Algebraic Structures


Discrete Mathematics I

11th lecture, December 16, 2018

Martin J. Dürst


© 2006-19 Martin J. Dürst Aoyama Gakuin University

Today's Schedule


Remaining Schedule

About makeup classes: The material in the makeup class is part of the final exam. If you have another makeup class at the same time, please inform the teacher as soon as possible.

補講について: 補講の内容は期末試験の対象。補講が別の補講とぶつかる場合には事前に申し出ること。


Final Exam・期末試験

Complete contents of lecture and handouts
Past exams: 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018
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Important Points about Final Exam


Leftovers of Last Lecture

Hasse diagrams, equivalence relations and order relations in matrix representation


Summary of Last Lecture

We defined the following properties of binary relations:

  1. Reflexive: xA:xRx; ∀xA: (x, x) ∈ R
  2. Symmetric: ∀x, yA: xRyyRx;
    x, yA: (x, y) ∈ R ⇔ (y, x) ∈ R
  3. Antisymmetric: ∀x, yA: xRyyRxx=y
  4. Transitive: ∀x, y, zA: xRyyRzxRz

A relation that is reflexive, antisymmetric, and transitive is a (partial) order relation.

A relation that is reflexive, symmetric, and transitive is an equivalence relation.

(Partial) order relations can be represented with Hasse diagrams.


Last Week's Homework:
Combinations of Properties of Relations@@@@ adjust updated problem text

Investigate all combinations of the four properties of relations introduced in this lecture (reflexive, symmetric, antisymmetric, transitive). For each combination, give an example relation on the minimum size set possible, or explain why such a combination is impossible. Present the 16 combinations in a table similar to the tables used in the homework of lecture 4. In the table, also include a column with the minimum size of the set on which the example relation is formed. Use {b}, {b, c}, and {b, c, d} for sets with one, two, and three elements, respectively.

Hint: Two combinations are impossible. One combination is possible with a relation on an empty set. One combination is possible with a relation on a set of size one. Four combinations are possible with a relation on a set of size two. The other combinations need a relation on a set of size three.


Homework Solution

re­flex­ive sym­met­ric an­ti­sym­met­ric tran­si­tive min­i­mum size minimal example
F F F F 3 {(b,c), (c,d), (c,a)}
F F F T 3 {(b,b), (b,c), (b,d), (c,b), (c,c), (c,d)}
F F T F 3 {(b,c), (c,d)}
F F T T 2 {(b,c)}
F T F F 2 {(b,c), (c,b)}
F T F T 3 {(b,b), (b,d), (d,b), (d,d)}
F T T F - impossible
F T T T 1 {}
T F F F 3 {(b,b), (c,c), (d,d), (b,c), (c,d), (c,b)}
T F F T 3 {(b,b), (c,c), (d,d), (b,c), (c,b), (b,d), (c,d)}
T F T F 3 {(b,b), (c,c), (d,d), (b,c), (c,d)}
T F T T 2 {(b,b), (c,c), (b,c)} (order relation)
T T F F 3 {(b,b), (c,c), (d,d), (b,c), (c,d), (c,b), (d,c)}
T T F T 2 {(b,b), (c,c), (b,c), (c,b)} (equivalence relation)
T T T F - impossible
T T T T 0 {} (order and equivalence relation)

Explanation why two combinations are impossible:
In both cases, the relations need to be symmetric and antisymmetric, but not transitive. In relations that are both symmetric and antisymmetric, all positions except those on the (main) diagonal are false. Such relations are automatically transitive (because they are their own transitive closure).


Algebraic Structure

Very general view on mathematical objects

An algebraic structure is a class of mathematical objects that all share the same general structure.

Properties shared by all algebraic structures are:


Previously Encountered Examples


Example of Algebraic Structure: Group


The Integers with Addition as a Group (ℤ, +)


The Reals with Multiplication as a Group (ℝ-{0}, ·)


The Positive Reals with Multiplication as a Group (ℝ+, ·)




Permutations as Exchanges


Composition of Permutations


Permutations as a Group


Group Theorem: Uniqueness of Identity

Existence of identity element (axiom): ∃eA: ∀bA: eb = b = be

Theorem: The identity element of a group is unique
(∃cA: ∀xA: cx = x) ⇒ c = e

Another way to express this: There is only one identity element
(|{c|cA, ∀xA: cx = x)}|= 1)


cx = x [inverse axiom, closure]

(cx)•x' = xx' [associativity axiom]

c•(xx') = xx' [inverse axiom, on both sides]

ce = e [identity axiom]

c = e Q.E.D. (similar proof for right idenity)


Group Theorem: Uniqueness of Inverse

Existence of an inverse (axiom): ∀bA: ∃b'A: bb' = e = b'•b

Theorem: Each inverse is unique
a, b∈A: (ab = eb=a')


ab = e [applying a'• on the left]

a'•(ab) = a'•e [associativity axiom]

(a'•a)•b = a'•e [inverse axiom]

eb = a'•e [identity axiom, on both sides]

b = a' Q.E.D. (similar proof for left inverse)


Group Theorem: Cancellation Law

Theorem: ∀a, b, c ∈A: (ac = bca=b)


ac = bc [applying c' on the right]

(ac)•c' = (bc)•c' [associativity]

a•(cc') = b•(cc') [inverse axiom, on both sides]

ae = be [identity axiom, on both sides]

a = b Q.E.D. (similar proof for left cancellation)


Group Isomorphism


Examples of Isomorphic Groups

G e a b
e e a b
a a b e
b b e a
K 0 2 1
0 0 2 1
2 2 1 0
1 1 0 2
H 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1


Cayley Tables


This Week's Homework

Deadline: December 19, 2019 (Thursday), 19:00.

Format: A4 single page (using both sides is okay; NO cover page), easily readable handwriting (NO printouts), name (kanji and kana) and student number at the top right

Where to submit: Box in front of room O-529 (building O, 5th floor)

Homework 1: Create a Cayley table of the symmetric group of order 3. Use lexical order for the permutations. Use the notation introduced in this lecture.

Homework 2: If we define isomorphic groups as being "the same", there are two different groups of size 4. Give an example of each group as a Cayley table. Hint: Check all the conditions (axioms) for a group. There will be a deduction if you use the same elements of the group as another student.



algebraic structure
群 (ぐん)
group theory
inverse element
inverse, reciprocal
symmetric group
Cayley table
multiplication table
九九 (表)
group isomorphism
row heading
column heading
lexical (or lexicographic(al)) order