DIM0121 – Fundamentos
Matemáticos da Computação II
Umberto Rivieccio
DIMAp – Departamento de Informática e Matemática Aplicada
urivieccio@dimap.ufrn.br
UFRN
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Course overview
1. Languages and algebras.
2. Subalgebras.
3. Homomorphisms and isomorphisms.
4. Congruences, quotient algebras and direct products.
5. Term algebras and equations.
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Bibliography
BURRIS, S. & Sankappanavar, H.P. A Course in Universal Algebra. The
Millenium Edition, 2000 (Ch. II).
http://math.hawaii.edu/~ralph/Classes/619/univ-algebra.pdf
SCHEINERMAN, E. R. Mathematics. A discrete Introduction. Cengage
Learning, 2013 (Ch. 8).
Umberto Rivieccio (DIMAp - UFRN)
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n-ary operations and languages
Umberto Rivieccio (DIMAp - UFRN)
(BS, II.1)
DIM0121 – FMC II – Algebra
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n-ary operations and languages
Umberto Rivieccio (DIMAp - UFRN)
(BS, II.1)
DIM0121 – FMC II – Algebra
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Algebras
(BS, II.1)
Definition
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DIM0121 – FMC II – Algebra
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Algebras
(BS, II.1)
Examples: groups (grupos), rings (anéis)
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DIM0121 – FMC II – Algebra
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Algebras
(BS, II.1)
Examples: groups (grupos), rings (anéis)
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Algebras
(BS, I.1)
Examples: lattices (reticulados)
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Algebras
(BS, I.1)
Examples: lattices (reticulados)
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Algebras
(BS, II.1)
Examples: Boolean algebras (álgebras booleanas)
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DIM0121 – FMC II – Algebra
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Groups
(BS, II.1 vs. Scheinerman, ch. 8)
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Groups
(BS, II.1 vs. Scheinerman, ch. 8)
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Groups
(Scheinerman, ch. 8)
Examples
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DIM0121 – FMC II – Algebra
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Groups
(Scheinerman, ch. 8)
Examples: Klein 4-group
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DIM0121 – FMC II – Algebra
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Groups
Examples: group of permutations
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DIM0121 – FMC II – Algebra
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Groups
Equational proofs
Lemma 1
Let G = hG , ·,−1 , 1i be a group. Then, for all a ∈ G ,
a = (a−1 )−1 .
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DIM0121 – FMC II – Algebra
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Groups
Equational proofs
Lemma 1
Let G = hG , ·,−1 , 1i be a group. Then, for all a ∈ G ,
a = (a−1 )−1 .
Proof
a=1·a
(G2)
·a
−1
= (a
−1 −1
)
· (a
−1
= (a
−1 −1
·1
= (a
−1 −1
= ((a
−1 −1
)
)
Umberto Rivieccio (DIMAp - UFRN)
)·a
(G3)
· a)
(G1)
(G3)
(G2).
)
DIM0121 – FMC II – Algebra
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Groups
Equational proofs
Lemma 2
Let G = hG , ·,−1 , 1i be a group. Then, for all a, b ∈ G ,
(a · b) · (b −1 · a−1 ) = 1.
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DIM0121 – FMC II – Algebra
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Groups
Equational proofs
Lemma 2
Let G = hG , ·,−1 , 1i be a group. Then, for all a, b ∈ G ,
(a · b) · (b −1 · a−1 ) = 1.
Proof
(a · b) · (b −1 · a−1 ) = a · (b · (b −1 · a−1 ))
= a · ((b · b
= a · (1 · a
=a·a
−1
−1
)·a
−1
)
−1
=1
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
)
(G1)
(G1)
(G3)
(G2)
(G3).
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Groups
Equational proofs
Lemma 3
Let G = hG , ·,−1 , 1i be a group. Then, for all a, b ∈ G ,
(a · b)−1 = b −1 · a−1 .
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DIM0121 – FMC II – Algebra
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Groups
Equational proofs
Lemma 3
Let G = hG , ·,−1 , 1i be a group. Then, for all a, b ∈ G ,
(a · b)−1 = b −1 · a−1 .
Proof
(a · b)−1 = (a · b)−1 · 1
= (a · b)
−1
· ((a · b) · (b
−1
= ((a · b)
= 1 · (b
=b
−1
Umberto Rivieccio (DIMAp - UFRN)
−1
·a
(G2)
−1
· (a · b)) · (b
·a
−1
·a
−1
−1
·a
))
−1
)
−1
DIM0121 – FMC II – Algebra
)
Lemma 2
(G1)
(G3)
(G2).
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Groups
(Scheinerman, ch. 8, sec. 40)
Exercises
Notation:
2A = P(A) and X ∆Y = (X − Y ) ∪ (Y − X ).
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DIM0121 – FMC II – Algebra
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Groups
(Scheinerman, ch. 8, sec. 40)
Exercises
Notation:
2A = P(A) and X ∆Y = (X − Y ) ∪ (Y − X ).
Exercise (on lattices): show that both definitions of lattices (as algebras
and as posets) are equivalent (see BS, I.1, p. 8).
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DIM0121 – FMC II – Algebra
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Subalgebras
(BS, II.2)
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DIM0121 – FMC II – Algebra
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Subalgebras
(BS, II.2)
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Subalgebras
(Scheinerman, ch. 8, sec. 42)
Example: subgroups
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DIM0121 – FMC II – Algebra
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Subalgebras
(Scheinerman, ch. 8, sec. 42)
Example: subgroups
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DIM0121 – FMC II – Algebra
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Subalgebras
(Scheinerman, ch. 8, sec. 42, p. 303)
Example: subgroups
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DIM0121 – FMC II – Algebra
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Subalgebras
(Scheinerman, ch. 8, sec. 42, p. 303)
Example: subgroups
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Subalgebras
(Scheinerman, ch. 8, sec. 42, p. 303)
Example: subgroups
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DIM0121 – FMC II – Algebra
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Subalgebras
(Scheinerman, ch. 8, sec. 42
Example: subgroups of a group of permutations
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Definition
Let G = hG , ·,−1 , 1i be a group and N ⊆ G a subuniverse of G. We say
that N is (the universe of) a normal subgroup iff, for all a, b ∈ G ,
a·b ∈N
Umberto Rivieccio (DIMAp - UFRN)
iff
b · a ∈ N.
DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Definition
Let G = hG , ·,−1 , 1i be a group and N ⊆ G a subuniverse of G. We say
that N is (the universe of) a normal subgroup iff, for all a, b ∈ G ,
a·b ∈N
iff
b · a ∈ N.
Lemma 4
A subgroup N ⊆ G is normal if and only if aN = Na for all a ∈ G , where
aN := {a · n : n ∈ N}
and
Na := {n · a : n ∈ N}.
Exercise: Prove the lemma (before looking at the next slide).
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
Thus (b · a−1 ) · a = b ∈ Na (justify).
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
Thus (b · a−1 ) · a = b ∈ Na (justify).
Hence, aN ⊆ Na (Exercise: show Na ⊆ aN).
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
Thus (b · a−1 ) · a = b ∈ Na (justify).
Hence, aN ⊆ Na (Exercise: show Na ⊆ aN).
Conversely, suppose aN = Na for all a ∈ G (H2). Let b · c ∈ N.
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
Thus (b · a−1 ) · a = b ∈ Na (justify).
Hence, aN ⊆ Na (Exercise: show Na ⊆ aN).
Conversely, suppose aN = Na for all a ∈ G (H2). Let b · c ∈ N.
Then b −1 · (b · c) = c ∈ b −1 N (justify).
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
Thus (b · a−1 ) · a = b ∈ Na (justify).
Hence, aN ⊆ Na (Exercise: show Na ⊆ aN).
Conversely, suppose aN = Na for all a ∈ G (H2). Let b · c ∈ N.
Then b −1 · (b · c) = c ∈ b −1 N (justify).
Using (H2) we get c ∈ Nb −1 , i.e. c = n · b −1 for some n ∈ N.
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DIM0121 – FMC II – Algebra
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
Thus (b · a−1 ) · a = b ∈ Na (justify).
Hence, aN ⊆ Na (Exercise: show Na ⊆ aN).
Conversely, suppose aN = Na for all a ∈ G (H2). Let b · c ∈ N.
Then b −1 · (b · c) = c ∈ b −1 N (justify).
Using (H2) we get c ∈ Nb −1 , i.e. c = n · b −1 for some n ∈ N.
Then c · b = (n · b −1 ) · b = n (justify).
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Subalgebras
Normal subgroups
Proof of Lemma 4 (with gaps)
Let N be normal (H1). Let b ∈ aN, i.e. b = a · n for some n ∈ N.
Then a−1 · b = a−1 · (a · n) = n (Exercise: justify this passage).
From a−1 · b ∈ N, by (H1), we get b · a−1 ∈ N.
Thus (b · a−1 ) · a = b ∈ Na (justify).
Hence, aN ⊆ Na (Exercise: show Na ⊆ aN).
Conversely, suppose aN = Na for all a ∈ G (H2). Let b · c ∈ N.
Then b −1 · (b · c) = c ∈ b −1 N (justify).
Using (H2) we get c ∈ Nb −1 , i.e. c = n · b −1 for some n ∈ N.
Then c · b = (n · b −1 ) · b = n (justify).
This means c · b ∈ N, as required.
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Subalgebras
(Scheinerman, ch. 8, sec. 42)
Subgroups: exercises
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Subalgebras
Subgroups: exercises
Exercise
Hint: use Lemma 4 when convenient.
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
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DIM0121 – FMC II – Algebra
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
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DIM0121 – FMC II – Algebra
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c.
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 .
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 . Since
(c −1 · (c · a)) · c −1 = a · c −1 and (c −1 · (a · c)) · c −1 = c −1 · a (justify), we have
a · c −1 = c −1 · a.
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 . Since
(c −1 · (c · a)) · c −1 = a · c −1 and (c −1 · (a · c)) · c −1 = c −1 · a (justify), we have
a · c −1 = c −1 · a. Hence c −1 ∈ C .
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 . Since
(c −1 · (c · a)) · c −1 = a · c −1 and (c −1 · (a · c)) · c −1 = c −1 · a (justify), we have
a · c −1 = c −1 · a. Hence c −1 ∈ C .
(iii). Let c, d ∈ C and a ∈ G . Then:
a · (c · d) = (a · c) · d = (c · a) · d = d · (c · a) = (d · c) · a = (c · d) · a
(justify each of the above equalities).
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
25 / 53
Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 . Since
(c −1 · (c · a)) · c −1 = a · c −1 and (c −1 · (a · c)) · c −1 = c −1 · a (justify), we have
a · c −1 = c −1 · a. Hence c −1 ∈ C .
(iii). Let c, d ∈ C and a ∈ G . Then:
a · (c · d) = (a · c) · d = (c · a) · d = d · (c · a) = (d · c) · a = (c · d) · a
(justify each of the above equalities).
(iv). C is normal: Let c · d ∈ C . Then (c · d) · d = d · (c · d) by def. of C .
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
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Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 . Since
(c −1 · (c · a)) · c −1 = a · c −1 and (c −1 · (a · c)) · c −1 = c −1 · a (justify), we have
a · c −1 = c −1 · a. Hence c −1 ∈ C .
(iii). Let c, d ∈ C and a ∈ G . Then:
a · (c · d) = (a · c) · d = (c · a) · d = d · (c · a) = (d · c) · a = (c · d) · a
(justify each of the above equalities).
(iv). C is normal: Let c · d ∈ C . Then (c · d) · d = d · (c · d) by def. of C . From
this we have ((c · d) · d) · d −1 = (d · (c · d)) · d −1 .
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
25 / 53
Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 . Since
(c −1 · (c · a)) · c −1 = a · c −1 and (c −1 · (a · c)) · c −1 = c −1 · a (justify), we have
a · c −1 = c −1 · a. Hence c −1 ∈ C .
(iii). Let c, d ∈ C and a ∈ G . Then:
a · (c · d) = (a · c) · d = (c · a) · d = d · (c · a) = (d · c) · a = (c · d) · a
(justify each of the above equalities).
(iv). C is normal: Let c · d ∈ C . Then (c · d) · d = d · (c · d) by def. of C . From
this we have ((c · d) · d) · d −1 = (d · (c · d)) · d −1 . Since ((c · d) · d) · d −1 = c · d
and (d · (c · d)) · d −1 = d · c (justify), we conclude c · d = d · c. So d · c ∈ C .
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
25 / 53
Subgroups
Exercise: the center of a group
The set C := {c ∈ G : a · c = c · a for all a ∈ G } is the universe of a
normal subgroup of every group G = hG , ·,−1 , 1i.
(i). 1 ∈ C , because 1 · a = a · 1 = a for all a ∈ G (justify).
(ii). For all c ∈ C and a ∈ G , we have c · a = a · c. Then
c −1 · (c · a) = c −1 · (a · c) and (c −1 · (c · a)) · c −1 = (c −1 · (a · c)) · c −1 . Since
(c −1 · (c · a)) · c −1 = a · c −1 and (c −1 · (a · c)) · c −1 = c −1 · a (justify), we have
a · c −1 = c −1 · a. Hence c −1 ∈ C .
(iii). Let c, d ∈ C and a ∈ G . Then:
a · (c · d) = (a · c) · d = (c · a) · d = d · (c · a) = (d · c) · a = (c · d) · a
(justify each of the above equalities).
(iv). C is normal: Let c · d ∈ C . Then (c · d) · d = d · (c · d) by def. of C . From
this we have ((c · d) · d) · d −1 = (d · (c · d)) · d −1 . Since ((c · d) · d) · d −1 = c · d
and (d · (c · d)) · d −1 = d · c (justify), we conclude c · d = d · c. So d · c ∈ C .
Exercise: prove (ii) using Lemma 1 and Lemma 3.
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Subalgebras
The subuniverse/subalgebra generated by a set
The above is well-defined because the arbitrary intersection of subuniverses
is a subuniverse (Exercise).
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Subalgebras
The subuniverse/subalgebra generated by a set
The above is well-defined because the arbitrary intersection of subuniverses
is a subuniverse (Exercise).
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Homomorphisms and isomorphisms
(BS, II.6)
An injective homomorphism is called an embedding (imersão).
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Homomorphisms and subuniverses
Umberto Rivieccio (DIMAp - UFRN)
(BS, II.6)
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Group isomorphism
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(Scheinerman, ch. 8, sec. 41)
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Group isomorphism
(Scheinerman, ch. 8, sec. 41)
Exercises
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Homomorphisms and subuniverses
Normal subgroups
Proposition
Let hA, ·,−1 , 1i and hB, ·,−1 , 1i be groups and let α : A → B be a
homomorphism. Then the set
N := {a ∈ A : α(a) = 1}
is the universe of a normal subgroup of A.
Exercise
Prove the above proposition.
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Homomorphisms
(BS, II.6)
Composition
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Homomorphisms
(BS, II.6)
Kernel (Núcleo)
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Homomorphisms
(BS, II.6)
Kernel (Núcleo)
Exercise. Show that:
1
ker(α) is an equivalence relation.
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Homomorphisms
(BS, II.6)
Kernel (Núcleo)
Exercise. Show that:
1
ker(α) is an equivalence relation.
2
ker(α) is compatible with any (n-ary) operation f :
for all a1 , . . . , an , b1 , . . . , bn ∈ A, if (a1 , b1 ), . . . , (an , bn ) ∈ ker(α),
then (f A (a1 , . . . , an ), f A (b1 , . . . , bn )) ∈ ker(α).
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Congruences
(BS, II.5)
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Congruences
(BS, II.5)
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Congruences
(BS, II.5)
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Let A = hA, ·,−1 , 1i be a group and B ≤ A a subgroup of A. Define the
relation:
θB := {(a, b) ∈ A × A : a · b −1 ∈ B}.
Then:
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Let A = hA, ·,−1 , 1i be a group and B ≤ A a subgroup of A. Define the
relation:
θB := {(a, b) ∈ A × A : a · b −1 ∈ B}.
Then:
1
θB is an equivalence relation (Lemma 42.6, p. 305).
Umberto Rivieccio (DIMAp - UFRN)
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Let A = hA, ·,−1 , 1i be a group and B ≤ A a subgroup of A. Define the
relation:
θB := {(a, b) ∈ A × A : a · b −1 ∈ B}.
Then:
1
θB is an equivalence relation (Lemma 42.6, p. 305).
2
If B is normal, then θB is a congruence of A and:
Umberto Rivieccio (DIMAp - UFRN)
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36 / 53
Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Let A = hA, ·,−1 , 1i be a group and B ≤ A a subgroup of A. Define the
relation:
θB := {(a, b) ∈ A × A : a · b −1 ∈ B}.
Then:
1
θB is an equivalence relation (Lemma 42.6, p. 305).
2
If B is normal, then θB is a congruence of A and:
3
B = {a ∈ A : (a, 1) ∈ θB }.
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Let A = hA, ·,−1 , 1i be a group and B ≤ A a subgroup of A. Define the
relation:
θB := {(a, b) ∈ A × A : a · b −1 ∈ B}.
Then:
1
θB is an equivalence relation (Lemma 42.6, p. 305).
2
If B is normal, then θB is a congruence of A and:
3
B = {a ∈ A : (a, 1) ∈ θB }.
Umberto Rivieccio (DIMAp - UFRN)
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36 / 53
Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Let A = hA, ·,−1 , 1i be a group and B ≤ A a subgroup of A. Define the
relation:
θB := {(a, b) ∈ A × A : a · b −1 ∈ B}.
Then:
1
θB is an equivalence relation (Lemma 42.6, p. 305).
2
If B is normal, then θB is a congruence of A and:
3
B = {a ∈ A : (a, 1) ∈ θB }.
Recall that normal means: for all a, b ∈ A,
a·b ∈B
Umberto Rivieccio (DIMAp - UFRN)
iff
b · a ∈ B.
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Proof of (2)
(θB is compatible with ·). Assume (a, b), (c, d) ∈ θB , i.e., a · b −1 ∈ B (2.1) and
c · d −1 ∈ B (2.2).
Since B is normal, (2.1) implies b −1 · a ∈ B (2.3).
Since B is a subuniverse, from (2.2) and (2.3) we get (c · d −1 ) · (b −1 · a) ∈ B
(2.4).
Using (G1), from (2.4) we get (c · d −1 ) · (b −1 · a) = (c · d −1 · b −1 ) · a ∈ B (2.5).
Since B is normal, (2.5) implies a · (c · d −1 · b −1 ) ∈ B.
Using (G1), from (2.5) we get a · (c · d −1 · b −1 ) = (a · c) · (d −1 · b −1 ) ∈ B (2.6).
Since d −1 · b −1 = (b · d)−1 (Lemma 3, slide 15), from (2.6) we get
(a · c) · (d −1 · b −1 ) = (a · c) · (b · d)−1 ∈ B, i.e., i.e., (a · c, b · d) ∈ θB .
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup
Proof of (2)
(θB is compatible with ·). Assume (a, b), (c, d) ∈ θB , i.e., a · b −1 ∈ B (2.1) and
c · d −1 ∈ B (2.2).
Since B is normal, (2.1) implies b −1 · a ∈ B (2.3).
Since B is a subuniverse, from (2.2) and (2.3) we get (c · d −1 ) · (b −1 · a) ∈ B
(2.4).
Using (G1), from (2.4) we get (c · d −1 ) · (b −1 · a) = (c · d −1 · b −1 ) · a ∈ B (2.5).
Since B is normal, (2.5) implies a · (c · d −1 · b −1 ) ∈ B.
Using (G1), from (2.5) we get a · (c · d −1 · b −1 ) = (a · c) · (d −1 · b −1 ) ∈ B (2.6).
Since d −1 · b −1 = (b · d)−1 (Lemma 3, slide 15), from (2.6) we get
(a · c) · (d −1 · b −1 ) = (a · c) · (b · d)−1 ∈ B, i.e., i.e., (a · c, b · d) ∈ θB .
(θB is compatible with −1 ). Assume (a, b) ∈ θB , i.e., a · b −1 ∈ B (2.1). Since B is
a subuniverse, (2.1) implies (a · b −1 )−1 ∈ B (2.7). From (2.7) and (Lemma 3,
slide 15) we get (a · b −1 )−1 = (b −1 )−1 · a−1 ∈ B (2.8). Finally, from (Lemma 1,
slide 13) and (2.8), we get (b −1 )−1 · a−1 = b · a−1 ∈ B, i.e. (b, a) ∈ θB .
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup: the congruence modulo n ∈ Z
Let Z = hZ, +, −, 0i be the additive group of integers and let n ∈ Z. The
subuniverse generated by {n} in Z is
B = {a ∈ Z : n divides a}.
(Notice: B is obviously normal because Z is commutative).
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Congruences
(Scheinerman, ch. 8, sec. 42)
Congruence modulo a subgroup: the congruence modulo n ∈ Z
Let Z = hZ, +, −, 0i be the additive group of integers and let n ∈ Z. The
subuniverse generated by {n} in Z is
B = {a ∈ Z : n divides a}.
(Notice: B is obviously normal because Z is commutative).
Then
θB = {(a, b) ∈ Z × Z : a + (−b) ∈ B}
= {(a, b) ∈ Z × Z : n divides a + (−b) ∈ B}
that is, θB is precisely the usual congruence modulo n on Z
(cf. Scheinerman, Exercise 42.15).
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Congruences
(Scheinerman, ch. 8, sec. 42)
Exercises
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Quotient algebras
Umberto Rivieccio (DIMAp - UFRN)
(BS, II.5)
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Quotient algebras
Umberto Rivieccio (DIMAp - UFRN)
(BS, II.5)
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Quotient algebras and homomorphisms
Umberto Rivieccio (DIMAp - UFRN)
DIM0121 – FMC II – Algebra
(BS, II.6)
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Quotient algebras and homomorphisms
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DIM0121 – FMC II – Algebra
(BS, II.6)
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Quotient algebras and homomorphisms
(BS, II.6)
Exercises
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Direct products
(BS, II.7)
(Produtos diretos)
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Direct products
(BS, II.7)
(Produtos diretos)
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Direct products
(BS, II.7)
(Produtos diretos)
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Direct products
(BS, II.7)
(Produtos diretos)
Umberto Rivieccio (DIMAp - UFRN)
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Direct products
(BS, II.7)
(Produtos diretos)
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Terms and term algebra
(BS, II.10)
Terms
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Terms and term algebra
(BS, II.10)
Terms: examples
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Terms and term algebra
(BS, II.10)
Term functions
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Terms and term algebra
(BS, II.10)
Term functions
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Terms and term algebra
(BS, II.10)
Term algebra
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Identities
(BS, II.11
Identities (identidades/equações)
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Identities
(BS, II.11
Identities (identidades/equações)
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Identities
(BS, II.11
Identities (identidades/equações)
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