NOTES ON THE FINITE LATTICE REPRESENTATION PROBLEM
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1 NOTES ON THE FINITE LATTICE REPRESENTATION PROBLEM WILLIAM J. DEMEO Notation and Terminology. Given a finite lattice L, the expression L = [H, G] means there exist finite groups H < G such that L is isomorphic to the interval {K H K G} in the subgroup lattice of G. A group G is called almost simple if G has a normal subgroup S G which is nonabelian, simple, and has trivial centralizer, C G (S) = (e). If H G, then the core of H in G, denoted core G (H), is the largest normal subgroup of G contained in H; it is given by core G (H) = ghg 1. A subgroup H G for which core G (H) = (e) is called core-free in G. g G In [8], Pálfy and Pudlák prove Theorem 1 (P 5, 1980). The following statements are equivalent: (A) Every finite lattice is isomorphic to the congruence lattice of a finite algebra. (B) Every finite lattice is isomorphic to an interval in the subgroup lattice of a finite group. Also noted in [8] is the important fact that (B) is equivalent to: (B ) Every finite lattice is isomorphic to the congruence lattice of a finite transitive G-set. There are many examples in the literature of the following situation: a specific finite lattice (or class of finite lattices) is considered, and it is shown that if such a lattice is an interval in the subgroup lattice of a finite group, then this group must be of a certain form, or must have certain properties, or must not have certain properties, etc. As the number of such results grows, it becomes increasingly useful to keep in mind the following: Lemma 1. Let G 1,..., G n be classes of groups and suppose that for each i {1,..., n} there exists a finite lattice L i with the property that L i = [H, G] only if G Gi. Then (B) is equivalent to (C) For every finite lattice L, there is a finite group G n G i such that L = [H, G]. Proof. Obviously, (C) implies (B). Assume (B) holds and let L be any finite lattice. Suppose G 1,..., G n and L 1,..., L n satisfy the hypothesis of the lemma. Construct a new lattice P = P(L, L 1,..., L n ) as shown in Figure 1 (a). By (B), there exist finite groups H G with P = [H, G]. Let K, K 1,..., K n be the subgroups of G which cover H and satisfy L = [K, G], and L i = [Ki, G], i = 1,..., n (Figure 1 (b)). Thus, L is an interval in the subgroup lattice of G, and, since L i = [Ki, G], we must have G G i, by hypothesis. This is true for all 1 i n, so G n G i, which proves that (B) implies (C). i=1 i=1 Date: 4 February
2 Figure 1. The parachute construction. G L L1 L2 L n... K L L1 L2 L n K n K 1... K 2 (a) (b) H Examples. Let G 1 denote the class of all finite groups that are not solvable. In [7], Pálfy describes a lattice L 1 with the property L 1 = [H, G] G G1. Let A n and S n denote the alternating and symmetric groups on n letters, and consider the class G 2 := {G G a finite group and ( n) G / {A n, S n }}. In his thesis [1], Alberto Basile proves a result (the last theorem of the appendix below) which implies that M 6 = [H, G] only if G G2. Given these examples and Lemma 1, we see that (B) holds if and only if for every finite lattice L there exist finite groups H < G such that L = [H, G] and G is not solvable, not alternating, and not symmetric. Now, assuming our goal is to solve the finite lattice representation problem, Lemma 1 suggests the following path to a solution: find examples of lattices L i which place restrictions on the G for which L i = [H, G] can hold, say G Gi, and eventually reach i G i = (at which point we are done). We would like to generalize Lemma 1 because it is much easier and more common to find a class of groups G i and a lattice L i with the following property: If L i = [H, G] with H core-free in G, then G Gi. ( ) This leads naturally to the following question: Given a class of groups G and a finite lattice L satisfying ( ), when can we safely drop the caveat with H core-free in G and get back to the hypothesis of Lemma 1? There is a very simple sufficient condition involving the complement of the class G, which we denote by G c := {G G is a group and G / G }. Lemma 2. Let G be a class of groups and L a finite lattice such that (1) L = [H, G] with H core-free G G, 2
3 and suppose G c is closed under homomorphic images (i.e., if G G c and N G, then G/N G c ). Then, (2) L = [H, G] G G. Proof. Suppose L satisfies (1) and G c is closed under homomorphic images. We prove (2) holds. Assume the contrary. Then there is a finite group G G c with L = [H, G]. Let N = core G (H). Then L = [H/N, G/N] and H/N is core-free in G/N so, by hypothesis (1), G/N G. But G/N G c, since G c is closed under homomorphic images. Examples. In the first example above, G 1 was the class of non-solvable groups, and so G1 c (the solvable groups) is closed under homomorphic images. In the second example, we took G 2 to be all non-alternating, non-symmetric groups, so G2 c = {A n n < ω} {S n n < ω}. This class is also closed under homomorphic images. It follows from Lemma 2 that the examples found by Pálfy and Basile do not require the core-free hypothesis. In contrast, consider the following result of Köhler [5]: If n 1 is not a power of a prime, then M n = [H, G] with H core-free G is subdirectly irreducible. Lemma 2 does not apply in this case since G c, the class of subdirectly reducible lattices, is obviously not closed under homomorphic images. 1 Though Lemma 2 seems like a useful observation, the last example above shows that a generalized version of Lemma 1 a version based on hypothesis ( ) would be much more powerful, as it would allow us to impose greater restrictions on G, such as those implied by the results of Köhler and many others. Surprisingly, the parachute construction used in the proof of Lemma 1 works in the more general case, with only a trivial modification to the hypotheses namely, the lattices L i cannot be two-element chains (which almost goes without saying in the present context). In the sequel, we let 2 denote a two-element chain. Lemma 3. Let G 1,..., G n be classes of groups and suppose that for each i {1,..., n} there is a finite lattice L i 2 which satisfies the following: If L i = [H, G] and H is core-free in G, then G Gi. ( ) Then (B) is equivalent to (C). Proof. Obviously, (C) implies (B). Assume (B) and let L be any finite lattice. Suppose G 1,..., G n and L 1,..., L n satisfy ( ) and L i 2 for all i. Note that there is no loss of generality in assuming that n 2. For if n = 1, just throw in one of the examples above to make n = 2. Call this additional class of groups G 2. Then, at the end of the argument, we ll have G G 1 G 2, and therefore, G G 1, which is the stated conclusion of the theorem in case n = 1. Construct the lattice P = P(L, L 1,..., L n ) as in the proof of Lemma 1. By (B) there exist finite groups H G with P = [H, G], and we can assume without loss of generality that H is 1 Every algebra, and in particular every group G, has a subdirect decomposition into subdirectly irreducibles, G G/N 1 G/N n. Thus, there will always be homomorphic images, G/N i, which are subdirectly irreducible. 3
4 Figure 2 G L i L j K i K j N i H N i K j core-free 2 in G. Let K, K 1,..., K n be the subgroups of G which cover H and satisfy L = [K, G], and L i = [Ki, G], 1 i n, as in Figure 1 (b). Thus, L is an upper interval in the subgroup lattice of G, and it remains to show that G n G i. i=1 Now, it is not immediately apparent that the subgroups K i are core-free in G. If they are, we are done, since this implies that, for each i {1,..., n}, G satisfies ( ) (with H replaced by K i in each case). We will give a direct proof of the fact that the subgroups K i are core-free in G. However, we note that this fact follows from Lemma 6 below, as well as from a more general result about L-P lattices. (See, e.g., Börner [2].) For each i {1,..., n}, let N i = core G (K i ). We prove that N i = (e) for all i. Suppose, on the contrary, that N i (e) for some i, and consider any K j with j i. (This is where we use n 2; though, if n = 1, we could have used K instead of K j, but then we would need to assume L 2.) A sketch of the part of the subgroup lattice under consideration is shown in Figure 2. Notice that N i K j = G. This is because N i is not below H, since H is core-free, so N i H = K i. Therefore, K i = N i H N i K j, so the subgroup N i K j is above both K i and K j, so it can only be the group G. Clearly 3 N i K j K j. Thus, by the standard isomorphism theorem, K j /N i K j = Ni K j /N i = G/N i. Under this correspondence, we have, in particular, [N i K j, K j ] H N i H = K i [N i, G]. Now look at Figure 2. By construction, K j covers H, so H is maximal in the lattice [N i K j, K j ]. On the other hand, [K i, G] = L i and we assumed at the outset that L i 2. Therefore, K i cannot be maximal in the interval [N i, G]. This contradiction proves the impossibility of a nontrivial core. That is, we must have core G (K i ) = (e) for all 1 i n, as claimed. 2 This is standard. For, if P = [H, G] with N := coreg(h) (e), then P = [H/N, G/N]. 3 Consider the canonical projection G G/Ni restricted to K j; this is a homomorphism of K j with kernel N i K j. 4
5 We prove a few more lemmas which lead to additional constraints on any group which has a non-trivial parachute lattice as an upper interval in its subgroup lattice. We will need the following basic theorem (see, e.g., page 122 of Rose [9]): Theorem 2 (Dedekind s rule). Let G be a group and let A, B and C be subgroups of G with A B. Then, (3) (4) A(C B) = AC B, (C B)A = CA B. and The next lemma is a minor generalization of a standard result (cf. Lemma 1.3 of [2], for example) and Lemma 5 shows why the standard result is a corollary. To state Lemma 4, we need some new notation. Let U and H be subgroups of a group, and let U 0 := U H. Suppose the set U H := {uh u U, h H} is a group. (Equivalently, U H = HU.) Consider the interval [U 0, U] := {V U 0 V U}. We define a new sublattice of [U 0, U] as follows: [U 0, U] H := {V [U 0, U] V H = HV }. That is, [U 0, U] H is the sublattice of [U 0, U] consisting of those V for which V H is a group. Lemma 4. Let U and H be subgroups of a group, and let U 0 := U H. If UH is a group, then the interval [H, UH] is isomorphic to [U 0, U] H. The inverse isomorphisms are given by (5) l 1 : [H, UH] X U X [U 0, U] H l 2 : [U 0, U] H V V H [H, UH]. Proof. Fix X [H, UH]. I claim that U X [U 0, U] H. Indeed, (U X)H = UH X (by Dedekind s rule (4), since H X) = HU X (since UH is a group) = H(U X) (by Dedekind s rule (3)). This proves U X [U 0, U] H. Also note that the first line above gives l 2 l 1 (X) = (U X)H = UH X = X, so l 2 l 1 is the identity on [H, UH]. Fix V [U 0, U] H. Then V H = HV implies V H [H, UH]. To complete the proof, note that l 1 l 2 is the identity on [U 0, U] H, since l 1 l 2 (V ) = V H U = V (H U) = V U 0 = V, by Dedekind s rule (3). Lemma 5. Let U and H be subgroups of a group, let U 0 := U H, and assume G := UH is a group. If U G and U 0 V U, then V H = HV if and only if H N G (V ). Proof. Obviously, if H N G (V ), then V H = HV (even when U G). Suppose V H = HV. We must show ( v V ) ( h H) hvh 1 V. Fix v V, h H. Then, hv = v h for some v V, h H, since V H = HV. Therefore, v h h 1 = hvh 1 = u for some u U, since H N G (U). This proves that hvh 1 V H U = V (H U) = V U 0 = V, as desired. 5
6 Remark 1. The careful reader will have noted that we only used the hypothesis H N G (U), but this is equivalent to the hypothesis U G when G = UH, as is the case above. Suppose U, V, and H are subgroups of a group. Then V is called H-invariant provided H normalizes V ; i.e., H N G (V ). If U 0 = H U, the sublattice of H-invariant subgroups in the interval [U 0, U] is sometimes denoted by [U 0, U] H. We are now ready to state the standard result that follows from Lemmas 4 and 5. Corollary 1 (cf. Lemma 1.3 of [2]). Let U and H be subgroups of a group, let U 0 := U H, and assume G := UH is a group. If U G, then the interval [H, G] is isomorphic to [U 0, U] H. The inverse isomorphisms are the ones given in (5). Proof. As Lemma 5 shows, if U G, then [U 0, U] H = {V [U 0, U] H N G (V )} = {V [U 0, U] V H = HV } = [U 0, U] H, so the result follows from Lemma 4. Next we prove that any group which has a nontrivial parachute lattice as an upper interval in its subgroup lattice must have some rather special properties. Lemma 6. Let P = P(L 1,..., L n ) with n 2 and L i > 2 for all i, and suppose P = [H, G], with H core-free in G. (i) If (e) N G, then NH = G. (ii) If M is a minimal normal subgroup of G, then C G (M) = (e). (iii) G is subdirectly irreducible. (iv) G is not solvable. Remark 2. If a subgroup M G is abelian, then M C G (M), so (ii) implies that a minimal normal subgroup of G must be nonabelian. Proof. (i) Let (e) N G. Then N H, since H is core-free in G. Therefore, H < NH. As in the proofs above, let K i denote the subgroups of G corresponding to the atoms of P. Then H is covered by each K i, so K j NH for some 1 j n. Suppose, by way of contradiction, that NH < G. By assumption, n 2 and L i > 2. Thus, for any i j, we have K i Y < Z < G for some subgroups Y and Z which satisfy (NH) Z = H and (NH) Y = G. Also, (NH)Y = NY is a group, so (NH)Y = NH Y = G. But then, by Dedekind s rule, we have Y = HY = ((NH) Z)Y = (NH)Y Z = G Z = Z, contrary to Y < Z. This contradiction proves that NH = G. (ii) If C G (M) (e), then (i) implies C G (M)H = G, since C G (M) N G (M) = G. Consider any H < K < G. Then (e) < M K < M (strictly, by Corollary 1). Now M K is normalized by H and centralized (hence normalized) by C G (M). (Indeed, C G (M) centralizes every subgroup of M.) Therefore, M K C G (M)H = G, contradicting the minimality of M. 6
7 (iii) We prove that G has a unique minimal normal subgroup. Let M be a minimal 4 normal subgroup of G and let N G be any normal subgroup not containing M. We show that N = (e). Since both subgroups are normal, the commutator 5 of M and N lies in the intersection M N, which is trivial by the minimality of M. Thus, M and N centralize each other. In particular, N C G (M) = (e), by (ii). (iv) Let M denote the commutator of M. As remarked above, M is nonabelian, so M (e). Also, M M G, and M is a characteristic subgroup of M (i.e., M invariant under Aut(M)). Therefore, M G, and, as M is a minimal normal subgroup of G, we have M = M. Thus, M is not solvable, so G is not solvable. Remark 3. It follows from (i) that, if P is a nontrivial parachute lattice with P = [H, G], where H is core-free, then core G (X) = (e) for every H X < G. This gives a second way to complete the proof of Lemma 3. Remark 4. Baddeley, Börner, Lucchini and others have already proved these statements (and more) for the more general case of quasiprimitive permutation groups. In particular, the proof of (i) above uses the same argument as that found in [2], where it is used to prove (Lemma 2.4): if L = [H, G] is an LP-lattice, then G must be a quasiprimitive permutation group. We remark that parachute lattices, in which the panels L i have L i > 2, are themselves LP-lattices, so Lemma 6 follows, as a special case, from the theorems Baddeley, Börner, Lucchini, et al. However, the parachute construction has a special advantage, besides providing a quick route to these results for this special case. It demonstrates a natural way to insert arbitrary finite lattices L i as upper intervals [K i, G] of Sub[G], with K i core-free in G (the italicized phrase being the key consequence of (i) above), so that once we prove special properties about groups G for which L i = [K i, G] (K i core-free), for some particular set of lattices, {L i }, then every finite lattice L must be an upper interval L = [K, G] for some G satisfying those same special properties. To summarize what we have thus far, the lemmas above imply that (B) holds if and only if every finite lattice is an interval [H, G], with H core-free in G, where (i) G is not solvable, not alternating, and not symmetric; (ii) G has a unique minimal normal subgroup M which satisfies MH = G and C G (M) = (e); in particular, M is nonabelian and core G (X) = (e) for all H X < G. Finally, we note that Theorem 4.3.A of Dixon and Mortimer [3] implies (iii) M = T 0 T r 1, where T i are simple minimal normal subgroups of M which are conjugate (under conjugation by elements of G). Thus, M is a direct power of a simple group T. Again, this reduction is similar to those made by Baddeley, Börner, Lucchini and others (see appendix), though the parachute construction enables us to add the non-alternating and nonsymmetric restrictions. To our knowledge, the non-solvable restriction isn t explicitly noted in 4 If G is simple, then M = G; minimal assumes nontrivial. 5 The commutator of M and N is the subgroup generated by the set {mnm 1 n 1 : m M, n N}. The commutator of M is the subgroup generated by {aba 1 b 1 : a, b M}. The n th degree commutator of M, denoted M (n), is defined recursively as the commutator of M (n 1). A group M is solvable iff M (n) = (e) for some n N. 7
8 previously published theorems, but it is a fairly easy consequence. Furthermore, as noted above, using the parachute we can add any and all restrictions on the group G that are found to hold for a particular finite lattice. Also, we arrive at these constrains on G in just a few pages, with what seems to us like less work than is required for the more general case. However, we should note that we have not considered the division into two cases (C) and (D), as described in the theorem of Börner stated in the appendix below. It seems that this is where much of the work lies. The division seems to be based on whether M H (e) (case (D)), which was dealt with by Lucchini in He proved that in this case M itself is simple, so G is an almost simple group. In case M H = (e) (case (C)), there are usually two further cases to consider: MH = G or MH < G. But, as we have seen above, the latter case does not occur for L-P lattices, in general, and parachute lattices in particular. The structure of the minimal normal subgroup. Item (iii) of the previous lemma states that M is a direct power of a minimal normal subgroup T M, and T is a nonabelian simple group. Precisely, M = T r = T 0 T r 1, where each T i is a simple minimal normal subgroup of M of the form T i = T g i (some g i G). 6 In fact, when C G (M) = (e), as is the case in our application, we can specify these conjugates more precisely. Let T be any minimal normal subgroup of M. Note that T is simple. (Why?) Let N = N H (T ) = {h H : T h = T } be the normalizer of T in H. Then we can prove the following Lemma 7. If H/N = {N, h 1 N,..., h k 1 N} is a full set of left cosets of N in H, then k = r and M = T 0 T r 1 = T T h 1 T h r 1. Proof. We start from the assumption that T is a minimal normal subgroup of M, and that M = T 0 T r 1, where each T i is a conjugate of T in G. ([3] ibid.) Let T = {T 0, T 1,..., T s 1 } (where, say, T 0 = T ) be a maximal set of conjugates of T such that the subgroup generated by T is a direct product of all members: T 0, T 1,..., T s 1 = T 0 T 1 T s 1. Let h H be arbitrary. If T h / T, then T h cannot intersect trivially with T i, otherwise, T 0, T 1,..., T s 1, T h = T 0 T 1 T s 1 T h, contradicting maximality of T. Furthermore, T h T i is a normal subgroup of T h, but the latter is simple, so T h T 0 T s 1. This holds for all h H, which proves that T 0 T s 1 is normalized by H (and, of course, by M), so T 0 T s 1 G. Therefore, T 0 T s 1 = M, by minimality of M. Now, if T h is not already in the list T 0, T 1,..., T s 1, then T h C M (T i ) for all i = 0, 1,..., s 1, since minimal normal subgroups centralize each other. Therefore, T h C M (T 0 T s 1 ) C G (M) = (e). This contradiction proves that every conjugate T h (h H) already appears in the list T. Thus, M = T 0 T 1 T k 1 = T h 0 T h 1 T h k 1 where {e = h 0, h 1,..., h k 1 } is a full set of N H (T )-coset representatives in H. 6 The reader is referred to Theorem 4.3.A of Dixon and Mortimer [3] for the proof of this general result. 8
9 Appendix: prior art Baddeley, Börner, Lucchini and others have reduced the problem to the two special cases described in the next theorem. Theorem (Börner [2], 1999). Statement (B) above is true if and only if at least one of the following is true: (C) For every finite lattice L there are finite groups H G such that L = [H, G] with the following properties: G has a unique minimal normal subgroup M, with M H = (e), MH = G, M is nonabelian, and if F denotes one of the simple direct factors of M and Q = N H (F ), then Q induces all inner automorphisms of F and Q is core-free in H. (D) For every finite lattice L which is generated by its coatoms, there exist finite groups H G such that L = [H, G], where G is almost simple and H is core-free in G. In [6], Pálfy had the following to say about statement (C): It is so restrictive that probably it can be proved to be false, and then the problem would be reduced to the case of almost simple groups. Then it would remain to do a case-by-case analysis, like it has been done for the alternating and symmetric groups by Alberto Basile [see theorem on next page]. For the remainder of the notes we borrow from Pálfy s nice description in [6] of some interesting results concerning the height-two modular lattices M n. Of particular interest to us is the theorem of Basile, which we already used above. If n = p k + 1 for some prime p and exponent k 1, then it is possible to find suitable intervals [H, G] = M n. Namely, let V be the 2-dimensional vector space over the Galois field F := GF (p k ) and take G = {x λx + v λ F, v V }, H = {x λx λ F }. Then every intermediate subgroup H K G has the form K = {x λx + v λ F, v U} for a suitable subspace U V, so [H, G] is just the lattice of subspaces of V. It had been conjectured that these were the only M n s occurring as intervals in subgroup lattices of finite groups. Then Walter Feit pointed out that this is not the case. In formulating his examples, we shall use the following notation: if p > 2 is prime and d divides (p 1)/2, then up to conjugacy there is a unique subgroup of order pd in the alternating group A p, which we denote by p d. Example (Feit [4], 1983). (1) M 7 = [31 5, A 31 ], the intermediate subgroups are the normalizer of 31 5 and six subgroups isomorphic to GL 5 (2). (2) M 11 = [31 3, A 31 ], the intermediate subgroups are the normalizer of 31 3 and ten subgroups isomorphic to PSL 3 (5). That these examples cannot be generalized was proved in Basile s thesis. 9
10 Theorem (Basile [1], 2001). If [H, G] = M n with G = S d or A d, then either n 3 or one of the following holds: n = 5, d = 13; n = 7, d = 31; n = 11, d = 31. References [1] Alberto Basile. Second maximal subgroups of the finite alternating and symmetric groups. PhD thesis, Australian National University, Canberra, April [2] Ferdinand Börner. A remark on the finite lattice representation problem. Contributions to general algebra, 11:5 38, Verlag Johannes Heyn, Klagenefurt. [3] John Dixon and Brian Mortimer. Permutation Groups. Springer, New York, [4] Walter Feit. An interval in the subgroup lattice of a finite group which is isomorphic to M 7. Algebra Universalis, 17: , [5] Peter Köhler. M 7 as an interval in a subgroup lattice. Algebra Universalis, 17: , [6] Peter Pál Pálfy. Groups and lattices. In Groups St. Andrews 2001 in Oxford, volume 2, pages [7] Peter Pál Pálfy. Intervals in subgroup lattices of finite groups. In Groups St Andrews 1993 in Galway, volume 2, pages [8] Peter Pál Pálfy and Pavel Pudlák. Congruence lattices of finite algebras and intervals in subgroup lattices of finite groups. Algebra Universalis, 11:22 27, [9] John S. Rose. A Course on Group Theory. Dover, New York,
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