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

I must show that the class of Abelian groups has the CEP and find an example of lattice that doesn't have the CEP.

For Abelian groups: I consider it doing by induction on number of elements in a group and use inductive proposition on quotient groups. This is just an idea and I don't know how to use it.

Lattices: I know that counterexample for lattices should be some non distributive lattice because all distributive are CEP. But I can't find any example of congruence in a sublattice that wouldn't be also a subset of a congruence of a lattice. Maybe my understanding of basic terms isn't correct...

Tnx in advance for any help


$M_3$ is a simple lattice, which means it has only two congruences. But if you omit one element you'll get a "square" $\mathcal P(\{0,1\})$, which has a non-trivial congruence.

Congruences in groups correspond to normal subgroups. So CEP property only means that every normal subgroup of $B$ is an intersection of a normal subgroup of $A$ with the subgroup $B$. In Abelian groups, every subgroup is normal; so in the Abelian case this property is trivial.

  • $\begingroup$ Every intersection will be a normal subgroup. But I don't see how we exhaust all possible normal subgroups of B using this argument. $\endgroup$ – Alvis Mar 27 '14 at 8:41
  • 1
    $\begingroup$ @AnuragSharma If $H$ is a subgroup of $B$, then $H$ is also subgroup of $A$. So we get $H=H\cap B$, i.e., we expressed $H$ as an intersection of a subgroup of $A$ and $B$. $\endgroup$ – Martin Sleziak Mar 27 '14 at 8:45
  • $\begingroup$ Oh. M really sorry for such a silly question. Thanks $\endgroup$ – Alvis Mar 27 '14 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.