi2 f1;2;. . .;tgand eachg2G, we have deg(Hi)ˆdeg(Hgi).

Because there exists a bijectionu:ND[Hi]ÿ!ND[Hgi] such thatu(Hxj)ˆHxgj , where ND[Hi]ˆ fHxj 2V(D(G))jHxj Hig forx2G;1jt. Therefore jND[Hi]j ˆ jND[Hgi]jandD(G) is (n1;n2;. . .;nt)ÿregular. p PROPOSITION 2.4. For any group G we have x(D(G))ˆv(D(G))ˆ 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)GˆPP, where P is a group of prime order.

(2)GˆPjQ, where P and Q are groups of prime order.

(3)GˆQ8, 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 ˆM, that is hxi is not adjacent to M and by completeness of D(G) we have Mˆ hxi. Let jxj ˆpn11 pnkk, where thepi's are distinct prime numbers,k1,ni1 and 1ik.Thenthereexistelementsxi2Gsuchthatjxij ˆpnii;1ikand Mˆ hxi ˆ hx1i hx2i hxki. Obviously there exists j2 f1;2;. . .;kg such thathxji<ÿj F(G). Again by completeness ofD(G) we haveMˆ hxjiand so Mis a cyclicpÿgroup, wherepˆpj. Now two cases hold:

Case 1. F(G)ˆ1. If jp(G)j ˆ1, then G is an elementary abelian pÿgroup, that is,G G

F(G)Zpn, for somen2. Ifn3, then clearly


D(G) is not complete. Hencenˆ2 and soGZpZp 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 G0, the derived subgroup ofG. So there exists a subgroup K ofGsuch thatGˆG0jK andjG0j andjKjare prime num- bers. HenceGhas type (2).

Case 2. F(G)6ˆ1. Let M1;M22 M(G). There exist prime numbers p1;p2 and positive integersn1;n2 such thatjM1j ˆpn11,jM2j ˆpn22. Since F(G)6ˆ1,p1ˆp2. This means that all maximal subgroups ofGare cyclic pÿgroup, for some primep. HencejGj ˆpn, 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) Zpn (ii) Zpnÿ1Zp

(iii) hx;ajxpˆ1ˆapnÿ1;axˆa1‡pnÿ2i;n3 (iv) The dihedral groupD2n;n3, of order 2n

(v) Generalized quaternion groupQ2nˆ hx;yjx2nÿ1ˆ1;y2ˆx2nÿ2, xyˆxÿ1i;n3

(vi) Semidihedral group SD2nˆ hx;ajx2ˆ1ˆa2nÿ1;axˆa2nÿ2ÿ1i;


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 sonˆ2. ThusGhas type (1).

Now suppose that (iii) holds. Then clearly hxi and hx;apnÿ2i 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. Ifnˆ3, thenGˆQ8andD(G)ˆK3so D(G) is complete. Whenn4, it suffices to consider two distinct vertices hyi andhy;x2nÿ3i 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


N .

PROOF. (1) Let H1;H2 be arbitrary vertices of V(D(G)). Since Hi<ÿj F(G), iˆ1;2, there exist Mi2 M(G) such that HiMi (may be M1ˆM2). ThereforehH1;M1i ˆGˆ hH2;M2iandH1M1M2H2, which means thatD(G) is connected andd(D(G))3.

(2) Let M1;M22 M(G) and take z2Gn(M1[M2). If hzi<ÿ4G, then we have a triangle in D(G) with vertices fM1;hzi;M2g, hence g(D(G))ˆ3. Thus let Gˆ hzi and jzj ˆp1n1 pknk, where the pi's are distinct prime numbers k2 and ni1 for 1ik. Hence GZp1n1...pknk. If k3, then jM(G)j 3 and clearly g(D(G))ˆ3. Now letkˆ2. Ifn1ˆ1 or n2ˆ1, 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))ˆ4.

(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:NG(H)j jH2V(D(G))g ‡1:

PROOF. (1) Let G :ˆ 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 ˆpq, whereqis a prime. Since Gis non-cyclic of prime power order, we haveGZpZq, 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 HgM 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 groupZmp, 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))ˆ1 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 P1, or G is a pÿgroup of order pn, where n is a pos- itive integer and G is isomorphic to one of the groups:

(1) ZpZpnÿ1, p is a prime number

(2) hx;ajxp ˆ 1ˆ apnÿ1;ax ˆa1‡pnÿ2i; p is a prime number and n3

(3) The dihedral group D2n;n3, of order2n (4) Generalized quaternion group Q2n;n3

(5) Semidihedral group SD2nˆ hx;ajx2 ˆ1ˆa2nÿ1;axˆa2nÿ2ÿ1i;


PROOF. First let G be a solvable group with Sylow basis fP1;P2g, 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 fP1g. IfH<P1, thenHF(G)P1and maximality ofF(G) inP1 implies that HF(G)ˆP1. Since GˆP1P2ˆHF(G)P2ˆHP2, and thus HˆP1 a contradiction. Therefore H<ÿj P1 and so hH;P1i ˆG, that is HP1.

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

F(G)Zmp, m2. By [10, Theorem 5.3.4] G has a cyclic maximal subgroupA. Let A :ˆ A

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

a2G andjAj ˆ pmÿ1. SinceG is an elementary abelianpÿgroup,mˆ2.

ThusjF(G)j ˆpnÿ2. IfK2V(D(G))n fAg, thenK<ÿj A, becauseAis cyclic.

ThereforehK;Ai ˆGand soKA. Thus ifGhas one of the types (1)-(5), theng(D(G))ˆ1.

Conversely suppose that g(D(G))ˆ1 and Dˆ 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 Aˆ 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 ˆpn, 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;. . .;Pl;Pg, l1.

Without loss of generality we can assume thatPiA;1il, (see [8]). If P1<ÿ4A, then P1 is not adjacent with A, a contradiction. Thus AˆP1, 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 GQ8orZp2 1p22 or Z2Z2, where p1;p2are distinct prime numbers.

(2)D(G)is bipartite if and only if GZpn1

1 pn22 where ni1;i2 f1;2g.

PROOF. (1) IfGis one of the groupsQ8 orZ2Z2, thenD(G)ˆK3. Also if GZp2

1p22, then D(G)ˆC4 where Cn 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)ˆC5 and V(D(G))ˆ fH1;H2;H3;H4;H5g and D(G) is:

H1H2H3H4H5H1. It is easy to see that jM(G)j ˆ2, say M(G)ˆ 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)ˆC3 soD(G) is complete and by Theorem 2.5,Gis isomorphic to one of the following:ZpZporZpjZq orQ8. ClearlyV(D(G))ˆ M(G). IfGZpZp, thenD(G)ˆKp‡1and so pˆ2. HenceGZ2Z2. Now suppose, if possible, thatGZpjZq. We can assume thatp6ˆqand so we haveV(D(G))ˆSylp(G)[Sylq(G). Thus jV(D(G))j ˆnp‡nqˆ1‡p>1‡2ˆ3 a contradiction. Therefore this case is impossible.


Now let D(G)ˆC4. Then jM(G)j ˆ2 and Gˆ hgi is a cyclic group.

Suppose that jgj ˆpn11pn22 pnkk, where the pi's are distinct primes and ni1;1ik. Since jM(G)j ˆ2 we conclude that kˆ2 and GZpn1

1pn22. LetV(D(G))ˆ fH1;H2;H3;H4gsuch thatM(G)ˆ fH1;H2g andD(G) is the cycle:H1H2H3H4H1. Ifn13 orn23, then D(G) contains K1;3 as a subgraph, a contradiction. Therefore (n1;n2)2 f(1;1);(1;2);(2;1);(2;2)g. If (n1;n2)2 f(1;1);(1;2);(2;1)g, thenjV(D(G))j 2 f2;3g. contradictingjV(D(G))j ˆ4. Thus (n1;n2)ˆ(2;2), that isGZp2


and the proof is complete.

(2) If D(G) is bipartite, it contains no odd cycle and this implies that jM(G)j ˆ2. 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 typesZp3

1p32,Z3Z3or S3, where p1 and p2are distinct primes and S3 is the symmetric group on three symbols.

(2)D(G)is4ÿregular if and only if GZp4

1p42. PROOF. (1) IfGZp3

1p32 , 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)ˆK3;3. If GˆS3 or GˆZ3Z3, thenD(G)ˆK4.

Conversely let D(G) be 3ÿregular. Thus jM(G)j 4. First let jM(G)j ˆ2. ThereforeGis cyclic with two distinct prime divisorsp1;p2. Note that the maximal subgroups have no common neighbors. So jV(D(G))j ˆ6 and thus GZp3

1p32. Now suppose that jM(G)j ˆ4. Then D(G)ˆK4and by Theorem 2.5, GPjQorGˆPP, whereP andQ are groups of prime order. IfGˆPP, thenD(G)ˆKp‡1 and sopˆ3,

Fig. 1.


that isGZ3Z3. IfGPjQ, thenV(D(G))ˆSylp(G)[Sylq(G) and obviously we havejPj ˆ3 andjQj ˆ2. ThusGS3.

Finally letjM(G)j ˆ3. By [2, Lemma 1]Gis a cyclic group of order pnqmrk, 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 ˆ6 and any two of the maximal subgroups of G have no common neighbors as shown in Figure 1. Now note that F(G)ˆMi\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) LetGZp4

1p42. ThenjV(D(G))j ˆ8 andjM(G)j ˆ2. SoD(G) has six non-maximal vertex and clearly it is 4ÿregular. In factD(G)ˆK4;4.

Conversely ifD(G) is 4ÿregular, thenjM(G)j 5. First letjM(G)j ˆ5.

SoD(G)ˆK5and by Theorem 2.5, we haveGPPorGPjQwhere PandQare groups of prime orders. By an easy inspection we can see that neither case can happen. We claim that ifjM(G)j ˆ4, 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)ˆMi\Mj, for eachMi;Mj2 M(G);1i;j4.

Since deg(H)ˆ4,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 ˆ3. Similar to the proof of (1), we havejV(D(G))j ˆ6 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 orderpnqmrk, wherep;q;r are distinct primes and n;m;k1 or G is a 2ÿgroup such that

F(G)G Z2Z2. 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 Mˆ hxi. Therefore F(G) is a maximal subgroup ofHand alsoHis maximal inMand so we have:

jG:F(G)j ˆ jG:MjjM:HjjH:F(G)j ˆ23; which is a contradiction.

Finally letjM(G)j ˆ2. 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. ThereforeGZp4

1p42andD(G)ˆK4;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.


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• In the second one (§ 4), probably the most interesting one, we will show that one can recover the metric structure of the graph of the stable model of a Mumford curve (i.e. a

Our main motivation is an application to the recent polynomial time primal- ity testing algorithm [AKS].. We retain the above notation throughout the

For example, the following result was proved: a cyclic group of prime order is conjugate to a linear automorphism of the plane if and only if it does not fix a curve of positive

Moreover, the dimension of the projective cover of the simple module associated to the conjugacy class of the p-regular element g ∈ G is equal to the number of conjugacy classes

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

More generally, a group G is said to be inertial of elementary type if there are normal subgroups F and A, with F A, such that jFj and jG : Aj are finite, A=F is abelian and