M n as a 0, 1-Sublattice of Con A Does not Force the Term Condition
Ross Willard · Proceedings of the American Mathematical Society · 1988
ABSTRACT. For every n> 3 there exists a finite nonabelian algebra whose congruence lattice has Mn as a 0, 1-sublattice. This answers a question of R. McKenzie and D. Hobby. DEFINITION 0.1. Suppose L and L1 are bounded lattices. A copy of L (in L1) is any sublattice L ' of L1 which is isomorphic to L. L ' is a 0, 1-copy of L if it includes the least and greatest elements of L1; in this case L ' is also called a 0, 1-sublattice of L1. DEFINITION 0. 2. For n> 1, Mn is the finite lattice of height 2 having exactly n atoms. For example, M6 is Suppose G is a group and N(G) is its lattice of normal subgroups. There is a trivial proof, using the commutator operation on normal subgroups, that if N(G) has a 0,1-copy of M3 then G is abelian. This same proof extends, via the general commutator theory of universal algebra, to any algebra A belonging to a variety whose congruence lattices satisfy the modular law. Here N(G) is replaced by Con A (the lattice of congruence relations of A), while 'abelian ' means the following 'term condition': DEFINITION 0. 3. An algebra A is abelian if for every n> 1, every (n + 1)-ary term t(x, Y1... Yn) in the language of A, and all a, b, c a.... acn dla... adn E Al tA (a, c) = tA (a, d) iff tA(bC) = tA(b, d). In their forthcoming book on tame congruence theory [1], R. McKenzie and D. Hobby ask whether the above phenomenon Con A having M3 as a 0, 1-sublattice Received by the editors September 28, 1987.