GROUPS WITH A MAXIMAL IRREDUNDANT 6-COVER #

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1 Communications in Algebra, 33: , 2005 Copyright Taylor & Francis, Inc. ISSN: print/ online DOI: /AB ROUPS WITH A MAXIMAL IRREDUNDANT 6-COVER # A. Abdollahi, M. J. Ataei, S. M. Jafarian Amiri, and A. Mohammadi Hassanabadi Department of Mathematics, University of Isfahan, Isfahan and Institute for Studies in Theoretical Physics and Mathematics (IPM), Tehran, Iran A cover for a group is a collection of proper subgroups whose union is the whole group. A cover is irredundant if no proper sub-collection is also a cover, and is called maximal if all its members are maximal subgroups. For an integer n>2, a cover with n members is called an n-cover. Also, we denote = n if has an n-cover and does not have any m-cover for each integer m<n. In this article, we completely characterize groups with a maximal irredundant 6-cover with core-free intersection. As an application of this result, we characterize the groups with = 6. The intersection of an irredundant n-cover is known to have index bounded by a function of n, though in general the precise bound is not known. We also prove that the exact bound is 36 when n is 6. Key Words: Covering groups by subgroups Mathematics Subject Classification: 20F INTRODUCTION AND RESULTS A covering or cover for a group is a collection of subgroups of whose union is. We use the term n-cover for a cover with n members. The cover is irredundant if no proper subcollection is also a cover, and is called maximal if all its members are maximal subgroups of. A cover is called a core-free intersection if the core of its intersection is trivial. Scorza (1926) determined the structure of all groups having an irredundant 3-cover with core-free intersection. Proposition 1.1 (Scorza, 1926). Let A i 1 i 3 be an irredundant cover with core-free intersection D for a group. Then D = 1 and C 2 C 2. reco (1956) characterized all groups having an irredundant 4-cover with core-free intersection. Received May 17, 2004; Revised August 18, 2004 # Communicated by A. Turull. Address correspondence to A. Abdollahi, Department of Mathematics, University of Isfahan, Isfahan , Iran; Fax: ; a.abdollahi@math.ui.ac.ir 3225

2 3226 ABDOLLAHI ET AL. Proposition 1.2 (reco, 1956). Let A i 1 i 4 be an irredundant cover with core-free intersection D for a groups. If the cover is maximal, then either (1) D = 1 and Sym 3 or C 3 C 3 ;or (2) D =2, =18 and embeds into Sym 3 Sym 3. If the cover is not maximal, then either (1) D = 1 and D 8,or C 4 C 2,or C 2 C 2 C 2 ;or (2) D =2 and D 8 C 2, where D 8 is the dihedral group of order 8. We shall frequently use the following lemma and we state it here for the reader s convenience. Lemma 1.3 (See Bryce et al., 1997, Lemma 2.2). Let = A i 1 i m be an irredundant covering of a group whose intersection of the members is D. (a) If p is a prime, x a p-element of and i x A i =n, then either x D or p m n. (b) j i A j = D for all i 1 2 m. (c) If i S A i = D and S =n, then i T A i D m n + 1 whenever T =n 1. (d) If is maximal and U is an abelian minimal normal subgroup of. Then, if i U A i =n, either U D or U m n. Neumann (1954) proved that if has an irredundant n-cover then the index of the intersection of the cover in is bounded by a function of n and Tomkinson (1987) improved that bound. We shall write f n for the largest index D over all groups having an irredundant n-cover with intersection D. Bryce et al. (1997) characterized groups with maximal irredundamt 5-cover with core-free intersection and as a result they obtained an exact value for f 5, namely f 5 = 16. If is a group having an n-cover but not any m-cover with m<n, then we write = n. It is clear that there is no group, with = 2 and Cohn (1994, Theorem 5) characterized groups with = 3 4 or 5. He also determined finite groups whose proper factor groups do not have a 6-cover but = 6 (see Cohn, 1994, Theorem 6). Further problems of a similar nature, with slightly different aspects, have been studied by many authors (see Bryce and Serena, 2001; Bryce et al., 1999, 1995; Cohn, 1994; reco, 1953; Jungnickel and Storme, 2001; Tomkinson, 1987). Here, we characterize all groups having a maximal irredundant 6-cover with core-free intersection. As an application of this characterization, we obtain a necessary and sufficient condition for groups with = 6. Tomkinson (1987), has shown that 36 f Here, we have an exact value for f 6, namely f 6 = 36. Our main results are as follows. Theorem A. A group having a maximal irredundant 6-cover with core-free intersection is nilpotent if and only if C 5 C 5 or C 3 C 3 C 3.

3 6-COVER ROUPS 3227 A finite group is called semisimple if it has not nontrivial normal abelian subgroups (see Robinson, 1982, p. 86 for further information on such groups). Theorem B. Semisimple groups do not have a maximal irredundant 6-cover with core-free intersection. Theorem C. Let be a group. Then has a maximal irredundant 6-cover with corefree intersection D if and only if satisfies one of the following properties. (1) D =1 and C 5 C 5 ; (2) D =1 and C 3 C 3 C 3 ; (3) D =1 and Sym 3 Sym 3 ; (4) D =1 and C 3 C 3 C 2 with Z = 1; (5) D =2 and C 3 3 C 2 with Z = 1; (6) D =1 and C 2 C 2 Sym 3 or C 2 0 where 0 = C 3 C 3 C 2 with Z 0 = 1; (7) D =1 and C 5 C 2 or C 5 C 4 and Z = 1; (8) D =2 and C 5 C 5 C 2 with Z = 1; (9) D =4 and C 5 C 5 C 4 with Z = 1. Theorem D. The largest index D over all groups having an irredundant 6-cover with intersection D is 36. Theorem E. Let be a finite group with > 5. Then = 6 if and only if has a factor group isomorphic to one of the following groups: (1) C 5 C 5, (2) D 10, the dihedral group of order 10, (3) H = a b a 5 = b 4 = 1 ba= a 2 b. Throughout, we write 6 to denote the class of all groups having a maximal irredundant 6-cover with core-free intersection D. For a group in 6, we assume that M i 1 i 6 is a maximal irredundant 6-cover, with core-free intersection D = 6 i=1 M i. We note that 6 -groups are finite. Also, we use the usual notations; for example, C n denotes the cyclic group of order n C n j is the direct product of j copies of C n ; the core of a subgroup H of is denoted by H and we write, H, to say that H is embedable in. 2. NILPOTENT 6 -ROUPS In this section we characterize all nilpotent 6 -groups. First, we need some preliminary lemmas. Lemma 2.1. Let be a 6 -group. Then (1) is not a 2-group. (2) If is a 3-group, then C 3 C 3 C 3. Moreover C 3 3 is a 6 -group.

4 3228 ABDOLLAHI ET AL. Proof. (1) Aiming for a contradiction, suppose that is a 2-group. Since the Frattini subgroup of, = 2 D, we have D. Therefore, D = 1 and is an elementary abelian 2-group. By parts (b) and (c) of Lemma 1.3, we have M i M j M k M t 2 whenever i j k and t are distinct. Therefore 32. Also 8, since otherwise does not have six distinct maximal subgroups. If =8, then M i M j =2, whenever i j and so for distinct i j and k, M i M j M k 2. This is a contradiction, since the cover is irredundant. If =16, then M i M j =4. We conclude that M i M j M k = 1 and the inclusion-exclusion principle gives 16, which is a contradiction. If =32, then for i j, M i M j =8 and so M i M j M k 8 for distinct i j k. Thus, M i M j M k 4. If M i M j M k 8, then M i M j M k =32, which contradicts the irredundancy of the cover. Therefore M i M j M k =4. Now, by parts (b) and (c) of Lemma 1.3, M i M j M k M t =2, where all indices are distinct. Therefore =31, the final contradiction. (2) Suppose that is a 3-group. A similar argument as in (1) gives that is an elemantary abelian 3-group and it follows from Lemma 1.3(a) that M i = 1 ( ) i T for every subset T with T =4. Therefore 81. Since C 3 C 3 has only four maximal subgroups, we have 27. Assume that =27. Then C 3 C 3 C 3. Now it is easy to see that (e.g. by AP, 2002) if = a b c, then the following maximal subgroups M 1 = c a 1 b M 2 = a 1 c b M 3 = a 1 c ab M 4 = b 1 c a M 5 = ac ab M 6 = a 1 c a 1 b form a maximal irredundant 6-cover with core-free intersection for. Now suppose, for a contradiction, that =81. Then M i M j =9 for i j. If M i M j M k =9 for some distinct i j k, then M i M j M k = M i M j, and so by, 27, a contradiction. Thus k L M k =3 for all L with L =3. Now, from inclusion-exclusion principal it follows that =77, the final contradiction. Proposition 2.2. Let be a 6 -group. Then is a p-group for a prime p if and only if C 5 C 5 or C 3 3. Proof. Let be a p-group (p a prime), having the property 6. D, since = p D. Therefore D = 1 and is an elementary abelian p-group. By Lemma 1.3(a), p 5 and by Lemma 2.1 we may assume that p 2 3. Hence assume p = 5. Then Lemma 1.3(a), yields that =25 and so C 5 C 5. The converse holds, since an elementary abelian group of order 25 contains six subgroups of order 5 and by (Cohn, 1994, Lemma 3) these provide a maximal irredundant 6-cover with core-free intersection for. Now Lemma 2.1(2) completes the proof. Proof of Theorem A. Suppose that is a nilpotent group. Firstly, by the hypothesis, we have that M i is normal in for all i and D = 1. Thus every prime divisor p of is at most five. Hence is a group

5 6-COVER ROUPS 3229 and by the hypothesis C 2 i C 3 j C 5 k. By Lemma 1.3, every non-trivial 5-element lies in exactly one M i. Assume that 5 divides. Then there is at most M i whose index is either 2 or 3. Thus, there are at least five M i s of index 5. Since i T M i = 1 for all T with T =5, we have that C 5 k. Now Proposition 2.2 implies that C 5 C 5. Therefore, we may assume that C 2 i C 3 j. Suppose, for a contradiction, that i>0. By a similar argument as above, there are at least two M i s whose indices are 2. If there are exactly two M i s, say M 1, M 2, of index 2, then 4 divides. On the other hand, 6 i=2 M i = 1 and so =2 3 j, a contradiction. Thus there are at least three M i s with index 2. If there are four M i s of index 2, then it follows from Lemma 1.3(c) that is a 2-group, contrary to Lemma 2.1(1). Thus, there are exactly three M i s, M 1, M 2, M 3 say, of index 2. Hence all 3-elements of lie in M 1 M 2 M 3. Therefore, by Lemma 1.3(a) M l contains no nontrivial 3-element for every l Since is abelian, it has a unique Sylow 2-subgroup. It follows that M 4 = M 5 = M 6, a contradiction. Hence, we may assume that C 3 j. Thus the result follows from Proposition SEMISIMPLE ROUPS In this section, we prove Theorem B and to do so throughout we let denote a semisimple 6 -group and M i 1 i 6 be a maximal irredudant 6-cover for, with core-free intersection D = 6 i=1 M i. Also, for each i, 1 i 6 we assume that M i = i such that We shall need the following lemmas. Lemma 3.1. M i M j M 1 whenever i j 2 and i j, and 2 = 3 = 4 = 5 = 6 = 5. Proof. If 1 2 4, then by the hypothesis M 1 M 2 Sym 4 Sym 4 If M 1 M 2 = 1, then is soluble, which it is not, since is semisimple. If M 1 M 2 1, then every 5-elements of lies in M 1 M 2. Now applying Lemma 1.3 with p = 5 and m = 6, we get that every 5-element lies in D. Since D = 1 is a 2 3 -group. Thus, is soluble, which again it is not. Now it follows from Lemma 3.1 of Tomkinson (1997), that 2 = 3 = 4 = 5 = 6 = 5 and M i M j M 1 for distinct i j 2. Lemma 3.2. D = M 2 M 3 M 4. Proof. From Lemma 1.3(b) and Lemma 3.1, we have D = M 1 M 2 M 3 M 4 M 5 = M 2 M 3 M 4 M 5 (I) and by Lemma 1.3(c), M 1 M 2 M 3 D = 3. Thus Lemma 3.1 gives M 1 M 2 M 3 = M 2 M 3 and so M 2 M 3 D 3 (II)

6 3230 ABDOLLAHI ET AL. On the other hand, Lemma 1.3(c) implies that M 1 M 2 M 3 M 4 D = 2 and so M 2 M 3 M 4 D 2. If M 2 M 3 M 4 D =2 then by (II), we have M 2 M 3 D =2. Therefore, M 2 M 3 M 4 = M 2 M 3. Of course, in this equation, we may replace M 4 by M 5 or M 6. Then, it follows from (I) that D = M 2 M 3 contradicting M 2 M 3 D =2. Thus M 2 M 3 M 4 D =1, as required. Remark 3.3. The only primitive subgroups of Sym 5 with order divisible by 5 are C 5 C 5 C 2 C 5 C 4, Alt 5 and Sym 5. Lemma 3.4. For every i 2 M i 1. Proof. If M i = 1 for some i 2, then is a primitive subgroup of Sym 5. Therefore, by Remark 3.3, the following case arise: (a) C 5 C 5 C 2,orC 5 C 4 ; (b) Alt 5 or Sym 5. But case (a) contradicts the semisimplicity of, and case (b) contradicts Lemma 7 of Cohn (1994). Lemma 3.5. There exists at least one i such that Alt M i 5 or Sym 5. Proof. Assume that for every i 1, Alt M i 5 or Sym 5. Thus, Remark 3.3 implies that is soluble for every i 1. Now by Lemma 3.2, M i On the other hand, M 2 M 3 M 4 = D = 1 M 2 M 3 M 4 Therefore is soluble, which is a contradiction. Lemma 3.6. Let U be a minimal normal subgroup of. Then U Alt 5. Proof. Since D = 1 U D. Thus U M i for some i 1 and so U M i. Therefore U M i = 1. But = is a primitive subgroup of Sym M i 5 whose order is divisible by 5, since by Lemma 3.1 M i = i = 5 and M i is a maximal subgroup of. On the other hand, has a minimal normal subgroup U = U M i U so, by Remark 3.3, U C M i 5 or Alt 5 and since is semisimple, U Alt 5. Proof of Theorem B. Suppose that is a semisimple 6 group. To obtain a contradiction, we distinguish two cases. Case 1. Suppose that there are two maximal subgroups, say M 2 M 3 such that M 2 M 3 1. Therefore, there exists a minimal normal subgroup N

7 6-COVER ROUPS 3231 which is contained in K = M 2 M 3. So every 5-element of N lies in K and then by Lemma 1.3(a) lies in D. By the core-freeness of D N is a normal 2 3 subgroup of, which is contrary to Lemma 3.6. Case 2. For every two distinct i j 2 M i M j = 1. By Lemma 3.4 every M i for i 2, contains a minimal normal subgroup N i of and by the hypothesis N i s are distinct. Now Lemma 3.6 implies that On the other hand, in this case, embeds in Sym 5 Sym 5, which is not possible. This completes the proof. 4. NON-SEMISIMPLE ROUPS According to Theorem B, to characterize all 6 -groups, we only need to consider non-semisimple 6 -groups and this is the aim of this section. Our main proof faces a number of cases, in each of which we have to consider certain subgroups of, Sym 3 Sym 3 Sym 3 Sym 4 Sym 4 and Sym 5 Sym 5. Lemma 4.1. (1) The following are not 6 -groups. (a) Sym 3 Sym 3 Sym 3 and C 2 C 3 Sym 3. (b) All subgroups of Sym 3 Sym 3 Sym 3 of orders 72 or 108. (c) All subgroups of Sym 4 Sym 4 of orders 48 or 96. (2) The following are 6 -groups; (d) C 5 C 5 C 4 with Z = 1. In this case, the intersection of a maximal irredundant cover which is core-free has size 2. (e) C 5 C 5 C 4 with Z = 1. In this case, the intersection of a maximal irredundant cover which is core-free has size 4. (3) A subdirect product of three symmetric groups Sym 3 is a 6 -group if and only if is isomorphic to one the following groups: C 3 C 3 C 2 Sym 3 Sym 3 C 3 3 C 2, where the size of the intersection of an arbitrary maximal irredundant 6-cover with core-free intersection is 1, 1, and 2, respectively. Proof. We have used the following function written with AP (2002) program to show this lemma. The input of the function is a group and the outputs are all the irredundant 6-covers with core-free intersection for, and if there is no such a cover for, the output is the empty list. f:= function() local S, M, n, C, i, T, Q, R; n:=size(); M:=MaximalSubgroups(); C:=Combinations(M,6); S:=[]; for i in [1..Size(C)] do if Size(Union(C[i]))=n then Add(S, C[i]); fi; od; T: = []; for i in [1..Size(S)] do if Size(Core(, Intersection(S[i])))=1 then Add(T,S[i]); fi; od; R:=[]; for i in [1..Size(T)] do

8 3232 ABDOLLAHI ET AL. Q:=Combinations(T[i],5); if (n in List(Q, i->size(union(i)))=false then Add(R,T[i]); fi; od; return R; end; Recall that a group is a subdirect product of a family of groups i i I if there exists a family of normal subgroups N i i I of such that i I N i = 1 and /N i i for all i I. One may use the following command in AP (2002) for the part (3) of the lemma, to obtain all subdirect products of three Sym 3. SubdirectProducts(1,2); First we obtain all subdirect products H i of two Sym 3, and then we find all subdirect products of each H i with one Sym 3. These last groups are all subdirect products of three Sym 3. Towards the proof of our main results here we need to consider a number of special situations which may arise in a 6 -group. Proposition 4.2. If is a 6 -group and has a minimal normal subgroup of order two, then D = 1 and C 2 0, where 0 = C 3 C 3 C 2 with Z 0 = 1 or C 2 C 2 Sym 3. Proof. Suppose that U is a minimal normal abelian subgroups of and U =2. Then by Lemma 1.3(d), U is not contained in at least two M i s, say U M 5 M 6. Since U is central, = M 5 U = M 6 U and M 5 M 6. The 5-elements of, if any, are in M 5 M 6 so, by Lemma 1.3(a), they are in D. It follows that is a 2 3 group, and so is soluble. Suppose that V is a minimal normal abelian subgroup of, contained in M 5 M 6. Note that V 4. (i) If V =2 then by Lemma 1.3(d), V M 3 M 4.So M 3 = M 4 =2 and so every 3-element of is in 6 i=3 M i and therefore by Lemma 1.3(a), is in D. Since D = 1 is a 2-group, a contradiction to Lemma 2.1. (ii) If V =3 then M i =3 for i = and = VM i for i = By the hypothesis embeds into Aut V C C V 2. Therefore = C V or C V =2. If = C V, then V is central and M i is normal in for i = 2 3 4, so D = 1 and is abelian, which contradicts Theorem A. Thus C V =2. We claim that V and U are contained in M 1. By Doerk and Hawkes (1992, Proposition 16.8, p. 59), C V = V M i for i = If V M 1, then = VM 1 and C V = V M 1. Therefore, every 2-element of C V is in 4 i=1 M i. Now if 1 v V and x is a 2-element of C V, then vx M 5 M 6. Therefore, x M 5 M 6 and so is in D. By Lemma 1.3(a), and core-freeness of D C V is an elementary abelian 3-group. This contradicts the fact that U is in C V, sov M 1. If U M 1, then = UM 1 and M 1 =2. Therefore, every 3-element of lies in M 1 M 5 M 6. An argument similar to that of previous paragraph shows that is a 2-group, which is contrary to Lemma 2.1(1), so U M 1.

9 6-COVER ROUPS 3233 Note that, by our assumptions in (ii), embeds in C 2 C 2 Sym 3 Sym 3 Sym 3, so since M 1 is a maximal subgroup of, we have M 1 =2 or 3. Now we distinguish two cases: Case 1. Suppose that M 1 =2. Then M 1 M 5 M 6 M 2 = 1, since otherwise has a normal subgroup of order 2, which contradicts M 3 =3. Therefore, embeds into C 2 C 2 C 2 Sym 3. If K = M 5 M 6 M 2 1 then, being normal, it contains a minimal normal subgroup, say W. Now, K M 3 M 4 D, sow is not in both M 3 M 4. Therefore W =3 yielding W M 1 as M 1 has index 2. However, this contradicts Lemma 1.3(d), which requires W 2. Therefore, K = 1 and embeds in C 2 C 2 Sym 3. Also, since is not nilpotent, we have Sym M i 3 for i = Now, as M 2 =3 M 2 =2or4. If M 2 =2 then =12 and C 2 Sym 3, which is impossible, since C 2 Sym 3 has just six maximal subgroups, which form a redundant 6-cover. If M 2 =4 then =24, from which it follows that C 2 C 2 Sym 3. Case 2. Suppose that M 1 =3. Then M 1 M 2 M 5 = 1. For, otherwise, has a minimal normal subgroup of order 3, which is not contained in M 6. This is impossible. Hence embeds in C 2 Sym 3 Sym 3. To finish (ii), assume that M 1. Then, embeds into C 2 C 3 Sym 3.Itis clear that 9 divides and so is a multiple of 18. If =18, then is abelian in this case. Thus =36 and = C 2 C 3 Sym 3. Since has a central normal subgroup of order 3, it has five normal maximal subgroups of indices 2 or 3 in the cover. Therefore is abelian, which is wrong. Consequently, M 1 is not normal in and so 36 divides. If =72, then = C 2 Sym 3 Sym 3 and by Lemma 4.1, such a group is not a 6 -group. Hence =36. But the group C 2 Sym 3 Sym 3 has one subgroup of order 36, which is a 6 -group and so C 2 0, where 0 = C 3 C 3 C 2 and Z 0 = 1. (iii) If V =4, then again by Lemma 1.3(d), V M i 1 i 4. So M i =4 for i = and 4. Now apply Lemma 3.2 of Tomkinson (1997) to obtain M i M j M 5 M 6 for any two distinct i j Since M i =4 for i = 1 2 U M 1 M 2, which contradicts the choice of U. Lemma 4.3. Let be a 6 -group. Suppose that has a normal subgroup of order 3 and none of order 2. Then C 3 C 3 C 2, C 3 3 Sym 3 Sym 3 or C 3 3 C 2, where the size of the intersection of an arbitrary maximal irredundant 6-cover with corefree intersection is 1,1,1, and 2, respectively. Proof. Suppose that U is a normal subgroup of order 3. Then U is contained in at most three subgroups of M i s, say M 1 M 2 or M 3. Therefore, = UM i and M i =3 for i = Now, we claim that 6 i=4 M i = 1or 3 i=1 M i = 1. Suppose that K = M 4 M 5 M 6 is nontrivial. Then there is a minimal normal subgroup V of contained in K. Therefore, V =3 by the hypothesis and Lemma 1.3(d), we have M i =3 for i = By applying Lemma 3.2 of Tomkinson (1997) we have, M i M j M 4 M 5 M 6 ( )

10 3234 ABDOLLAHI ET AL. It follows that T = M 1 M 2 M 3 is trivial. For, otherwise T is a 2-group, since every 3-element of T lies in at least four subgroups by. Therefore, has a normal subgroup of order 2 contained in T, which is a contradiction. Suppose that K = 1. If M i for some i 4 5 6, then U is a central subgroup and so M j for all i Therefore is a 3-group and so C 3 3, by Lemma 2.1(2). Therefore, we may assume that M i for all i 4 5 6, which yields Sym M i 3 for all i Hence is a subdirect product of three symmetric groups Sym 3. Now Lemma 4.1(3) completes this part of the proof. If K 1, then T = 1 and by a similar argument the proof will complete. Lemma 4.4. If is a 6 -group and has a minimal normal subgroup of order 4, then is a subgroup of Sym 4 Sym 4. Proof. Suppose that U is a minimal normal subgroup of order 4. Then U is not contained in at least four M i s, say, U M 3 M 4 M 5 M 6.So = UM i and M i =4 for i = By Proposition 4.2 and Lemma 4.3, does not have a normal subgroup of order 2 or 3. If M 3 M 4 is nontrivial, then has a minimal normal subgroup of order 4 contained in M 3 M 4 and by Lemma 3.2 of Tomkinson (1997), we have M i M j M 3 M 4 for any two distinct i j For each i 3 4, we have M i M 5 M 6 = 1, since otherwise it contains a minimal normal subgroup of order 4, which contradicts Lemma 1.3(d). Therefore, M 5 M 6 = 1 and the proof is complete. Proposition 4.5. There is no 6 -group with a minimal normal subgroup of order 4. Proof. Suppose that is a 6 -group having a minimal normal subgroup U of order 4 and U M 3 M 4 M 5 M 6.So = UM i and M i =4 for i = It follows that C U = U M i and every 3-element of C U lies in 6 i=3 M i and so lies in D. Thus C U is an elementary abelian 2-group and =2 m 3 for some positive integer m. embeds into Sym 4 Sym 4 by Lemma 4.4. Obviously, Sym 4 is not a 6 -group. Therefore, has at least two minimal subgroups of order 4 and so 16 divides. Thus is a multiple of 48. By corefreeness = U H such that H C U = 1 and H is a subgroup of Sym 3. Also, C U = U M i = M i M j for any two distinct i j and so C U =16. Thus = 48 or 96. On the other hand by Lemma 4.1 such a group is not a 6 -group, which is a contradiction. Proof of Theorem C. Suppose that is a 6 -group. By Theorem B, we may assume that is not semisimple. So has a minimal normal abelian subgroup U. Now by Lemma 1.3, U 5 and we deduce from Proposition 4.5 that U 4. If U =2, then Proposition 4.2 completes this part of the proof. Now we may assume that has no normal subgroup order 2. If U =3, then Lemma 4.3 completes this part of the proof. If U =5 then U is not contained in five of the subgroups M i s, say for i Thus M i =5 = UM i and C U = U M i for i 2. Now it is clear that C U is an elementary abelian 5-group and embeds into C U

11 6-COVER ROUPS 3235 Aut U C 4. Note that if Z 1, by our argument we may choose U so that U Z, which implies = U M i for each i 2. If C U = U, then is a primitive subgroup of Sym 5. But by Remark 3.3 and Lemma 7 of Cohn (1994), the only primitive subgroups of Sym 5 which have the property 6 are C 5 C 2 and C 5 C 4. Now suppose that C U U. Then M i M j = 1 for i>j 2 and so C U C 5 C 5.IfZ 1, then as we mentioned above, = U M i for each i 2. It follows that C 5 C 5. Thus assume that Z = 1. Therefore, is isomorphic to C 5 C 5 C 2 or C 5 C 5 C 4. This completes the proof in one direction. The converse is clear (see parts (d) and (e) of Lemma 4.1). 5. THE EXACT VALUE FOR f 6 Proof of Theorem D. By Corollary 3.2 of Tomkinson (1987), f Suppose, contrary to what we want to show, that is a group with an irredundant 6-cover with core-free intersection D such that D > 36. By Theorem C is not maximal. Suppose is chosen from among such 6-covers of with as many maximal subgroups as possible. Let be a cover of got from by replacing one of its non-maximal sub-groups by a maximal subgroup containing it. Let D be the intersection of and note that D D. is redundant; for, otherwise, D = D by Lemma 1.3(b), and so D = 1, while has more maximal subgroups than does. It follows that we may write = 6 i=1 A i, where A 1 is not maximal and if A 1 is a maximal subgroup containing it, then = A 1 A 2 A 3 A 4 A 5 A 6 is redundant as a cover of. If is an irredundant union of five subgroups in, then we may suppose that = A 1 A 2 A 3 A 4 A 5 since A 1 is certainly essential. If D 1 = A 1 A 2 A 3 A 4 A 5, then it follows from Theorem 1.1 of Bryce et al. (1997) that D But D 1 D 2 by Lemma 1.3(c). Then D 32, a contradiction. Now, if is an irredundant union of four subgroups in, we may suppose that = A 1 A 2 A 3 A 4 (1) If D 1 = A 1 A 2 A 3 A 4 by Proposition 1.2, D 1 9, where equality holds only if A 1 A i = D 1 2 i 4. If we have equality, then it follows that A 1 = A 1 D 1 A 1 A 5 A 1 A 6 (2) If this cover is irredundant, then A 1 D 9, and since A 1 3 D 27, a contradiction. Hence (2) is redundant. Now we distinguish three cases. Case 1. If A 1 = D 1 A 1 A 5 A 1 A 6 is irredundant, then by Proposition 1.1, A 1 D =4 and so D 12, which is a contradiction.

12 3236 ABDOLLAHI ET AL. Case 2. If A 1 = A 1 A 1 A 5 A 1 A 6 is irredundant and D 2 = A 1 A 1 A 5 A 1 A 6, then Proposition 1.1 yields that A 1 D 2 =4. On the other hand, A 1 D 1 divides D 1 =9 which implies that A 1 D 1 =3. It follows that A 1 D = A 1 D 1 D 2 =12. So D 36, which is not. Case 3. If A 1 = A 1 D 1 A 1 A 5 is irredundant, then A 1 D =4 and so D 12, which is impossible. Suppose that D 1 =8, and (1) is not maximal. Then by Proposition 1.2, there exists at least one i 2 3 4, such that A i is not maximal. Then A i 4. On the other hand, A 1 =2, since is nilpotent. Now, we have A D 1 1 A i =4or8.If A 1 A i =4, then A i =4, which yields that A 1 A i = A i. This contradicts the irredundance of (1). Hence A 1 A i =8. Therefore, since D 1 A 1 A i and D 1 =8, we have D 1 = A 1 A i. Now Lemma 1.3(c) implies that A 1 D 1 =2and so D 1 =4, a contradiction. So D 1 =6and then D 36 by Lemma 2.1 of Tomkinson (1987). Hence (1) is redundant. Now we may assume that This cover is certainly irredundant, so Now we have = A 1 A 2 A 3 A 1 =2 and D 1 = A 1 A 2 A 3 = A 2 A 3 D 1 =4 A 1 = A 1 D 1 A 1 A 4 A 1 A 5 A 1 A 6 (3) If this cover is irredundant, then by Theorem 1.1 of Bryce et al. (1997), A 1 D 16 and so D 32. Therefore (3) is redundant and again we have three cases. Case 1. If A 1 = D 1 A 1 A 4 A 1 A 5 A 1 A 6 is irredundant, then A 1 D 9 and D 18, a contradiction. Case 2. If A 1 = A 1 A 1 A 4 A 1 A 5 A 1 A 6 is irredundant, then F = A 1 A 4 A 5 A 6 and A 1 F 9. Now by Lemma 1.3 F D 2. Therefore, D = A 1 A 1 F F D = 36, which is not possible. Case 3. If A 1 = A 1 D 1 A 1 A 4 A 1 A 5 is irredundant, then A 1 D 9 and finally D = A 1 A 1 D 2 9 = 18. This final contradiction completes the proof of Theorem D. Proof of Theorem E. Suppose that = 6. Then, we may consider = 6 i=1 M i, where M i s are maximal subgroups of and D = 6 i=1 M i. Therefore, has a maximal irredundant 6-cover with core-free intersection D. It follows from D

13 6-COVER ROUPS 3237 Theorem C that is isomorphic to one the following groups, where if occurs, D the underlying group is assumed to be centerless: (1) C 3 3, (2) C 5 C 5, (3) C 2 C 2 Sym 3, (4) C 2 C 3 C 3 C 2, (5) Sym 3 Sym 3, (6) C 3 C 3 C 2, (7) C 3 3 C 2, (8) D 10 = C 5 C 2, (9) H = C 5 C 4, (10) C 5 C 5 C 2, (11) C 5 C 5 C 4. But and C 2 C 2 Sym 3 = C 2 C 3 C 3 C 2 = Sym 3 Sym 3 = 3 C 3 3 = C 3 3 C 2 = C 3 2 C 2 = 4 By the proof of Theorem C, the groups (10) and (11) in the above list have a factor group isomorphic to groups D 10 and H respectively. This completes the proof of the part only if. To prove the converse, suppose that has a factor group isomorphic to one of the groups listed in the statement of the theorem. Now Theorem 6 and Lemma 1.2 of Cohn (1994) yield that 6. On the other hand > 5, so = 6 as required. ACKNOWLEDMENT This research was in part supported by a grant from IPM (No ). REFERENCES Bryce, R. A., Serena, L. (2001). A note on minimal coverings of groups by subgroups. J. Austral. Math. Soc. 71: Bryce, R. A., Fedri, V., Serena, L. (1995). A Hughes-like property for finite groups. Proc. of the Edinburgh Math. Soc. 38: Bryce, R. A., Fedri, V., Serena, L. (1997). Covering group with subgroups. Bull. Austral. Math. Soc. 55: Bryce, R. A., Fedri, V., Serena, L. (1999). Subgroup coverings of some linear groups. Bull. Austral. Math. Soc. 60: Cohn, J. H. E. (1994). On n-sum groups. Math. Scand. 75: Doerk, K., Hawkes, T. (1992). Finite Soluble roups. Berlin, New York: Walter de ruyter. The AP roup, AP-roups, Algorithms, and programming, Version 4.3; 2002, (

14 3238 ABDOLLAHI ET AL. reco, D. (1953). Su alcuni gruppi finiti che sono somma di cinque sottagruppi. Rend. Sem. Mat. Univ. Padova 22: reco, D. (1956). Sui gruppi che sono somma di quattro o cinque sottagruppi. Rend. Accad. Sci. Fis. Math. Napoli (4)23: Jungnickel, D., Storme, L. (2001). Packing and covering groups with subgroups. J. Algebra 239: Neumann, B. H. (1954). roups covered by finitely many cosets. Publ. Math. Debrecen 3: Robinson, D. J. S. (1982). A Course in the Theory of roups. New York: Springer-Verlag. Scorza,. (1926). I gruppi che possono pensarsi come somma di tre loro sottogruppi. Boll. Un. Mat. Ital. 5: Tomkinson, M. J. (1987). roups covered by finitely many cosets or subgroups. Comm. Algebra 15: Tomkinson, M. J. (1997). roups as the union of proper subgroups. Math. Scand. 81: