# Deducing a lattice is modular.

This is (the second part of) Exercise 2.6.4 of Howie's "Fundamentals of Semigroup Theory". The first part is here.

## The Details.

Definition 1: A lattice $$(L, \le, \wedge, \vee)$$ is modular if, for all $$a, b, c$$ in $$L$$, $$a\le c\implies (a\vee b)\wedge c=a\vee(b\wedge c).$$

Definition 2: The set of congruences on a semigroup $$S$$ is denoted $$\mathcal C(S)$$.

Proposition 1.8.3 (of Howie's book): Let $$\mathcal K$$ be a sublattice of the lattice $$(\mathcal C(S), \subseteq, \cap, \vee)$$ of congruences of a semigroup $$S$$, and suppose that $$\rho\circ\sigma=\sigma\circ\rho$$ for all $$\rho, \sigma$$ in $$\mathcal K$$. Then $$\mathcal K$$ is a modular lattice.

Definition 3: We define, for an arbitrary equivalence $$E$$ on a semigroup $$S$$, $$E^{\flat}=\{(a, b)\in S\times S : (\forall x, y\in S^1) (xay, xby)\in E\}$$ as the largest congruence on $$S$$ contained in $$E$$.

Lemma 1: Let $$S$$ be a semigroup, let $$\lambda$$ be a right congruence on $$S$$ contained in $$\mathcal L$$ and let $$\rho$$ be a left congruence on $$S$$ contained in $$\mathcal R$$. Then $$\lambda\circ\rho=\rho\circ\lambda$$.

## The Question:

Deduce (from Lemma 1), using Proposition 1.8.3, that the sublattice $$[1_S, \mathcal{H}^{\flat}]$$, consisting of all congruences on $$S$$ contained in $$\mathcal H$$, is modular.

## My Attempt:

I haven't got anywhere. It should be a simple application of the proposition.

• Any congruence contained in $\mathcal H$ is both a left congruence and a right congruence, and is contained in both $\mathcal L$ and $\mathcal R$. Hence any two congruences contained in $\mathcal H$ permute, according to the Lemma. The Proposition now gives modularity. – Keith Kearnes May 22 '17 at 13:27
Any congruence contained in $\mathcal{H}$ is both a left congruence and a right congruence, and is contained in both $\mathcal{L}$ and $\mathcal{R}$. Hence any two congruences contained in $\mathcal{H}$ permute, according to the Lemma. The Proposition now gives modularity.