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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

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$M_3$ 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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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