Lattice Theory Foundations

Distributive and Modular Lattices: the M5 and N5 Criteria

Q: Is every modular lattice distributive on finite subsets?

No. M5 is finite, modular and not distributive, so finiteness provides no rescue. The relationship between the two conditions is not affected by cardinality at all — both are equational, hence inherited by every subalgebra regardless of size.

Q: Are subgroup lattices always modular?

Only for abelian groups, and more generally for groups in which all subgroups are normal. The lattice of normal subgroups of any group is modular. The full subgroup lattice of a non-abelian group need not be: the symmetric group on three letters has a subgroup lattice containing a pentagon.

Q: How hard is it to search for M5 and N5 in practice?

For a finite lattice it is a finite search over five-element subsets, so it is decidable but grows quickly. In practice one checks the identity directly on triples, which is cubic rather than quintic. For infinite lattices the criterion is a theoretical characterisation rather than an algorithm, and one verifies the identity instead.

Two five-element lattices decide everything. M5 and N5 are the complete obstruction set for distributivity and modularity respectively, and no third example is needed.

Engineering · Mathematics4 min readKV-MATH-0205

Learning objectives

State the distributive and modular laws and verify their self-duality.
Show that distributivity implies modularity and that the converse fails.
Recognise M5 and N5 and explain what each one witnesses.
Apply the forbidden-sublattice criteria to classify a given lattice.
Connect these conditions to congruence-distributive and congruence-modular varieties.

01The two laws

Distributivity asserts that meet distributes over join; modularity is a weakened form conditioned on a comparability hypothesis.

Distributive: x ∧ (y ∨ z) ≈ (x ∧ y) ∨ (x ∧ z)
Modular: x ≤ z ⟹ x ∨ (y ∧ z) = (x ∨ y) ∧ z

The modular law can be written as a genuine identity by replacing x with x ∧ z, which makes the comparability automatic: (x ∧ z) ∨ (y ∧ z) ≈ ((x ∧ z) ∨ y) ∧ z. Both conditions are therefore equational and define varieties.

Both laws are self-dual, though for different reasons. In the distributive case the dual identity — join distributing over meet — is a consequence of the stated one, not an independent axiom; proving this is a standard exercise using absorption. In the modular case the dual statement is literally the same statement read upside down.

02Distributive implies modular

Assume distributivity and x ≤ z. Then x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) by the dual distributive law, and x ∨ z = z because x ≤ z. So x ∨ (y ∧ z) = (x ∨ y) ∧ z, which is modularity.

Key resultThe implication is strict

M5, the diamond with a bottom, a top and three pairwise incomparable middle elements, is modular but not distributive. It therefore witnesses that modularity is genuinely weaker, and it is the reason the forbidden-sublattice criterion for distributivity needs two excluded configurations rather than one.

The chain of implications runs: distributive ⟹ modular ⟹ arbitrary lattice, with both implications strict. Every chain is distributive; the subgroup lattice of an abelian group is modular; the subgroup lattice of a general group need not be even that.

03The two obstructions

M5 and N5
Lattice	Shape	Modular?	Distributive?	Witnesses
M5 (the diamond)	0 < a, b, c < 1, middle elements pairwise incomparable	Yes	No	Modularity without distributivity
N5 (the pentagon)	0 < a < b < 1 and 0 < c < 1, with c incomparable to a and b	No	No	Failure of modularity

In M5, take x = a, y = b, z = c. Then a ∧ (b ∨ c) = a ∧ 1 = a, whereas (a ∧ b) ∨ (a ∧ c) = 0 ∨ 0 = 0. Distributivity fails at a single triple. Modularity survives because no triple of M5 satisfies the comparability hypothesis in a way that produces a counterexample.

In N5, take x = a, z = b so that x ≤ z, and y = c. Then a ∨ (c ∧ b) = a ∨ 0 = a, while (a ∨ c) ∧ b = 1 ∧ b = b, and a ≠ b. Modularity fails.

04The characterisation theorems

ProcedureClassifying a lattice by forbidden sublattices

in: lattice L → out: distributive / modular / neither

input: lattice L
search for a sublattice of L isomorphic to N5
if found: L is not modular, hence not distributive → STOP
search for a sublattice of L isomorphic to M5
if found: L is modular-or-not, but definitely not distributive
if not found and no N5 either: L is distributive
if N5 absent but M5 present: L is modular and not distributive

Completeness: these are the only obstructions — the theorems assert no third forbidden configuration exists. Caveat: the search is over sublattices, meaning subsets closed under the parent's operations, not merely sub-posets of the right shape.

CautionThe shape must be a sublattice, not a sub-poset

Finding five elements arranged in the pentagon pattern under the inherited order proves nothing. The subset must be closed under the parent lattice's join and meet. A common failure is to spot a pentagon-shaped sub-poset whose join in the parent escapes the five chosen elements — that configuration does not witness non-modularity.

05Why this matters downstream

These conditions reappear immediately as conditions on congruence lattices, and that is where their real weight lies in universal algebra.

Congruence-distributive varieties
Every member has a distributive congruence lattice. Lattices themselves qualify. Jónsson's lemma applies, giving strong control over subdirectly irreducibles.
Congruence-modular varieties
Weaker, but enough to support the commutator theory and the centre. Groups, rings and modules all qualify.
Mal'cev characterisations
Each condition is equivalent to the existence of certain terms — Jónsson terms for distributivity, Day terms for modularity — which converts a lattice-theoretic condition into a checkable syntactic one.

Frequently asked

Is every modular lattice distributive on finite subsets?

No. M5 is finite, modular and not distributive, so finiteness provides no rescue. The relationship between the two conditions is not affected by cardinality at all — both are equational, hence inherited by every subalgebra regardless of size.

Are subgroup lattices always modular?

Only for abelian groups, and more generally for groups in which all subgroups are normal. The lattice of normal subgroups of any group is modular. The full subgroup lattice of a non-abelian group need not be: the symmetric group on three letters has a subgroup lattice containing a pentagon.

How hard is it to search for M5 and N5 in practice?

For a finite lattice it is a finite search over five-element subsets, so it is decidable but grows quickly. In practice one checks the identity directly on triples, which is cubic rather than quintic. For infinite lattices the criterion is a theoretical characterisation rather than an algorithm, and one verifies the identity instead.

Sources and further reading

S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, Millennium Edition (a corrected re-typesetting of Springer GTM 78, 1981).
G. Grätzer, Universal Algebra, 2nd edition, Springer.
R. McKenzie, G. McNulty and W. Taylor, Algebras, Lattices, Varieties, Volume I.

Original KEVOS® explanatory article. Written from the topic map of the cited works; no text is reproduced from them.