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CEP for (distributive) lattices and groups?

CEP for Abelian groups and latticesLattices are congruence-distributiveCEP for Abelian groups and latticesEvery group and ring is congruence-permutable , but not necessarily congruence-distributiveProof that the lattice of fully invariant congruences is a sublattice of the lattice of all congruences$mathbfN_5$ as a congruence latticeSemilattices are congruence-semi-distributivePrime ideal theorem for modular lattices?If a variety of algebras V has the CEP then $HS(K) = SH(K), K subseteq V$Jónsson's lemma : A variety generated by finitely many finite lattices has finitely many subvarieties.Is there any duality for distributive lattices?

1

$begingroup$

An algebra $A$ has the congruence extension property (CEP) if for every $Ble A$ and $thetainoperatornameCon(B)$ there is a $varphiinoperatornameCon(A)$ such that $theta =varphicap(Btimes B)$. 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?

group-theory lattice-orders universal-algebra congruence-relations

share|cite|improve this question

edited Mar 17 at 0:33

Santana Afton

2,9932629

asked Mar 17 at 0:05

ljubinaaljubinaa

274

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the definition of $operatornameCon(A)$?
  $endgroup$
  – Santana Afton
  Mar 17 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 $Ntriangleleft H$ then there exists $Mtriangleleft G$ such that $Mcap H=N$. 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 at 0:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SantanaAfton: A congruence on $A$ is an equivalence relation on $A$ which, when viewed as a subset of $Atimes A$, is also a subalgebra of $Atimes A$ (with the induced structure); they are the objects that play the role of normal subgroups for groups and ideals for rings, to define quotients. $mathrmCon(A)$ is the collection of all congruences on $A$.
  $endgroup$
  – Arturo Magidin
  Mar 17 at 0:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  related (possibly duplicate) question: link
  $endgroup$
  – Eran
  Mar 18 at 18:24
  

  
  
  
  

add a comment | 

1

$begingroup$

An algebra $A$ has the congruence extension property (CEP) if for every $Ble A$ and $thetainoperatornameCon(B)$ there is a $varphiinoperatornameCon(A)$ such that $theta =varphicap(Btimes B)$. 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?

group-theory lattice-orders universal-algebra congruence-relations

share|cite|improve this question

edited Mar 17 at 0:33

Santana Afton

2,9932629

asked Mar 17 at 0:05

ljubinaaljubinaa

274

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the definition of $operatornameCon(A)$?
  $endgroup$
  – Santana Afton
  Mar 17 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 $Ntriangleleft H$ then there exists $Mtriangleleft G$ such that $Mcap H=N$. 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 at 0:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SantanaAfton: A congruence on $A$ is an equivalence relation on $A$ which, when viewed as a subset of $Atimes A$, is also a subalgebra of $Atimes A$ (with the induced structure); they are the objects that play the role of normal subgroups for groups and ideals for rings, to define quotients. $mathrmCon(A)$ is the collection of all congruences on $A$.
  $endgroup$
  – Arturo Magidin
  Mar 17 at 0:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  related (possibly duplicate) question: link
  $endgroup$
  – Eran
  Mar 18 at 18:24
  

  
  
  
  

add a comment | 

$begingroup$

An algebra $A$ has the congruence extension property (CEP) if for every $Ble A$ and $thetainoperatornameCon(B)$ there is a $varphiinoperatornameCon(A)$ such that $theta =varphicap(Btimes B)$. 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?

group-theory lattice-orders universal-algebra congruence-relations

share|cite|improve this question

edited Mar 17 at 0:33

Santana Afton

2,9932629

asked Mar 17 at 0:05

ljubinaaljubinaa

274

$endgroup$

share|cite|improve this question

edited Mar 17 at 0:33

Santana Afton

2,9932629

asked Mar 17 at 0:05

ljubinaaljubinaa

274

share|cite|improve this question

edited Mar 17 at 0:33

Santana Afton

2,9932629

asked Mar 17 at 0:05

ljubinaaljubinaa

274

edited Mar 17 at 0:33

Santana Afton

2,9932629

edited Mar 17 at 0:33

Santana Afton

2,9932629

asked Mar 17 at 0:05

ljubinaaljubinaa

274

asked Mar 17 at 0:05

ljubinaaljubinaa

274

1 Answer
1

active

oldest

votes

3

$begingroup$

The variety of all groups does not have the property. Congruences correspond to normal subgroups, so you are essentially asking whether if $Hleq G$, and $Ntriangleleft H$, does there always exist an $Mtriangleleft G$ such that $Mcap H= N$. 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 $0vee y=y$ with $xvee y=1$; similarly, it must identify $0vee z = z$ with $xvee z = 1$. Thus, $y$, $z$, and $1$ are identified in $Psi$. That means that $zwedge 1 = z$ must be identified with $zwedge y = 0$, 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.

share|cite|improve this answer

edited Mar 17 at 7:27

answered Mar 17 at 1:39

Arturo MagidinArturo Magidin

265k34590920

$endgroup$

add a comment | 

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3

$begingroup$

The variety of all groups does not have the property. Congruences correspond to normal subgroups, so you are essentially asking whether if $Hleq G$, and $Ntriangleleft H$, does there always exist an $Mtriangleleft G$ such that $Mcap H= N$. 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 $0vee y=y$ with $xvee y=1$; similarly, it must identify $0vee z = z$ with $xvee z = 1$. Thus, $y$, $z$, and $1$ are identified in $Psi$. That means that $zwedge 1 = z$ must be identified with $zwedge y = 0$, 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.

share|cite|improve this answer

edited Mar 17 at 7:27

answered Mar 17 at 1:39

Arturo MagidinArturo Magidin

265k34590920

$endgroup$

add a comment | 

3

$begingroup$

The variety of all groups does not have the property. Congruences correspond to normal subgroups, so you are essentially asking whether if $Hleq G$, and $Ntriangleleft H$, does there always exist an $Mtriangleleft G$ such that $Mcap H= N$. 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 $0vee y=y$ with $xvee y=1$; similarly, it must identify $0vee z = z$ with $xvee z = 1$. Thus, $y$, $z$, and $1$ are identified in $Psi$. That means that $zwedge 1 = z$ must be identified with $zwedge y = 0$, 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.

share|cite|improve this answer

edited Mar 17 at 7:27

answered Mar 17 at 1:39

Arturo MagidinArturo Magidin

265k34590920

$endgroup$

add a comment | 

$begingroup$

The variety of all groups does not have the property. Congruences correspond to normal subgroups, so you are essentially asking whether if $Hleq G$, and $Ntriangleleft H$, does there always exist an $Mtriangleleft G$ such that $Mcap H= N$. 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 $0vee y=y$ with $xvee y=1$; similarly, it must identify $0vee z = z$ with $xvee z = 1$. Thus, $y$, $z$, and $1$ are identified in $Psi$. That means that $zwedge 1 = z$ must be identified with $zwedge y = 0$, 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.

share|cite|improve this answer

edited Mar 17 at 7:27

answered Mar 17 at 1:39

Arturo MagidinArturo Magidin

265k34590920

$endgroup$

share|cite|improve this answer

edited Mar 17 at 7:27

answered Mar 17 at 1:39

Arturo MagidinArturo Magidin

265k34590920

answered Mar 17 at 1:39

Arturo MagidinArturo Magidin

265k34590920

answered Mar 17 at 1:39

Arturo MagidinArturo Magidin

265k34590920