Core Structure Theory
The Subalgebra Lattice Sub(A) is Algebraic
The theorem that the subuniverses of any algebra form an algebraic lattice, and the converse showing that every algebraic lattice arises this way.
Learning objectives
- Prove that Sub(A) is a complete lattice and identify its operations
- Identify the compact elements as the finitely generated subuniverses
- State the representation theorem for algebraic lattices
The lattice
The subuniverses of A, ordered by inclusion, form a complete lattice Sub(A) in which the meet of a family is its intersection and the join is Sg of its union.
Completeness follows from the one-sided criterion: A is the greatest subuniverse, and arbitrary intersections of subuniverses are subuniverses, so all infima exist. That is enough.
The join of two subuniverses is Sg of their union, not the union itself. For subgroups of a group this is the familiar fact that a union of two subgroups is a subgroup only when one contains the other. The asymmetry between meet and join is permanent and is the main source of difficulty in computing these lattices.
Algebraicity
Every element of Sub(A) is the join of the compact elements below it, and the compact elements are precisely the finitely generated subuniverses.
The converse of compactness also holds: a compact element of Sub(A) must be finitely generated, since it is the join of the finitely generated subuniverses below it and compactness forces one of them to suffice.
The representation theorem
Every algebraic lattice is isomorphic to Sub(A) for some algebra A.
The construction takes the compact elements of the given lattice as generators and introduces, for each finite join relation among them, an operation witnessing it. The resulting algebra has the prescribed lattice of subuniverses.
Birkhoff–Frink for Sub and Grätzer–Schmidt for Con both say the same thing: algebraicity is the only constraint. Neither lattice carries hidden structure beyond being algebraic, which is a strong and slightly deflating result — it says these invariants are as unconstrained as they could be.
Worked examples
| Algebra | <strong>Sub</strong>(<strong>A</strong>) | Notes |
|---|---|---|
| Cyclic group of order p | Two-element chain | Only the trivial subgroup and the whole group |
| Cyclic group of order pn | Chain of length n+1 | Subgroups are linearly ordered |
| Klein four-group | M5 | Three subgroups of order two, pairwise meeting trivially |
| Vector space of dimension 2 | Modular, not distributive | Contains M5 from any three distinct lines |
| Free semigroup on one generator | Complicated | Subsemigroups of the positive integers under addition |
Because the Klein four-group's subgroup lattice is M5, and M5 is modular but not distributive, no group with that subgroup lattice can have a distributive subgroup lattice. Ore's theorem sharpens this: a group has a distributive subgroup lattice exactly when it is locally cyclic.
Frequently asked questions
Is Sub(A) ever distributive?
Yes — for cyclic groups, and more generally for locally cyclic groups by Ore's theorem. But it is not distributive in general, and not even modular in general.
Does Sub(A) determine A?
No. Many non-isomorphic algebras share a subalgebra lattice; the two-element chain arises from every simple algebra with no proper non-trivial subalgebras.
Source. S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, The Millennium Edition — a corrected re-typesetting of Springer-Verlag Graduate Texts in Mathematics 78 (1981). Section II.3, book pages 33-34.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates and reorganises mathematical results; it is not a reproduction of the source text.