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?
  • $\begingroup$ What is the definition of $\operatorname{Con}(A)$? $\endgroup$ – Santana Afton Mar 17 '19 at 0:30
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin Mar 17 '19 at 0:36
  • $\begingroup$ @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 $A$. $\endgroup$ – Arturo Magidin Mar 17 '19 at 0:38
  • $\begingroup$ related (possibly duplicate) question: link $\endgroup$ – 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.

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.

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