An Investigation into the An Investigation into the Structure of DigroupsStructure of Digroups

David WhiteDavid White

Bowdoin CollegeBowdoin CollegeBrown University SUMSBrown University SUMS

March 8, 2008March 8, 2008

IntroductionIntroduction

Digroups and the Coquecigrue Digroups and the Coquecigrue ProblemProblem

Definition of a Digroup and an Definition of a Digroup and an ExampleExample

Trivial DigroupTrivial Digroup

GroupsGroups

CoquecigrueCoquecigrue

LieGroups

Linear Lie Digroups

????????

LinearLie Racks

Lie Racks

Racks

Lie

Algebras

Split Leibniz

Algebras

Leibniz

Algebras

What is a Digroup?What is a Digroup?

A digroup is a set G with two binary A digroup is a set G with two binary operations, operations, and and , a unary operation , a unary operation -1-1 and a 1 which satisfies:and a 1 which satisfies:

G1. (G, G1. (G, ) and (G, ) and (G, ) are both ) are both semigroups (i.e. semigroups (i.e. and and are are associative)associative)

G2* x G2* x (x (x z) = (x z) = (x x) x) z zG5. 1 G5. 1 x = x = x x = x = x 1 1G6. x G6. x x x-1-1 = 1 = x = 1 = x-1-1 x x

Inverses and IdentitiesInverses and Identities

In a digroup there can be multiple identity In a digroup there can be multiple identity elements. The set of identities are defined elements. The set of identities are defined

E = {e E = {e G| e G| e x = x} x = x}

The set of inverses are defined The set of inverses are defined

J = {xJ = {x-1-1| x | x G} G}

This can be defined with respect to any This can be defined with respect to any identity.identity.

Digroup ExampleDigroup Example 00 11 22 33 44 55

00 11 00 11 00 00 11

11 00 11 00 11 11 00

22 44 22 44 22 22 44

33 55 33 55 33 33 55

44 22 44 22 44 44 22

55 33 55 33 55 55 33

00 11 22 33 44 55

00 11 00 33 22 55 44

11 00 11 22 33 44 55

22 11 00 33 22 55 44

33 00 11 22 33 44 55

44 00 11 22 33 44 55

55 11 00 33 22 55 44

Identities: {1, 3, 4} 1 is a bar unit: 1-1

= 1-1 = {0, 1}

3 is not a bar unit: 3-1 = {2,3} but 3-1

= {3, 5}

Trivial DigroupsTrivial DigroupsEvery element is a Every element is a bar unitbar unit

There is one trivial digroup of There is one trivial digroup of every orderevery order 00 11 22 nn

00 00 11 22 nn

11 00 11 22 nn

22 00 11 22 nn

nn 00 11 22 nn

00 11 22 nn

00 00 00 00 00

11 11 11 11 11

22 22 22 22 22

nn nn nn nn nn

GroupsGroupsExample - 5

00 11 22 33 44

00 33 00 11 44 22

11 00 11 22 33 44

22 11 22 44 00 33

33 44 33 00 22 11

44 22 44 33 11 00

00 11 22 33 44

00 33 00 11 44 22

11 00 11 22 33 44

22 11 22 44 00 33

33 44 33 00 22 11

44 22 44 33 11 00

Note that the two operations are the same.

Fact: A digroup is a group iff Fact: A digroup is a group iff = = iff the bar unit is the only identity iff |E| = 1iff the bar unit is the only identity iff |E| = 1

ContentsContents

SubdigroupsSubdigroups

Lagrange’s TheoremLagrange’s Theorem

CommutantCommutant

Abelian DigroupsAbelian Digroups

Digroup ClassificationDigroup Classification

SubdigroupsSubdigroupsDefinition:Definition: We call a subset H of a digroup G a subdigroup if H has the structure of a digroup under the operations of G. Theorem : In order that H be a subdigroup of digroup G, it is necessary and sufficient that (1) there exists e H such that e is a bar-unit of H, and (2) for all f, g, l, m, n H the elements f e, g−1 l, and m n−1 belong to H.Theorem: Let G be a digroup. Consider any nonempty subset S E such that S contains a bar unit. Define H G as

H = J e

1 J e

2

J e

3 … J e

n , where n = |S| and e i

S. H is a subdigroup if S. H is a subdigroup if

J e

1 J e

2

J e

3 … J e

n = J e

1

J e

2 J e

3

… J e

n

Lagrange CorrespondenceLagrange Correspondence

Lemma Lemma : J is a group: J is a group

Lemma Lemma : If G is a right group, : If G is a right group, G = J G = J E E (J x E, (J x E, ) )Lemma:Lemma: |G| = |J| * |E| |G| = |J| * |E|

Lagrange’s Theorem for Digroups: Lagrange’s Theorem for Digroups: Let G Let G be digroup (G, be digroup (G, , , ) and H ) and H be a be a subdigroup ofsubdigroup of G. G.

| H| | H| || |G| and [J:J |G| and [J:JHH] | [G: H] ] | [G: H] |E |EHH| | |E|.| | |E|.

CommutantCommutantThere are four potential candidates There are four potential candidates for the commutant of a digroup Gfor the commutant of a digroup G

1.1. C1 = {x C1 = {x G| g G| g x = x x = x g g g g G} G}

2.2. C2= {x C2= {x G| g G| g x = x x = x g g g g G} G}

3.3. C3= {x C3= {x G| g G| g x = x x = x g g g g G} G}

4.4. C4= {x C4= {x G| g G| g x = x x = x g g g g G} G}

Abelian DigroupsAbelian Digroups 00 11 22 33 44 55

00 11 00 11 00 00 11

11 00 11 00 11 11 00

22 33 22 33 22 22 33

33 22 33 22 33 33 22

44 55 44 55 44 44 55

55 44 55 44 55 55 44

00 11 22 33 44 55

00 11 00 33 22 55 44

11 00 11 22 33 44 55

22 11 00 33 22 55 44

33 00 11 22 33 44 55

44 00 11 22 33 44 55

55 11 00 33 22 55 44

A digroup is abelian iff the transpose of the Cayley table is the Cayley table is the Cayley table Cayley tableThe definition of abelian is x y = y y = y x for all x, y x for all x, y G G

Digroup ClassificationsDigroup Classifications

There are primarily 4 types of digroupsThere are primarily 4 types of digroups

1.1. GroupsGroups

2.2. Trivial DigroupsTrivial Digroups

3.3. Principal DigroupsPrincipal Digroups

4.4. Permuted DigroupsPermuted Digroups

Partitions/ Inverse SetsPartitions/ Inverse Sets

For a digroup D, let |E| = m and |J| = n . For a digroup D, let |E| = m and |J| = n . The columns of the The columns of the Cayley table are Cayley table are broken into m partitions with n elements broken into m partitions with n elements in each partition that are in each column in each partition that are in each column m times.m times.

Similarly the rows of the Similarly the rows of the Cayley table Cayley table are broken into m partitions with n are broken into m partitions with n elements in each partition that are in elements in each partition that are in each row m times.each row m times.

ExampleExample 00 11 22 33 44 55

00 11 00 11 00 00 11

11 00 11 00 11 11 00

22 44 22 44 22 22 44

33 55 33 55 33 33 55

44 22 44 22 44 44 22

55 33 55 33 55 55 33

00 11 22 33 44 55

00 11 00 33 22 55 44

11 00 11 22 33 44 55

22 11 00 33 22 55 44

33 00 11 22 33 44 55

44 00 11 22 33 44 55

55 11 00 33 22 55 44

D2 (6, 2)Partitions: {1,0}, {2,3}, {4,5} {1,0},{2,4},{3,5}Rows vs. Columns

Principal DigroupsPrincipal Digroups

The Principal Digroup of order k based The Principal Digroup of order k based upon a group G is notated:upon a group G is notated:

D 1 ( k, J )

The principal digroup is the digroup whose partitions for are the same as those for

We say this is the Principal Digroup of order k based upon the group J.

ExampleExample 00 11 22 33 44 55

00 11 00 11 00 00 11

11 00 11 00 11 11 00

22 33 22 33 22 33 22

33 22 33 22 33 22 33

44 55 44 55 44 55 44

55 44 55 44 55 44 55

00 11 22 33 44 55

00 11 00 33 22 55 44

11 00 11 22 33 44 55

22 11 00 33 22 55 44

33 00 11 22 33 44 55

44 00 11 22 33 44 55

55 11 00 33 22 55 44

D1 ( 6, 2)Partitions: {1,0}, {2,3}, {4,5} {1,0}, {2,3}, {4,5}

Arranging a Principal DigroupArranging a Principal Digroup

00 11 22 33 44 55

00 11 00 33 22 55 44

11 00 11 22 33 44 55

22 11 00 33 22 55 44

33 00 11 22 33 44 55

44 00 11 22 33 44 55

55 11 00 33 22 55 44

00 11 22 33 44 55

11 00 11 22 33 44 55

33 00 11 22 33 44 55

44 00 11 22 33 44 55

00 11 00 33 22 55 44

22 11 00 33 22 55 44

55 11 00 33 22 55 44

00 11

11 00 11

00 11 00

22 33

22 33 22

33 22 33

44 55

44 44 55

55 55 44

2

D1 ( 6, 2)

a1 ee1 a2 ee1 an ee1 a1

eem

a2

eem

an

eem

a1 ee1

J J J eem

a2 ee1

an ee1

a1

eem

JJ

J eema2

eem

an

eem

Therefore the number of Principle Digroups with |J||J|=n , |E|= m is going to be =n , |E|= m is going to be the number of ways that the number of ways that you can arrange J, or the you can arrange J, or the number of groups of order number of groups of order n.n.

Theorem: For any group G of order n, and for any multiple mn of n, there exists a digroup D such that the order of D is mn and the inverse set J of D is isomorphic to G. (every possible |E| has a Principal digroup) Theorem: Two Digroups ( D, , ) and ( D , , ) with non-isomorphic J must be non-isomorphic, and with isomorphic J must be isomorphic. (there is one principal digroup for every distinct J)Result: The number of principal digroups of order k with |E|= m is the number of groups of order k/m.In other words, every non-group, non-trivial digroup has In other words, every non-group, non-trivial digroup has a a structure that is the same for every particular basis particular basis group J.group J.

Digroup Digroup orderorder

|E||E| Names

1212 11 12, 2 x 6 , A4 , D6 , T

22 D1 ( 12, 6), D1 ( 12,

D3 )

Therefore the number of Principle Digroups with |J||J|=n , |E|= m is going to be =n , |E|= m is going to be the number of ways that the number of ways that you can arrange J, or the you can arrange J, or the number of groups of order number of groups of order n :n :

Permuted DigroupsPermuted DigroupsSince all the Since all the structures are the same for a particular J basis group all that is left is to consider digroups that have structures with different partitions.The Permuted Digroup of order k based The Permuted Digroup of order k based upon a group J is notated:upon a group J is notated:

D i ( k, J ) where i > 1The permuted digroups are the digroup

whose partitions for are different from those for .

ExampleExample 00 11 22 33 44 55

00 11 00 33 22 55 44

11 00 11 22 33 44 55

22 11 00 33 22 55 44

33 00 11 22 33 44 55

44 00 11 22 33 44 55

55 11 00 33 22 55 44

D2 ( 6, 2)Partitions: {1,0}, {2,3}, {4,5} {1,0},{2,4},{3,5}

00 11 22 33 44 55

00 11 00 11 00 00 11

11 00 11 00 11 11 00

22 44 22 44 22 22 44

33 55 33 55 33 33 55

44 22 44 22 44 44 22

55 33 55 33 55 55 33

The number of Permuted The number of Permuted DigroupsDigroups

The complete structure of The complete structure of is the same for all non-trivial, non-group digroups. Therefore, the permuted digroups are the only remaining possible digroups that remain to be classified. How many of them are there?

Theorem: Consider a digroup ( D, , ) with |D|= n. The number of permuted digroups with |E| = n/p is

The ClassificationThe ClassificationDigroupsDigroups

OrderOrder |E||E| NameName

22 11 2

22 trivialtrivial

33 11 3

33 trivialtrivial

44 11 4 , 2 x 2

22 D1 ( 4, 2)

44 trivialtrivial

55 11 5

55 trivialtrivial

66 11 6 , D3

22 D1 ( 6, 3)

33 D1 ( 6, 2), D2 ( 6, 2),

66 trivialtrivial

OrderOrder |E||E| NameName

77 11 7

77 trivialtrivial

88 11 8 , 4 x 2 , 2 x

2 x 2 , D4 , Q

22 D1 ( 8, 4), D1 ( 8, 2 x 2 )

44 D1 ( 8, 2), D2 ( 8, 2 )

88 trivial

44 trivialtrivial

99 11 9

33 D1 ( 9, 3)

99 trivial

1010 11 10 , D5

22 D1 ( 10, 5)

55 D1 ( 10, 2) , D2 ( 10, 2 x

2 ), D3 ( 10, 2 x

2 )

OrderOrder |E||E| NameName

1010 1010 trivial

1111 11 11

1111 trivial

1212 11 12, 2 x 6 , A4 , D6 , T

22 D1 ( 12, 6), D1 ( 12,

D3 )

33 D1 ( 12, 4),

D1 ( 12, 4), D2 ( 12, 2

x 2 ),

D2 ( 12, 2 x 2 ),

44 D1 ( 12, 3), D2 ( 12,

3 ),

99 66 D1 ( 12, 2), D2 ( 12,

D3 ),

D2 ( 12, 2)

1212 trivial

1313 11 13

1313 trivial

ConclusionConclusionWork still to be done in the study of digroups:Work still to be done in the study of digroups:

The number of digroups of form The number of digroups of form D i ( k, J ) where i > 1, where |J| is compositeSylow and Cauchy CorrespondencesSylow and Cauchy CorrespondencesClass Equation?Class Equation?Decomposition TheoremDecomposition TheoremThe full solution to the Coquecigrue problem The full solution to the Coquecigrue problem

This research was conducted collaboratively with Kyle Prifogle, This research was conducted collaboratively with Kyle Prifogle, Andrew Magyar, and William Young as a part of Wabash Andrew Magyar, and William Young as a part of Wabash College’s Summer Institute in Algebra (WSIA) under the College’s Summer Institute in Algebra (WSIA) under the advisement of JD Phillips and funded by the NSF.advisement of JD Phillips and funded by the NSF.

Thank you for your time!Thank you for your time!

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