By Stankey Burris, H. P. Sankappanavar

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2 We would like to note that in 1953 J´onsson improved on Birkhoff’s result above by showing that one could derive the so-called Arguesian identity for lattices from congruencepermutability. In §12 we will concern ourselves again with congruence-distributivity and permutability. References 1. 2. 3. 4. G. Birkhoff [3] G. T. Schmidt [1963] B. J´onsson [1953] P. Pudl´ak [1976] Exercises §5 1. 2). 2. 2). 3. Show that the normal subgroups of a group form an algebraic lattice which is modular. 4. Show that every group and ring is congruence-permutable, but not necessarily congruencedistributive.

Then clearly fB,b (a1 , . . , an ) ∈ C({a1 , . . , an }), hence for X ⊆ A, Sg(X) ⊆ C(X). On the other hand C(X) = {C(B) : B ⊆ X and B is finite} and, for B finite, C(B) = {fB,b (a1 , . . , an ) : B = {a1 , . . , an }, b ∈ C(B)} ⊆ Sg(B) ⊆ Sg(X) imply C(X) ⊆ Sg(X); hence C(X) = Sg(X). Thus LC = Sub(A), so Sub(A) ∼ = L. 2 The following set-theoretic result is used to justify the possibility of certain constructions in universal algebra—in particular it shows that for a given type there cannot be “too many” algebras (up to isomorphism) generated by sets no larger than a given cardinality.

2. Let P be a poset such that A exists for every subset A, or such that exists for every subset A. Then P is a complete lattice. A Proof. Suppose A exists for every A ⊆ P. Then letting Au be the set of upper bounds of A in P, it is routine to verify that Au is indeed A. The other half of the theorem is proved similarly. 2 In the above theorem the existence of ∅ guarantees a largest element in P, and likewise the existence of ∅ guarantees a smallest element in P. 2 would be to say that P is complete if it has a largest element and the inf of every nonempty subset exists, or if it has a smallest element and the sup of every nonempty subset exists.

### A Course in Universal Algebra by Stankey Burris, H. P. Sankappanavar

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