# CEP for (distributive) lattices and groups?

An algebra $$A$$ has the congruence extension property (CEP) if for every $$B\le A$$ and $$\theta\in\operatorname{Con}(B)$$ there is a $$\varphi\in\operatorname{Con}(A)$$ such that $$\theta =\varphi\cap(B\times B)$$. A class $$K$$ of algebras has the CEP if every algebra in the class has the CEP.

• Does the class of all lattices has CEP?
• Does the class of all groups has CEP?
• Does the class of all distributive lattices has CEP?
• What is the definition of $\operatorname{Con}(A)$? Commented Mar 17, 2019 at 0:30
• Since in groups congruences correspond to normal subgroups, what you are asking for groups would be that given a group $G$, and a subgroup $H$, if $N\triangleleft H$ then there exists $M\triangleleft G$ such that $M\cap H=N$. It is now easy to see that this does not hold, for example by taking $G=A_5$ which is simple, but all of whose proper subgroups that are not of prime order are not simple. Commented Mar 17, 2019 at 0:36
• @SantanaAfton: A congruence on $A$ is an equivalence relation on $A$ which, when viewed as a subset of $A\times A$, is also a subalgebra of $A\times A$ (with the induced structure); they are the objects that play the role of normal subgroups for groups and ideals for rings, to define quotients. $\mathrm{Con}(A)$ is the collection of all congruences on $A$. Commented Mar 17, 2019 at 0:38
• related (possibly duplicate) question: link
– Eran
Commented Mar 18, 2019 at 18:24

The variety of all groups does not have the property. Congruences correspond to normal subgroups, so you are essentially asking whether if $$H\leq G$$, and $$N\triangleleft H$$, does there always exist an $$M\triangleleft G$$ such that $$M\cap H= N$$. This does not hold; for example, if $$G=A_5$$, $$H=A_4$$, and $$N$$ is a nontrivial proper normal subgroup of $$A_4$$, then you cannot find any appropriate $$M$$.

The variety of all latices does not have the property either. Let $$L$$ be the nondistributive lattice $$M_3$$, with elements $$0$$, $$1$$, $$x$$, $$y$$, and $$z$$ (where the join of any distinct elements of $$\{x,y,z\}$$ is $$1$$, and the meet is $$0$$). Let $$M$$ be the sublattice $$\{0,x,1\}$$. Then let $$\Phi$$ be the congruence in $$M$$ that identifies $$0$$ and $$x$$.

Let $$\Psi$$ be a congruence on $$L$$ that identifies $$0$$ and $$x$$. Then it must identify $$0\vee y=y$$ with $$x\vee y=1$$; similarly, it must identify $$0\vee z = z$$ with $$x\vee z = 1$$. Thus, $$y$$, $$z$$, and $$1$$ are identified in $$\Psi$$. That means that $$z\wedge 1 = z$$ must be identified with $$z\wedge y = 0$$, so all of $$0$$, $$z$$, $$y$$, and $$1$$ are identified in $$\Psi$$. That means that $$x$$ is identified with $$1$$ as well, so that $$\Phi$$ is a proper subcongruence of $$\Psi|_M$$. Thus, $$M_3$$ does not have the congruence extension property.

In fact, I believe (but don’t have access to my textbooks right now) that the Congruence Extension Property precisely characterizes the distributive lattices among all latices (see for instance the opening line of this paper ) which would give an affirmative answer to your final question.

If a lattice $$L$$ has the pentagon $$N_5$$ as a sublattice ($$N_5=\{0,a,b,c,1\}$$, with $$0, $$0, $$a\wedge c=b\wedge c=0$$ and $$a\vee c=b\vee c=1$$), then $$L$$ doesn't have the CEP, because the chain $$\{0,a,b,1\}$$ is a sublattice where $$\theta(0,a)=\{0,a\}^2\cup\{(b,b),(1,1)\},$$ but in $$N_5$$, if $$0\theta a$$ then $$c\theta 1$$, and so $$0\theta b$$, and so $$\theta_L(0,a)\supseteq\{0,a,b\}^2.$$ So there is no congruence of $$L$$ that restricts to $$\theta(0,a)$$ in $$\{0,a,b,1\}$$.

This together with the answer of Arturo Magidin shows that a lattice has the CEP iff it is distributive.

This is a complement to the previous answers, in what concerns lattices.

In his answer, Arturo Magidin showed that $$M_3$$ doesn't have the CEP and hinted that a lattice is distributive iff it has the CEP.

Later, Hugo showed that $$N_5$$ doesn't have the CEP either, whence non-distributive lattices don't have the CEP, claiming that with that the hint of Arturo Magidin was proven.
This was certainly an overlook of the problem since it still remains to show that distributive lattices do have the CEP.

A variety is said to have Equationally Definable Principal Congruences (EDPC, for short) if there is a conjunction $$\Phi(w,x,y,z)$$ of finitely many equations on four variables such that for every algebra $$\mathbf A$$ in that variety and every $$a,b,c,d \in A$$ $$(c,d) \in \Theta(a,b) \iff \mathbf A \vDash \Phi(a,b,c,d).$$ In this question there is an answer (by Berci) in which it is shown that if a variety has EDPC then it has the CEP.

In this question there is an answer of mine showing that the variety of distributive lattices has EDPC.

Hence we can conclude that distributive lattices have the CEP and that the property of having the CEP characterizes distributive lattices among all lattices as suggested.

As a side note, the formula that is being asked to prove in the last linked question is that, in a distributive lattice $$(c,d) \in \Theta(a,b) \iff (a \wedge b \wedge c = a \wedge b \wedge d \;\text{and}\; a \vee b \vee c = a \vee b \vee d).$$ A similar and equivalent formula is, supposing $$a \leq b$$, $$(c,d) \in \Theta(a,b) \iff (a \wedge c = a \wedge d \;\text{and}\; b \vee c = b \vee d),$$ where the equivalence comes from the fact that in a lattice (distributive or not), $$a \sim b$$ iff $$a \wedge b \sim a \vee b$$, for any congruence $$\sim$$.