# 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 $\operatorname{Con}(A)$? – Santana Afton Mar 17 '19 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. – Arturo Magidin Mar 17 '19 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 $A$. – Arturo Magidin Mar 17 '19 at 0:38
• related (possibly duplicate) question: link – Eran Mar 18 '19 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.