module over a ring is modular. As a special case, the lattice of subgroups of an abelian group is modular. The lattice of normal subgroups of a group is modular. But in general the lattice of all subgroups of a group is not modular. For an example, the lattice of subgroups of the dihedral group of order 8 is not modular. The smallest non-modular lattice is the "pentagon" lattice N5 consisting of five elements 0, 1, x, a, b such that 0 < x < b < 1, 0 < a < 1, and a is not comparable to x or to b. For this lattice, x ∨ (a ∧ b) = x ∨ 0 = x < b = 1 ∧ b = (x ∨ a) ∧ b holds, contradicting the modular law. Every non-modular lattice contains a copy of N5 as a sublattice. Properties Every distributive lattice is modular. proved that, in every finite modular lattice, the number of join-irreducible elements equals the number of meet-irreducible elements. More generally, for every , the number of elements of the lattice that cover exactly other elements equals the number that are covered by exactly other elements. A useful property to show that a lattice is not modular is as follows: A lattice is modular if and only if, for any , Sketch of proof: Let G be modular, and let the premise of the implication hold. Then using absorption and modular identity: c = (c∧b) ∨ c = (a∧b) ∨ c = a ∧ (b∨c) = a ∧ (b∨a) = a For the other direction, let the implication of the theorem hold in G. Let a,b,c be any elements in G, such that c ≤ a. Let x = (a∧b) ∨ c, y = a ∧ (b∨c). From the modular inequality immediately follows that x ≤ y. If we show that x∧b = y∧b, x∨b = y∨b, then using the assumption x = y must hold. The rest of the proof is routine manipulation with infima, suprema and inequalities. Diamond isomorphism theorem For any two elements a,b of a modular lattice, one can consider the intervals [a ∧ b, b] and [a, a ∨ b]. They are connected by order-preserving maps φ: [a ∧ b, b] → [a, a ∨ b] and ψ: [a, a ∨ b] → [a ∧ b, b] that are defined by φ(x) = x ∨