Download First Semester Abstract Algebra for Undergraduates

First Semester Abstract Algebra for
Undergraduates
Lecture notes by:
Khim R Shrestha, Ph. D.
Assistant Professor of Mathematics
University of Great Falls
Great Falls, Montana
Contents
1 Introduction to Groups
1
2 Groups
2
3 Finite Groups; Subgroups
4
4 Cyclic Groups
8
5 Permutation Groups
12
6 Isomorphisms
17
7 Cosets and Lagrange’s Theorem
21
8 External Direct Products
24
9 Normal Subgroups and Factor Groups
27
10 More on Group Homomorphisms
30
11 Introduction to Rings
34
12 Integral Domains
36
13 Ideals and Factor Rings
39
14 Ring Homomorphisms
42
ii
Chapter 1
Introduction to Groups
Homework: # 1, 3, 5, 7.
1
Chapter 2
Groups
Homework: # 1, 3, 5(a, d), 7, 9, 15, 19, 23, 25.
Definition 2.1. Let G be a set. A binary operation on G is a function ∗ : G×G → G.
The set G together with the binary operation ∗ will be written (G, ∗).
Definition 2.2. Let G be a set together with a binary operation ∗. We say (G, ∗) is
a group if
i. a ∗ b ∈ G for all a, b ∈ G. (Closure Property) (This is redundant)
ii. (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ G. (Associative Property)
iii. There is an element e ∈ G such that a ∗ e = a = e ∗ a for every a ∈ G. (Identity)
iv. For every a ∈ G there is an element b ∈ G such that a ∗ b = e = b ∗ a. (Inverse)
Elementary Properties of Groups
Theorem 2.1. The identity element of a group is unique.
Proof. Suppose that a group G has two identity elements e and e0 . Then a ∗ e =
e ∗ a = a and a ∗ e0 = e0 ∗ a = a for all a ∈ G. Now, e = e ∗ e0 = e0 .
Theorem 2.2. Let G be a group and a, b, c ∈ G. Then
i. a ∗ b = a ∗ c implies b = c. (left cancellation)
ii. b ∗ a = c ∗ a implies b = c. (right cancellation)
Proof. Let a0 be the inverse of a. Multiply on the left in (i) and on the right in (ii)
by a.
Theorem 2.3. The inverse of each element in a group is unique.
2
CHAPTER 2. GROUPS
3
Proof. Let G be a group and a ∈ G. Suppose that b and c are inverses of a, that is,
a ∗ b = b ∗ a = e and a ∗ c = c ∗ a = e. Now,
b = b ∗ e = b ∗ (a ∗ c) = (b ∗ a) ∗ c = e ∗ c = c.
Hence the inverse is unique.
Analogy of operators
∗
Multiplication
Addition
a∗b
ab
a+b
Identitye
1 or e
0
−1
Inverse
a
−a
n
a ∗ a ∗ · · · ∗ a(n a’s)
a
a + a + · · · + a that is na
ab−1
a−b
Theorem 2.4. For group elements a and b, (ab)−1 = b−1 a−1 .
Proof. We have (ab)(ab)−1 = e and (ab)(b−1 a−1 ) = a(bb−1 )a−1 = aea−1 = aa−1 = e.
Thus (ab)−1 and b−1 a−1 are both inverses of ab. Since the inverse is unique, we have
to have (ab)−1 = b−1 a−1 .
Definition 2.3. A group (G, ∗) is called an abelian group if a ∗ b = b ∗ a for all
a, b ∈ G.
Example 2.1. Prove that a group G is abelian if and only if (ab)−1 = a−1 b−1 .
Proof. Suppose that G is abelian. Then, (ab)−1 = b−1 a−1 = a−1 b−1 . Conversely,
assume that (ab)−1 = a−1 b−1 . This implies (ab)(a−1 b−1 ) = e. Multiply from right by
ba. We get,
(ab)(a−1 b−1 )(ba) = e(ba)
=⇒ ab = ab
Example 2.2. Let G be a group and (b−1 )n = (bn )−1 , where n is a positive integer.
Proof. bn (bn )−1 = e. And, bn (b−1 )n = (bb · · · b)(b−1 b−1 · · · b−1 ) = e. By the uniqueness
of the inverse (bn )−1 = (b−1 )n .
Chapter 3
Finite Groups; Subgroups
Homework: # 2, 4, 6, 12, 18, 20, 22, 26, 42, 44.
Definition 3.1. The number of elements of a group G is called the order of G and
is denoted by |G|.
Definition 3.2. The order of a group element g, denoted by |g|, is the smallest
positive integer n such that g n = 3 (or ng = 0 if the operation is addition). If there
is no such integer, then we say that the element has infinite order, that is, |g| = ∞.
Example 3.1. Take G = (Z, +). Then |G| = ∞. Also, the order of any element
k ∈ Z has infinite order.
Example 3.2. Consider U (10) = {1, 3, 7, 9} under multiplication modulo 10. So,
|U (10)| = 4. The order of the elements: |1| = 1, |3| = 4, |7| = 4, |9| = 2.
Example 3.3. Consider Z10 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} under addition modulo 10.
Then |Z10 | = 10. The order of the elements:
|0| = 1
|1| = 10
|2| = 5
|3| = 10
|4| = 5
|5| = 2
|6| = 5
|7| = 10
|8| = 5
|9| = 10
Definition 3.3. Let H be a subset of a group G. If H itself is a group under the
operation of G, then we say that H is a subgroup of G and write H ≤ G (or H < G
if H is a proper subgroup.)
The set {e} is a subgroup called trivial subgroup.
Example 3.4. SL2 (R) < GL2 (R).
Example 3.5. Z10 under addition modulo 10 is not a subgroup of Z under addition.
4
CHAPTER 3. FINITE GROUPS; SUBGROUPS
5
Subgroup Tests
Theorem 3.1. Let H be a non-empty subset of a group G. Then H is a subgroup of
G if and only if ab−1 ∈ H for all a, b ∈ H.
Proof. Let H ≤ G and a, b ∈ H. Since H is a group in itself, b−1 ∈ H and hence
ab−1 ∈ H.
Conversely, assume that ab−1 ∈ H for all a, b ∈ H. We want to show that H is a
group.
i. Associativity: H is associative because G is.
ii. Identity: We have ab−1 ∈ H for every a, b ∈ H. Take b = a. We get e = aa−1 ∈
H.
iii. Inverse: Let x ∈ H. Since e ∈ H, x−1 = ex−1 ∈ H.
iv. Closure: Let x, y ∈ H. Since y = (y −1 )−1 and y −1 ∈ H (by iii.), xy =
x(y −1 )−1 ∈ H.
Example 3.6. Let G be an abelian group with identity e. Then H = {x ∈ G : x2 =
e} is a subgroup of G.
Proof. Since e2 = e, e ∈ H. So H is nonempty. Let a, b ∈ H. Then by Theorem 3.1
it suffices to show that ab−1 ∈ H. For this we have to show that (ab−1 )2 = e. Since
G is abelian, (ab−1 )2 = (ab−1 )(ab−1 ) = a2 (b−1 )2 = e(b−1 )2 = (b2 )−1 = e−1 = e.
Corollary 3.2. Let H be a nonempty subset of a group G. Then H is a subgroup of
G if and only if a, b ∈ H and a−1 ∈ H whenever a, b ∈ H.
Proof. If H is a subgroup and a, b ∈ H then ab ∈ H and a−1 ∈ H.
To prove the other way around, it suffices to show that ab−1 ∈ H whenever
a, b ∈ H. Assume that a, b ∈ H. Then by hypothesis, b−1 ∈ H and ab−1 ∈ H.
Example 3.7. Let G be an abelian group and H = {x ∈ G : |x| is finite }. Then H
is a subgroup of G.
Proof. |e| = 1 implies e ∈ H. Let a, b ∈ H and let |a| = m and |b| = n. Now, since
G is abelian, (ab−1 )mn = amn (b−1 )mn = (am )n ((bn )m )−1 = en (em )−1 = e. Thus |ab−1 |
is finite. Hence ab−1 ∈ H.
Example 3.8. Let G be an abelian group and H and K be subgroups of G. Then
HK = {hk : h ∈ H, k ∈ K} is a subgroup of G.
CHAPTER 3. FINITE GROUPS; SUBGROUPS
6
Proof. e = ee implies e ∈ HK. So HK 6= ∅. Let a = h1 k1 and b = h2 k2 be
any two elements of HK. Using the abelian property of G, ab = (h1 k1 )(h2 k2 ) =
−1 −1
(h1 h2 )(k1 k2 ) ∈ HK. Similarly, a−1 = (h1 k1 )−1 = k1−1 h−1
1 = h1 k1 ∈ HK. Hence by
Corollary 3.2, HK is a subgroup.
Theorem 3.3. Let H be a nonempty subset of a group G. If
i. H is finite and
ii. H is closed under the operation of G,
then H is a subgroup of G.
Proof. Let a, b ∈ H. Since H is closed, ab ∈ H. It suffices to show that a−1 ∈ H.
Since H is finite, the elements in the sequence a, a2 , a3 , · · · ∈ H(H being closed) will
repeat after finite number of elements. Take any two elements so that am = an .
Suppose m > n. Then am−n = e. If m − n = 1, then a = e. This implies that
a−1 = a ∈ H. If m − n ≥ 2, aam−n−1 = am−n = e. Hence a−1 = am−n−1 ∈ H. We are
done.
Theorem 3.4. Let G be a group and a ∈ G. Then hai = {an : n ∈ Z} is a subgroup
of G.
Notation: For n > 0, a−n = (a−1 )n .
Proof. Since a ∈ hai, hai =
6 ∅. Let am , an ∈ hai. Then am (an )−1 = am−n ∈ hai. Hence
hai is a subgroup.
hai is called a cyclic subgroup.
Definition 3.4. A group G is called cyclic group if G = hai for some a ∈ G. The
element a is called a generator of G and G is said to be generated by a.
Remark 3.1. The cyclic groups are automatically abelian.
Example 3.9. Consider U (10) = {1, 3, 7, 9}. Then h7i = {7, 9, 3, 1} = U (10). Similarly find h3i and h9i.
Example 3.10. Take Z10 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Find h4i.
Solution: h4i = {4, 8, 2, 6, 0}.
Example 3.11. Z = h−1i = h1i. Therefore Z under addition is a cyclic group.
Definition 3.5. Let G be a group. The set
Z(G) = {a ∈ G : ax = xa for all x in G}
is called the center of G.
CHAPTER 3. FINITE GROUPS; SUBGROUPS
7
Theorem 3.5. Z(G) is a subgroup of G.
Proof. e ∈ Z(G) 6= ∅. Let a, b ∈ Z(G). Then (ab)x = a(bx) = a(xb) = (ax)b =
(xa)b = x(ab) =⇒ ab ∈ Z(G). a ∈ Z(G) =⇒ ax = xa for all x ∈ G. =⇒ xa−1 =
a−1 x =⇒ a−1 ∈ Z(G). We are done.
Definition 3.6. Let G be a group and g ∈ G. The centralizer of g is the set
C(g) = {x ∈ G : xg = gx}.
Theorem 3.6. The centralizer C(g) of a group element g ∈ G is a subgroup of G.
Proof. e ∈ C(g) 6= ∅. If ab ∈ C(g) then (ab)g = a(bg) = a(gb) = (ag)b = (ga)b =
g(ab) =⇒ ab ∈ C(g). ag = ga =⇒ ga−1 = a−1 g =⇒ a−1 ∈ C(g).
Chapter 4
Cyclic Groups
Homework: #2, 4, 8(a), 10, 11, 12, 18, 22.
Definition 4.1. A group G is called cyclic if G is generated by one element, that is,
G = {an : n ∈ Z}.
We write G = hai.
Example 4.1. Z = h1i = h−1i.
Example 4.2. Z10 = h1i = h3i = h7i = h9i.
Quiz problem: Is U (10) cyclic? If it is find a generator.
Theorem 4.1. Let G be a group and a ∈ G.
i. If a has infinite order, then ai = aj if and only if i = j.
ii. If |a| = n then hai = {e, a, a2 , a3 , · · · , an−1 } and ai = aj if and only if n|(i − j).
Proof.
i. ai = aj =⇒ ai a−j = e =⇒ ai−j = e. Since a has infinite order i − j = 0 =⇒
i = j. Other implication is obvious.
ii. Let ak ∈ hai. By the division algorithm, there exist integers q and r such that
k = qn + r where 0 ≤ r < n. So ak = aqn+r = aqn ar = (an )q ar = eq ar = ar ∈
{e, a, a2 , · · · , an−1 } =⇒ hai ⊂ {e, a, a2 , · · · , an−1 }. The other way around is
for free.
Let ai = aj =⇒ ai−j = e. i − j = qn + r, 0 ≤ r < n. Now, e = ai−j = aqn+r =
aqn ar = (an )q ar = ar =⇒ r = 0 =⇒ i − j = qn =⇒ n|(i − j).
Coversely, suppose n|(i − j) =⇒ i − j = nq for some q ∈ Z =⇒ ai−j = anq =
e =⇒ ai = aj .
8
CHAPTER 4. CYCLIC GROUPS
9
Corollary 4.2. For any group element a, |a| = |hai|.
Corollary 4.3. Let G be a group and let a ∈ G with |a| = n. If ak = e, then n|k.
Proof. Write k = nq + r with 0 ≤ r < n. Then e = ak = anq+r = (an )qar = ar =⇒
r = 0 =⇒ k = nq =⇒ n|q.
Theorem 4.4. Let a be a group element with |a| = n and k be a positive integer.
n
Then hak i = hagcd(n,k) i and |ak | = gcd(n,k)
.
Proof. Let gcd(n, k) = d. Then k = dr for some integer r. ak = adr = (ad )k =⇒
hak i ⊆ had i. There exist s, t ∈ Z such that d = ns + kt. So ad = ans+kt = ans akt =
(an )s (ak )t = (ak )t ∈ hak i =⇒ had i ⊆ hak i. Thus we have hak i = had i.
n
Since (ad ) d = an = e =⇒ |ad | ≤ nd . For any positive integer i < nd =⇒ id <
n =⇒ (ad )i = adi 6= e. Therefore |ad | = nd . Now |ak | = |hak i| = |had i| = |ad | = nd .
If |a| = 20, then ha6 i = ha2 i, ha11 i = hai.
Corollary 4.5. Let G be a finite cyclic group. The order of an element in G divides
|G|.
Proof. Let G = hai and |G| = |a| = n. Let hak i ∈ G. Then |ak | =
n
n
and gcd(n,k)
|n.
gcd(n,k)
Corollary 4.6. Let |a| = n. Then
1. hai i = haj i if and only if gcd(n, i) = gcd(n, j).
2. |ai | = |aj | if and only if gcd(n, i) = gcd(n, j).
Proof.
1. hai i = haj i =⇒ |hai i| = |haj i| =⇒
gcd(n, j).
n
gcd(n,i)
=
n
gcd(n,j)
=⇒ gcd(n, i) =
Conversely, gcd(n, i) = gcd(g, j) =⇒ hagcd(n,i) i = hagcd(n,j) i =⇒ hai i = haj i.
2. |ai | = |aj | if and only if
n
gcd(n,i)
=
n
gcd(n,j)
if and only if gcd(n, i) = gcd(n, j).
Corollary 4.7. Let |a| = n. Then hai = haj i if and only if gcd(n, j) = 1. Also,
|a| = |haj i| if and only if gcd(n, j) = 1.
Corollary 4.8. Zn = hki if and only if gcd(n, k) = 1.
CHAPTER 4. CYCLIC GROUPS
10
Classification of Subgroups of Cyclic Groups
Theorem 4.9 (Fundamental Theorem of Cyclic Groups).
i. Every subgroup of a cyclic group is cyclic.
ii. If |hai| = n, then order of any subgroup of hai divides n.
iii. If |hai| = n, then for each positive divisor k of n there is exactly one subgroup
of hai of order k.
Proof.
i. Let H be a subgroup of a cyclic group hai. If H = {e}. Then H is cyclic
automatically. So let H 6= {e}. Let m be the smallest positive integer such that
am ∈ H. We show that H = ham i. Clearly, ham i ⊆ H.
Next, let at ∈ H. By division algorithm, t = qm + r, 0 ≤ r < m. Now
at = aqm+r = aqm ar =⇒ ar = at a−qm = at (am )−q ∈ H =⇒ r = 0 =⇒ at =
amq = (am )q ∈ ham i. Thus H ⊆ ham i. Hence H = ham i.
ii. Let H be any subgroup of hai. Then H = ham i. Therefore |H| = |ham i| =
n
=⇒ |H| divides n.
gcd(n,m)
n
n
iii. Let k|n. Then ha k i is a subgroup or order k. To see this, |ha k i| =
n
= k.
n/k
n
gcd(n, n
)
k
=
Let H be any subgroup of hai and |H| = k. By (i), H = ham i. Now k = |H| =
n
=⇒ gcd(n, m) = nk . Therefore H = ham i = hagdc(n,m) i =
|ham i| = gcd(n,m)
han/k i.
Example 4.3. The list of subgroups of Z20 is
order
order
order
order
order
order
1
2
4
5
10
20
h20i =
h10i =
h5i =
h4i =
h2i =
h1i =
{0}
{0, 10}
{0, 5, 10, 15}
{0, 4, 8, 12, 16}
{0, 2, 4, 6, 8, 10, 12, 14, 16, 18}
{0, 1, 2, 3, · · · , 19}.
Explanation: Since 4|20, Z20 = h1i has a subgroup of order 4 which is generated by
120/4 = 15 = 5.
Question 4.1. Does Z24 have a subgroup of order 9?
CHAPTER 4. CYCLIC GROUPS
11
Answer: No because Z24 is a cyclic group, the order of it’s subgroup must divide
24 but 9 - 24.
= 3. So the
Example 4.4. The subgroup of order 8 in Z24 is generated by 24
8
subgroup of order 8 is h3i = {0, 3, 6, 9, 12, 15, 18, 21}. The other generators are 9, 15,
and 21 because gcd(8, 3) = 1 and 9 = 3 · 3. Similarly others.
Euler φ function
φ(1) = 1 and for n > 1, φ(n) = number of positive integers less than n and relatively
prime to n.
Theorem 4.10. The number of generators in a cyclic group of order n is φ(n).
Proof. Let hai = {e, a, a2 , · · · , an−1 } be a cyclic group of order n. Then we know that
hai = haj i if and only if gcd(n, j) = 1. It follows that # generators = φ(n).
Theorem 4.11. Let G be a cyclic group of order n and d|n. Then the number of
elements of order d in G is φ(d).
Proof. d|n =⇒ there is exactly one subgroup of order d in G, say it is
hai = {e, a, a2 , · · · , ad−1 }
Then all other elements of order d are in hai and they generate the same subgroup.
Therefore
# of elements of order d = # of generators of a subgroup of order d = φ(d).
Corollary 4.12. In a finite group G, the number of elements of order d is a multiple
of φ(d).
Proof. If there is no element of order d then there is nothing to prove. If a ∈ G has
order d then hai has φ(d) elements of order d. If all the elements of order d are in
hai, we are done. If there is another element b ∈ G of order d and b ∈
/ hai, then hbi
contains φ(d) elements of order d. Note that hai and hbi have no elements of order
d in common because if c is an element of order d that is in both hai and hbi, then
hai = hci = hbi =⇒ b ∈ hai. Thus we have 2φ(d) elements of order d. Continuing in
this fashion, we see that the number of elements of order d is multiple of φ(d).
Chapter 5
Permutation Groups
Homework: # 1, 2, 6, 8, 16, 24, 28
Definition 5.1. A permutation of a set A is a bijective function from A to itself. The
set of all permutations of A forms a group under composition. The group is called
the permutation group of A.
Example 5.1. Let A = {1, 2, 3}. Then α : A → A defined by α(1) = 2, α(2) =
3, α(3) = 1 (also show this by a diagram) is a permutation of A.
A common way to write this permutation is
1
2
3
···
α=
α(1) α(2) α(3) · · ·
n
α(n)
So the α in the example is written
1 2 3
2 3 1
1 2 3
1 3 2
α=
If
β=
then the composition βα is given by
βα =
1 2 3
3 2 1
To see this βα(1) = β(α(1)) = β(2) = 3 and so on.
Symmetric Group S3
S3 is a group of permutation group of A = {1, 2, 3}. So
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
S3 =
,
,
,
,
,
1 2 3
2 3 1
3 1 2
1 3 2
2 1 3
3 2 1
= {e, α, α2 , β, αβ, α2 β}
12
CHAPTER 5. PERMUTATION GROUPS
13
Symmetry Group Sn
The set of all permutations of {1, 2, 3, · · · , n} is called the symmetric group of degree
n and is denoted by Sn . An element of Sn looks like
1
2
···
n
α=
α(1) α(2) · · · α(n)
Order of Sn is n!. Sn is an-Abelian.
Example 5.2. D4 is a subgroup of S4 .
Cycle Notation
The permutation
α=
1 2 3 4 5 6
2 4 3 1 6 5
can be written in cycle notation as
α = (1, 2, 4)(3)(5, 6) = (124)(3)(56)
(1, 2, 4), (3) and (5, 6) are called the cycles. Then number of elements in a cycle is
called the length of the cycle. So (a1 , a2 , · · · , am ) is called a cycle of length m or an
m-cycle.
Example 5.3. Let
1 2 3 4 5
1 2 3 4 5
α=
and β =
3 2 1 5 4
4 5 1 3 2
In cycle notations α = (13)(2)(45) and β = (143)(25). The cycle notation for the
composition is:
αβ = (13)(2)(45)(143)(25) = (1524)(3)
and
βα = (143)(25)(13)(2)(45) = (1)(2534)
They are written as the multiplication of disjoint cycles.
It is preferred not to write cycles that have only one entry. Then the permutations
in above example are written as
α = (13)(45)
αβ = (1524)
β = (143)(25)
βα = (2534)
The identity permutation ε consists of only one cycle with one entry.
CHAPTER 5. PERMUTATION GROUPS
14
Properties of Permutations
Theorem 5.1. Every permutation of a finite set can be written as a product of disjoint
cycles.
Proof. Let A = {1, 2, · · · , n} and α be a permutation of A. Let a ∈ A. Since A
is finite, the sequence a, α(a), α2 (a), α3 (a), · · · must repeat. Let m be the smallest
positive integer such that αm (a) = a. Such m exist because if αi (a) = αj (a), i > j,
then αi−j (a) = a. Hence one of the cycles of α is (a, α(a), α2 (a), · · · , αm−1 (a)).
If there is an element b ∈ A not appearing in this cycle, we similarly construct a
cycle of α containing b, say (b, α(b), α2 (b), · · · , αk−1 (b)). These two cycles are disjoint
because αi (a) = αj (b), i ≥ j( we can assume this) =⇒ b = αi−j (a).
Continue this process until the set A is exhausted.
Theorem 5.2. Disjoint cycles of a permutation commute.
Proof. Let α be a permutation. It’s enough to consider that α has only two disjoint
cycles, say
α = (a1 , a2 , · · · , am )(b1 , b2 , · · · , bn )
Let α1 = (a1 , a2 , · · · , am ) and α2 = (b1 , b2 , · · · , bn ). We want to show that α1 α2 =
α2 α1 . Here
α1 α2 (ai ) = α1 (α2 (ai )) = α1 (ai ) = ai+1
α2 α1 (ai ) = α2 (α1 (ai )) = α2 (ai+1 ) = ai+1
where 1 ≤ i ≤ m and am+1 = a1 , and
α1 α2 (bj ) = α1 (α2 (bj )) = α1 (bj+1 ) = bj+1
α2 α1 (bj ) = α2 (α1 (bj )) = α2 (bj ) = bj+1
where 1 ≤ j ≤ n and bn+1 = b1 . For any element x not in the list a1 , · · · , am , b1 , · · · , bn
we have
α1 α2 (x) = α2 α1 (x) = x.
Therefore, α1 α2 = α2 α1 .
Theorem 5.3. A cycle of length n has order n.
Proof. Let α = (a1 , a2 , · · · , an ) be a cycle of length n. Then
(
ai+j
if i + j ≤ n
αj (ai ) =
ai+j−n if i + j > n.
It follows that αn (ai ) = ai+n−n = ai and for j < n,
(
ai+j
if i + j ≤ n =⇒ ai+j 6= ai
αj (ai ) =
ai+j−n if i + j > n =⇒ ai+j−n 6= ai .
Thus α has order n.
CHAPTER 5. PERMUTATION GROUPS
15
Theorem 5.4. The order of a permutation of a finite set is the least common multiple
of the lengths of its disjoint cycles.
Proof. We prove this theorem for a permutation which is a multiplication of two
disjoint cycles. Other permutations are similarly handled.
Let α be a permutation of a finite set which is a multiplication of two disjoint cycles
β and γ of length m and n, respectively. Let lcm(m, n) = l. Then l = mx = ny for
some positive integers x and y. Now,
αl = (βγ)l = β l γ l = (β m )x (γ n )y = ε
For k < l either m - k or n - k. Assume m - k. Then k = qm + r, 0 ≤ r < m. It
follows that
αk = (βγ)k = β k γ k = β qm+r γ k = β r γ k 6= ε.
Therefore the order of α is l.
Example 5.4. S5 has 5! = 120 elements. They have the following cycle structure.
Cycle Structure
5
4, 1
3, 2
3, 1, 1
2, 2, 1
2, 1, 1, 1
1, 1, 1, 1, 1
Order of the Elements
5
4
6
3
2
2
1
Number of elements
24
30
20
20
15
10
1
Theorem 5.5. Every permutation in Sn , n > 1, is a product of 2-cycles.
Proof. An identity permutation can be written as (12)(21). Since a permutations can
be written as a product of disjoint cycles, it suffices to show that a cycle is a product
of 2-cycles. Let (a1 a2 · · · ak ) be a k-cycle. Then
(a1 a2 · · · ak ) = (a1 ak )(a1 ak−1 )(a1 ak−2 ) · · · (a1 a3 )(a1 a2 )
= (a1 a2 )(a2 a3 )(a3 a4 ) · · · (ak−1 ak ).
Example 5.5. (12345) = (15)(14)(13)(12) = (12)(23)(34)(45) = (54)(53)(52)(51)
Lemma 5.6. If β1 β2 · · · βr = ε, where β’s are 2-cycles, then r is even.
Proof. Take for granted.
Theorem 5.7. If a permutatin α can be expressed as a product of even (odd) number
of 2-cycles, then every decomposition of α into a product of 2-cycles must have even
(odd) number of 2-cycles.
CHAPTER 5. PERMUTATION GROUPS
16
Proof. If α = β1 β2 · · · βr = γ1 γ2 · · · γs be two decompositions of α into the product of
2-cycles. Then
ε = γ1 γ2 · · · γs βr βr−1 · · · β2 β1
On the right is the product of r + s 2-cycles. By the lemma r + s is even. Thus r and
s are either both even or both odd.
Definition 5.2. A permutation that can be expressed as a product of an even number
of 2-cycles is called an even permutation. A permutation that can be expressed as a
product of an odd number of 2-cycles is called an odd permutation.
Theorem 5.8. The set of even permutations in Sn forms a subgroup of Sn .
Proof. Let H be a subset of Sn consisting of even permutations. Then ε ∈ H 6= ∅. If
α, β ∈ H, then αβ −1 ∈ H. (Show how!) So H is a subgroup.
Definition 5.3. The group of even permutations of n objects is denoted by An and
is called the alternating group of degree n.
Theorem 5.9. For n > 1, An has order n!/2.
Proof. If α is an odd permutation, then (12)α is even permutation and if α 6= β are
two distinct odd permutations then (12)α 6= (12)β are two distinct even permutations.
It follows that,
number of even permutations ≥ number of odd permutations
By the similar argument with even and odd switched we get,
number of even permutations ≤ number of odd permutations
Thus we have
number of even permutations = number of odd permutations = n!/2.
Chapter 6
Isomorphisms
Homework: # 4, 10, 22, 48, 50
Definition 6.1. Let (G, ∗) and (G0 , ◦) be two groups. A group homomorphism from
G to G0 is a mapping φ : G → G0 that preserves the group operation. That is,
φ(a ∗ b) = φ(a) ◦ φ(b)
for all a, b ∈ G.
If a homomorphism φ : G → G0 is bijective, then it is called an isomorphism and we
say that the groups G and G0 are isomorphic.
Example 6.1. Let G = hai = {an : n ∈ Z} be an infinite cyclic group. Then G ≈ Z.
The map φ : G → Z given by φ(ak ) = k is an isomorphis. To see this,
1. φ(ak al ) = φ(ak+l ) = k + l = φ(ak ) + φ(al ). So φ is a homomorphism.
2. φ(ak ) = φ(al ) =⇒ k = l =⇒ ak = al . So φ is one-to-one.
3. For k ∈ Z there is ak ∈ G such that φ(ak ) = k. So φ is onto.
Hence φ is an isomorphism.
Example 6.2. Any finite cyclic group of order n is isomorphic to Zn . U (10) and
U (5) are isomorphic to Z4 .
Example 6.3. The mapping φ : (R, +) → (R, +) given by φ(x) = x2 is not an
isomorphism because φ(x + y) = (x + y)2 and φ(x) + φ(y) = x2 + y 2 are not equal for
all x, y ∈ R.
Example 6.4. U (10) 6≈ U (12), where U (10) = {1, 3, 7, 9} and U (12) = {1, 5, 7, 11}.
Observe that x2 = 1 for every x ∈ U (12). Now φ(9) = φ(3 · 3) = φ(3)φ(3) = 1 and
φ(1) = φ(1 · 1) = φ(1)φ(1) = 1. Thus φ(9) = φ(1) and hence no homomorphism can
be one-to-one.
17
CHAPTER 6. ISOMORPHISMS
18
Example 6.5. U (8) = {1, 3, 5, 7} and U (12) = {1, 5, 7, 11} are isomorphic. φ(1) =
1, φ(3) = 5, φ(5) = 7, φ(7) = 11 is an isomorphism. φ is automatically bijective. Show
that it is a homomorphism.
Theorem 6.1 (Caley’s Theorem). Evert group is isomorphic to a group of permutations.
Proof. Let G0 = {Tg : g ∈ G and Tg : G → G is defined by Tg (a) = ga for all a ∈ G}.
1. Tg is a permutation of G.
(a) Tg (a) = Tg (b) =⇒ ga = gb =⇒ a = b =⇒ Tg is one-to-one.
(b) For every a ∈ G there is g −1 a ∈ G such that Tg (g −1 a) = g(g −1 a) = a =⇒
Tg is onto.
Hence Tg is a permutation of G.
2. G0 is a group under function composition.
(a) Closure: Tg Th (a) = (gh)a = Tgh (a) for all a ∈ G =⇒ Tg Th = Tgh ∈ G0 for
all g, h ∈ G.
(b) Associativity: Tg (Th Tk ) = Tg Thk = Tg(hk) = T(gh)k = Tgh Tk = (Tg Th )Tk .
(c) Identity: Te Tg = Teg = Tg = Tge = Tg Te =⇒ Te is the identity in G0 .
(d) Inverse: Tg Tg−1 = Tgg−1 = Te =⇒ (Tg )−1 = Tg−1 ∈ G0 .
3. φ : G → G0 defined by φ(g) = Tg is an isomorphism.
(a) φ is a homomorphism: φ(gh) = Tgh = Tg Th = φ(g)φ(h) for all g, h ∈ G.
(b) φ is One-to-one: φ(g) = φ(h) =⇒ Tg = Th =⇒ Tg (e) = Th (e) =⇒ ge =
he =⇒ g = h.
(c) φ is onto: For every Tg ∈ G0 automatically g ∈ G and φ(g) = Tg .
Properties of Isomorphisms
Theorem 6.2. Let φ : G → G0 be a group homomorphism. Then
1. If e is the identity of G, then φ(e) is the identity of G0 .
2. For every a ∈ G, φ(a−1 ) = φ(a)−1 .
3. For every a ∈ G, φ(an ) = (φ(a))n .
Proof.
CHAPTER 6. ISOMORPHISMS
19
1. Let the identity of G0 be e0 . Then φ(e) = φ(e) = φ(e)φ(e) =⇒ (φ(e))−1 φ(e) =
φ(e) =⇒ e0 = φ(e).
2. e0 = φ(e) = φ(aa−1 ) = φ(a)φ(a−1 ) =⇒ φ(a−1 ) = φ(a)−1 .
3. For n > 0, φ(an ) = φ(aa · · · a) = φ(a)φ(a) · · · φ(a) = (φ(a))n (Use induction).
For n = 0, it is obvious. For n < 0, φ(an ) = φ((a−1 )−n ) = (φ(a−1 ))−n =
(φ(a)−1 )−n = φ(a)n .
Theorem 6.3. Let φ : G → G0 be an isomorphism. Then
1. For a, b ∈ G, ab = ba if and only if φ(a)φ(b) = φ(b)φ(a).
2. G = hai if and only if G0 = hφ(a)i.
3. |a| = |φ(a)| for all a ∈ G. (isomorphisms preserve orders)
Proof.
1. ab = ba =⇒ φ(ab) = φ(ba) =⇒ φ(a)φ(b) = φ(b)φ(a).
Conversely, φ(a)φ(b) = φ(b)φ(a) =⇒ φ(ab) = φ(ba) =⇒ ab = ba. The last
implication follows because φ is one-to-one.
2. Let G = hai. Then φ(a) ∈ G0 =⇒ hφ(a)i ⊆ G0 . Let b ∈ G0 . Since φ is onto,
there is an ∈ hai such that φ(an ) = b =⇒ b = φ(an ) = (φ(a))n ∈ hφ(a)i =⇒
G0 ⊆ hφ(a)i. Thus G0 = hφ(a)i.
Conversely, let G0 = hφ(a)i. Since a ∈ G, hai ⊆ G. Let x ∈ G. Since
φ(x) ∈ hφ(a)i, φ(x) = φ(a)m = φ(am ) for some m. Since φ is one-to-one x =
am =⇒ x ∈ hai =⇒ G ⊆ hai. Thus G = hai.
3. Let |a| = n. Then φ(a)n = φ(an ) = φ(e) = e0 . If for 0 < i < n, φ(a)i = e0 , then
φ(ai ) = φ(e) =⇒ ai = e which is a contradiction. Hence φ(a)i 6= e0 . Hence
|φ(a)| = n.
Theorem 6.4. Suppose that φ : G → G0 is an isomorphism. Then
1. φ−1 : G0 → G is an isomorphism.
2. If H is a subgroup of G, then φ(H) = {φ(h) : h ∈ H} is a subgroup of G0 .
Proof.
CHAPTER 6. ISOMORPHISMS
20
1. Let a0 , b0 ∈ G0 . Then there exist a, b ∈ G such that φ(a) = a0 and φ(b) =
b0 .φ−1 (a0 b0 ) = φ−1 (φ(a)φ(b)) = φ−1 (φ(ab)) = ab = φ−1 (a0 )φ−1 (b0 ). Since φ is
bijective, φ−1 is bijective. Hence φ−1 is an isomorphism.
2. e0 = φ(e) ∈ φ(H) 6= ∅. Let h0 , k 0 ∈ φ(H). Then there exist h, k ∈ H such that
φ(h) = h0 and φ(k) = k 0 . Now h0 k 0−1 = φ(h)φ(k)−1 = φ(h)φ(k −1 ) = φ(hk −1 ) ∈
φ(H). Hence φ(H) is a subgroup of G0 .
Definition 6.2. An automorphism of a group G is an isomorphism from G to itself.
Example 6.6. φ : R2 → R2 defined by (a, b) 7→ (a, −b) is an automorphism of the
group R2 under componentwise addition.
We denote the set of all automorphisms of a group G by Aut(G).
Theorem 6.5. Let G be a group. Aut(G) is a group under function composition.
Proof.
1. Closure: Let φ, ψ ∈ Aut(G) then φ ◦ ψ ∈ Aut(G).
2. Associative: For any φ, ψ, χ ∈ Aut(G), (φ ◦ ψ) ◦ χ = φ ◦ (ψ ◦ χ).
3. Identity: Id : G → G, Id (g) = g for all g ∈ G is the identity in Aut(G).
4. Inverse: For every φ ∈ Aut(G), φ−1 ∈ Aut(G).
Therefore Aut(G) is a group.
Chapter 7
Cosets and Lagrange’s Theorem
Homework: # 2, 6, 8, 16, 30, 40.
Definition 7.1. Let G be a group and H be a subgroup of G. For a ∈ G, the set
aH = {ah : h ∈ H}
is called the left coset of H in G containing a. The set
Ha = {ha : h ∈ H}
is called the right coset of H in G containing a. The element a is called the coset
representative of aH and Ha.
Example 7.1. Let H = {0, 2, 4, 6, 8} be a subgroup of Z10 . Let’s determine all the
cosets of H in Z10 .
0 + H = 2 + H = 4 + H = 6 + H = 8 + H = {0, 2, 4, 6, 8} = H
1 + H = 3 + H = 5 + H = 7 + H = 9 + H = {1, 3, 5, 7, 9}
So there are only two left cosets H and 1 + H.
Example 7.2. Let G = S3 and H = {(1), (12)}. Let’s find all the cosets of H in G.
Recall that
S3 = {(1), (12), (23), (13), (123), (132)}.
(1)H
(12)H
(13)H
(23)H
(123)H
(132)H
= {(1), (12)}
= {(12), (1)}
= {(13), (123)}
= {(23), (132)}
= {(123), (13)}
= {(132), (23)}
We see that (1)H = (12)H, (13)H = (123)H, (23)H = (132)H. There are only 3
cosets.
21
CHAPTER 7. COSETS AND LAGRANGE’S THEOREM
22
Properties of Cosets
Theorem 7.1. Let H be a subgroup of a group G, and let a, b ∈ G. Then,
1. a ∈ aH.
Proof. a = ae ∈ aH.
2. aH = H if and only if a ∈ H.
Proof. Since a ∈ aH, aH = H =⇒ a ∈ H. Conversely, let a ∈ H. Then for
any h ∈ H, ah ∈ H =⇒ aH ⊆ H. Again for any h ∈ H, a−1 h ∈ H =⇒ h =
(aa−1 )h = a(a−1 h) ∈ aH =⇒ H ⊆ aH. Hence H = aH.
3. (ab)H = a(bH) and H(ab) = (Ha)b.
Proof. For all h ∈ H, (ab)h = a(bh) =⇒ (ab)H = a(bH).
4. aH = bH if and only if a ∈ bH.
Proof. Assume that aH = bH. Since a ∈ aH, a ∈ bH. Conversely, let a ∈
bH =⇒ a = bh for some h ∈ H =⇒ aH = (bh)H = b(hH) = bH.
5. either aH = bH or aH ∩ bH = ∅.
Proof. If aH ∩ bH 6= ∅ then x ∈ aH ∩ bH =⇒ aH = xH = bH.
6. aH = bH if and only if a−1 b ∈ H.
Proof. aH = bH if and only if H = a−1 (bH) = (a−1 b)H =⇒ a−1 b ∈ H.
7. |aH| = |bH|.
Proof. ah →
7 bh is a one-to-one and onto map from aH to bH. To see this
bh1 = bh2 =⇒ h1 = h2 =⇒ ah1 = ah2 . By definition it is clearly onto.
8. aH = Ha if and only if H = aHa−1 .
Proof. aH = Ha if and only if (aH)a−1 = (Ha)a−1 if and only if aHa−1 =
H.
9. aH is a subgroup of G if and only if a ∈ H.
CHAPTER 7. COSETS AND LAGRANGE’S THEOREM
23
Proof. If a ∈ H, then aH = H. So aH is a subgroup. Conversely, suppose
that aH is a subgroup. Then the identity e ∈ aH =⇒ e = ah for some
h ∈ H =⇒ a = h−1 ∈ H.
Theorem 7.2 (Lagrange’s Theorem). Let G be a finite group and H is a subgroup
of G, then |H| divides |G|. Moreover the number of distinct left (right) cosets of H
in G is |G|/|H|.
Proof. Let a1 H, a2 H, · · · , ak H be the distinct left cosets of H in G. For any a ∈ G, aH
is one of the cosets in the list, say aH = ai H. Then a ∈ ai H. Therefore
G = a1 H ∪ · · · ∪ ak H
Since these cosets are mutually disjoint,
|G| = |a1 H| + · · · + |ak H| = k|H|
Thus |H| divides |G|. From this it also follows that
k = the number of distinct left (right) cosets = |G|/|H|.
Definition 7.2. The index of a subgroup H in G is the number of distinct left cosets
of H in G and is denoted by |G : H| ( also [G : H])
Corollary 7.3. If H is a subgroup of a finite group G, then |G : H| = |G|/|H|.
Corollary 7.4. Let G be a finite group and a ∈ G. Then |a| divides |G|.
Proof. Since hai is a subgroup of G, |hai| divides |G|. Since |a| = |hai|, |a| divides
|G|.
Corollary 7.5. A group of prime order is cyclic.
Proof. Let |G| be prime and a ∈ G, a 6= e. Then |hai| > 1 and |hai| divides |G|. Since
|G| is prime |hai| = |G|. Thus we must have G = hai.
Corollary 7.6. Let G be a finite group and a ∈ G. Then a|G| = e.
Proof. Since |a| divides |G|, |G| = |a|k for some k ∈ Z. Therefore a|G| = a|a|k = ek =
e.
Theorem 7.7 (Fermat’s Little Theorem). For every integer a and every prime p,
ap
mod p = a mod p.
Proof. Write a = pq + r, 0 ≤ r < p. Then a mod p = r and ap mod p = rp mod p.
Let us show that rp mod p = r. Then we will be done. Since gcd(r, p) = 1, r ∈
U (p) = {1, 2, · · · , p − 1}. By the previous corollary rp−1 mod p = 1. Therefore rp
mod p = r.
Chapter 8
External Direct Products
Homework: 4, 8, 9, 10, 22
Definition 8.1. Let G1 , G2 , · · · , Gn be a finite collection of groups. The external
direct product of G1 , G2 , · · · , Gn is the set
G1 ⊕ G2 ⊕ · · · ⊕ Gn = {(g1 , g2 , · · · , gn ) : gi ∈ Gi },
where (g1 , g2 , · · · , gn )(g10 , g20 · · · , gn0 ) = (g1 g10 , g2 g20 , · · · , gn gn0 ).
It is clear that |G1 , G2 , · · · , Gn | = |G1 ||G2 | · · · |Gn |.
Theorem 8.1. The external direct product of groups G1 , G2 , · · · , Gn is a group.
Proof. By definition it is closed. Associativity follows from
((g1 , g2 , · · · , gn )(h1 , h2 , · · · , hn ))(k1 , k2 , · · · , kn )
=(g1 h1 , g2 h2 , · · · , gn hn )(k1 , k2 , · · · , kn )
=((g1 h1 )k1 , (g2 h2 )k2 , · · · , (gn hn )kn )
=(g1 (h1 k1 ), g2 (h2 k2 ), · · · , gn (hn kn ))
=(g1 , g2 , · · · , gn )(h1 k1 , h2 k2 , · · · , hn kn )
=(g1 , g2 , · · · , gn )((h1 , h2 , · · · , hn )(k1 , k2 , · · · , kn )).
(e1 , e2 , · · · , en ) is the identity. Inverse of (g1 , g2 , · · · , gn ) is (g1−1 , g2−1 , · · · , gn−1 ).
According to this definition R2 = R ⊕ R and R3 = R ⊕ R ⊕ R.
Example 8.1. U (5) = {1, 2, 3, 4} and U (8) = {1, 3, 5, 7}.
U (5) ⊕ U (8) = {(1, 1), (1, 3), (1, 5), (1, 7), (2, 1), (2, 3), (2, 5), (2, 7),
(3, 1), (3, 3), (3, 5), (3, 7), (4, 1), (4, 3), (4, 5), (4, 7)}
For instance, (3, 3)(3, 3) = (4, 1) and inverse of (3, 5) is (2, 5).
Example 8.2. A group of order 4 is isomorphic to Z4 or Z2 ⊕ Z2 .
24
CHAPTER 8. EXTERNAL DIRECT PRODUCTS
25
Properties of External Direct Products
Theorem 8.2. Let G1 , G2 , · · · , Gn be finite groups and (g1 , g2 , · · · , gn ) ∈ G1 ⊕ G2 ⊕
· · · ⊕ Gn . Then |(g1 , g2 , · · · , gn )| = lcm(|g1 |, |g2 |, · · · , |gn |).
Proof. Let |(g1 , g2 , · · · , gn )| = m and lcm(|g1 |, |g2 |, · · · , |gn |) = l. We want to show
that m = n. |(g1 , g2 , · · · , gn )| = m =⇒ (g1m , g2m , · · · , gnm ) = (e1 , e2 , · · · , en ) =⇒
|gi | divides m for every i = 1, 2, · · · , n. This implies m is a multiple of |g1 |, |g2 |, · · · , |gn | =⇒
l ≤ m. On the other hand, lcm(|g1 |, |g2 |, · · · , |gn |) = l =⇒ l is multiple of |g1 |, |g2 |, · · · , |gn |.
Therefore (g1 , g2 , · · · , gn )l = (g1l , g2l , · · · , gnl ) = (e1 , e2 , · · · , en ) =⇒ m ≤ l. Thus
m = l.
Example 8.3. Determine the number of elements of order 3 in Z6 ⊕ Z12 .
Solution: If (a, b) ∈ Z6 ⊕ Z12 has |(a, b)| = 3, then lcm(|a|, |b|) = 3.
Case i. |a| = 3, |b| = 1. There are two choices for a, namely, 2 and 4 and one choice
for b namely 0. This gives 2 elements of order 3.
Case ii. |a| = 3, |b| = 3. There are two choices for a and two choices for b, namely,
4 or 8. This gives 4 elements of order 3.
Case iii. |a| = 1, |b| = 3. There is one choice for a and two choices for b. This gives
2 elements of order 3.
Therefore, there are 8 elements of order 3 in Z6 ⊕ Z12 .
Example 8.4. Determine the number of cyclic subgroups of order 10 in Z30 ⊕ Z60 .
Example 8.5. Determine the number of cyclic subgroups of order 10 in Z25 ⊕ Z30 .
Solution: First, let us determine the number of elements of order 10 in Z25 ⊕ Z30 .
If (a, b) ∈ Z25 ⊕ Z30 , then lcm(|a|, |b|) = 10.
Case 1. |a| = 1 and |b| = 10. There is only one choice for a, namely 0. By Theorem
4.11, there are φ(10) = 4 choices for b, namely 3, 9, 21, and 27. This gives 4 elements
of order 10.
Case 2. |a| = 5 and |b| = 2 or 10. There are φ(5) = 4 choices for a and φ(2) + φ(10) =
1 + 4 = 5 choices for b. This gives 20 elements of order 10.
Thus, there are 24 elements of order 10 in Z25 ⊕ Z30 . Since each cyclic subgroup
of order 10 contains φ(10) = 4 elements of order 10, there must be 24/4 = 6 cyclic
subgroups of order 10.
Example 8.6. Does Z25 ⊕ Z30 have a subgroups isomorphic to Z5 ⊕ Z3 ?
CHAPTER 8. EXTERNAL DIRECT PRODUCTS
26
Solution: Yes, it is h6i ⊕ h10i.
Theorem 8.3. Let G and H be finite cyclic groups. Then G ⊕ H is cyclic if and only
if gcd(|G|, |H|) = 1.
Proof. Let |G| = m, |H| = n. Assume that G ⊕ H is cyclic, say G ⊕ H = h(g, h)i.
Let gcd(m, n) = d. Observe that (g, h)mn/d = ((g m )n/d , (hn )m/d ) = (e, e) =⇒ mn =
|(g, h)| ≤ mn/d =⇒ d = 1.
Conversely, assume that gcd(m, n) = 1. Let G = hgi and H = hhi. Then
mn
= mn = |G ⊕ H|. This implies that (g, h) generates
|(g, h)| = lcm(m, n) = gcd(m,n)
G ⊕ H.
Corollary 8.4. Let G1 , G2 , · · · , Gn be finite cyclic groups. Then G1 ⊕ G2 ⊕ · · · ⊕ Gn
is cyclic if and only if gcd(|Gi |, |Gj |) = 1 when i 6= j.
Proof. Use induction.
Corollary 8.5. Let m = n1 n2 · · · nk . Then Zm is isomorphic to Zn1 ⊕Zn−2 ⊕· · ·⊕Znk
if and only if gcd(ni , nj ) = 1 when i 6= j.
Proof. Follows directly from Corollary 8.4.
We have Z2 ⊕ Z3 ≈ Z6 . Similarly, Z2 ⊕ Z2 ⊕ Z3 ⊕ Z5 ≈ Z2 ⊕ Z6 ⊕ Z5 ≈ Z2 ⊕ Z30 .
But Z2 ⊕ Z6 6≈ Z12 .
Chapter 9
Normal Subgroups and Factor
Groups
Homework: #2, 4, 12, 14, 18
Normal Subgroups
Definition 9.1. A subgroup H of a group G is called a normal subgroup of G, written
H C G, if aH = Ha for all a ∈ G.
Question 9.1. What is the meaning of aH = Ha ?
Answer: For every h ∈ H there is h0 ∈ H such that ah = h0 a.
Theorem 9.1. A subgroup H of G is normal in G if and only if xHx−1 ⊆ H for all
x ∈ G.
Proof. Assume that H C G. Let x ∈ G. We want to show that xHx−1 ⊆ H. Let
xhx−1 ∈ xHx−1 . Since xH = Hx, xh = h0 x for some h0 ∈ H =⇒ xhx−1 = h0 ∈ H.
Conversely, assume that xHx−1 ⊆ H for all x ∈ G. Since x−1 ∈ G, it follows
that x−1 Hx ⊆ H. xHx−1 ⊆ H =⇒ xH ⊆ Hx and x−1 Hx ⊆ H =⇒ Hx ⊆ xH.
Therefore xH = Hx for all x ∈ G.
Example 9.1. Let H = {0, 3, 6, 9} and G = Z12 . Then for any a ∈ G, a+H = H +a.
Hence H C G. In general, every subgroup of an Abelian group is normal.
Example 9.2. The center Z(G) of a group G is normal. It is obvious that aZ(G) =
Z(G)a for all a ∈ G.
Example 9.3. The alternating group An of even permutations is a normal subgroup
of Sn . If α ∈ Sn is an even permutation, then αAn = An α (why?). Same is true if α
is odd.
Example 9.4. Let H C G and K be any subgroup of G. Then HK = {hk : h ∈
H, k ∈ K} is a subgroup of G.
27
CHAPTER 9. NORMAL SUBGROUPS AND FACTOR GROUPS
28
Solution: e = ee ∈ HK =⇒ HK 6= ∅. Let a = h1 k1 and b = h2 k2 ∈ HK
−1 −1
where h1 , h2 ∈ H and k1 , k2 ∈ K. Now, ab−1 = (h1 k1 )(k1−2 h−1
2 ) = h1 (k1 k2 )h2 =
−1
−1
h1 h0 (k1 k2 ) for some h0 ∈ H. Thus, ab−1 = (h1 h0 )(k1 k2 ) ∈ HK. Hence HK is a
subgroup of G.
Example 9.5. Recall that GL(2, R) is a group of 2 × 2 invertible matrices, that is,
the matrices with non-zero determinant and SL(2, R), the set of 2 × 2 matrices with
determinant 1 is a subgroup of GL(2, R). Notice that the operation here is matrix
multiplication.
Solution: For every x ∈ GL(2, R) we want to show that xSL(2, R)x−1 ⊆ SL(2, R).
Let xsx−1 ∈ xSL(2, R)x−1 . det(xsx−1 ) = det x · det s · det x−1 = det x · det x−1 =
1 =⇒ xsx−1 ∈ SL(2, R). By Theorem 9.1, SL(2, R) C GL(2, R).
Factor Groups
If H is a normal subgroup of a group G, by definition, the left cosets of H in G and
right cosets of H in G are same. So the set of left cosets is same as the set of right
cosets. That is,
{aH : a ∈ G} = {Ha : a ∈ G}
The set of left cosets of H in G which we denote by
G/H = {aH : a ∈ G}
Theorem 9.2. Let H be a normal subgroup of a group G. Then the G/H is a group
under the operation (aH)(bH) = (ab)H.
Proof.
i. We have to show that the operation is well defined. Let aH = a0 H and bH =
b0 H. We show that
(aH)(bH) = (a0 H)(b0 H), that is (ab)H = (a0 b0 )H.
aH = a0 H and bH = b0 H =⇒ a = a0 h1 and b = b0 h2 . Now, (ab)H =
(a0 h1 b0 h2 )H = (a0 h1 b0 )(h2 H) = (a0 h1 b0 )H = (a0 h1 )(b0 H) = (a0 h1 )(Hb0 ) =
(a0 h1 )(Hb0 ) = a0 (h1 H)b0 = a0 Hb0 = a0 b0 H. This proves that the operation is
well defined.
ii. Associativity is automatic: (aHbH)cH = (ab)HcH = (ab)cH = a(bc)H =
aH(bc)H = aH(bHcH).
iii. eH = H is the identity.
CHAPTER 9. NORMAL SUBGROUPS AND FACTOR GROUPS
29
iv. a−1 H is the inverse of aH.
Example 9.6. Let 5Z = {0, ±5, ±10, · · · }. Clearly 5Z is a subgroup of Z. Moreover
it is a normal subgroup. The left cosets of 5Z in Z are
0 + 5Z = {· · ·
1 + 5Z = {· · ·
2 + 5Z = {· · ·
3 + 5Z = {· · ·
4 + 5Z = {· · ·
, −10, −5, 0, 5, 10, · · · }
, −9, −4, 1, 6, 11, · · · }
, −8, −3, 2, 7, 12, · · · }
, −7, −2, 3, 8, 13, · · · }
, −6, −1, 4, 9, 14, · · · }
These are the only cosets because for any k ∈ Z, by division algorithm, k = 5q +r, 0 ≤
r < 5. Hence k + 5Z = r + 5q + 5Z = r + 5Z. Thus k + Z is one the above left cosets.
Its Cayley table is,
0 + 5Z
1 + 5Z
2 + 5Z
3 + 5Z
4 + 5Z
0 + 5Z
0 + 5Z
1 + 5Z
2 + 5Z
3 + 5Z
4 + 5Z
1 + 5Z
1 + 5Z
2 + 5Z
3 + 5Z
4 + 5Z
0 + 5Z
2 + 5Z
2 + 5Z
3 + 5Z
4 + 5Z
0 + 5Z
1 + 5Z
Clearly, Z/5Z is isomorphic to Z5 .
Example 9.7. In general, for any n > 0, Z/nZ ≈ Zn .
3 + 5Z
3 + 5Z
4 + 5Z
0 + 5Z
1 + 5Z
2 + 5Z
4 + 5Z
4 + 5Z
0 + 5Z
1 + 5Z
2 + 5Z
3 + 5Z
Chapter 10
More on Group Homomorphisms
Definition 10.1. A homomorphism φ from a group G to a group G0 is a mapping
from G into G0 such that
φ(ab) = φ(a)φ(b), for all a, b ∈ G.
Definition 10.2. The kernel of a group homomorphism φ : G → G0 is the set
ker φ = {g ∈ G : φ(g) = e0 }
where e0 is the identity of G0 .
Example 10.1. If φ : G → G0 is an isomorphism, then ker φ = {e}.
Example 10.2. Let φ : GL(2, R) → R∗ be defined by φ(A) = det A, where GL(2, R)
is a group of 2 × 2 invertible matrices under matrix multiplication and R∗ is a group
of nonzero real numbers under multiplication. Then
φ(AB) = det(AB) = det A det B = φ(A)φ(B)
implies φ is homomorphism. And,
ker φ = {A ∈ GL(2, R) : φ(A) = det A = 1} = SL(2, R).
Example 10.3. The set R[x] of all polynomials with real coefficients is a group under
addition. The derivative
Dx : R[x] → R[x] defined by Dx (f ) =
df
dx
is a homomorphism. the kernel of Dx is
df
ker Dx = f ∈ R[x] : Dx (f ) =
= 0 = All constant polynomials = R.
dx
Example 10.4. The kernel of φ : Z → Z10 , n 7→ n mod 10 is h10i.
30
CHAPTER 10.
MORE ON GROUP HOMOMORPHISMS
31
Properties of Homomorphisms
Theorem 10.1. Let φ : G → G0 be a group homomorphism. Then
i. ker φ is a subgroup of G.
ii. If H is normal subgroup of G then φ(H) is a normal subgroup of φ(G).
Proof.
i. Since φ(e) = e0 , e ∈ ker φ =⇒ ker φ 6= ∅. Let a, b ∈ ker φ. Then
φ(ab−1 = φ(a)φ(b−1 ) = e0 φ(b)−1 = φ(b)−1 = e0−1 = e0
implies ab−1 ∈ ker φ. Hence ker φ is subgroup of G.
ii. Recall that φ(H) is a subgroup of φ(G) and φ(G) is a subgroup of G0 . Let
φ(g) ∈ φ(G) be an arbitrary element. For every φ(h) ∈ φ(H),
φ(g)φ(h)φ(g)−1 = φ(ghg −1 ) ∈ φ(H)
because ghg −1 ∈ gHg −1 ⊆ H. Therefore φ(H) is normal in φ(G).
Left Cosets of the Kernel
Theorem 10.2. Let φ : G → G0 be a group homomorphism. Then
i. φ(a) = φ(b) if and only if a ker φ = b ker φ.
ii. If φ(g) = g 0 , then φ−1 (g 0 ) = {x ∈ G : φ(x) = g 0 } = g ker φ.
Proof.
i. Recall that the coset aH = bH if and only if a−1 b ∈ H. Here φ(a) = φ(b) ⇔
φ(a)−1 φ(b) = e0 ⇔ φ(a−1 b) = e0 ⇔ a−1 b ∈ ker φ. Hence φ(a) = φ(b) if and only
if a−1 b ∈ ker φ if and only if a ker φ = b ker φ.
ii. Let x ∈ φ−1 (g 0 ) =⇒ φ(x) = g 0 = φ(g) =⇒ x ker φ = g ker φ =⇒ x ∈
g ker φ =⇒ φ−1 (g 0 ) ⊆ g ker φ.
Again, take gk ∈ g ker φ where k ∈ ker φ. Now φ(gk) = φ(g)φ(k) = φ(g) =
g 0 =⇒ gk ∈ φ−1 (g 0 ) =⇒ g ker φ ⊆ φ−1 (g 0 ).
From the theorem above it is clear that if φ is a homomorphism from a group G
onto G0 then all the left cosets of ker φ in G are exactly φ−1 (g 0 ), g 0 ∈ G0 .
CHAPTER 10.
MORE ON GROUP HOMOMORPHISMS
32
Theorem 10.3. If | ker φ| = n then φ : G → φ(G) is a n-to-1 mapping, that is, for
every g 0 ∈ φ(G), |φ−1 (g 0 )| = n.
Proof. By Theorem 10.2 for every g 0 ∈ φ(G), φ−1 (g 0 ) is a left coset of ker φ in G.
Therefore |φ−1 (g 0 )| = | ker φ| = n.
Theorem 10.4. The group homomorphism φ : G → G0 is one-to-one if and only if
ker φ = {e}.
Proof. This theorem is just a particular case of Theorem 10.3 with n = 1. However
we can give an independent proof.
Assume that φ is one-to-one. Let g ∈ ker φ. Then φ(g) = e0 = φ(e) =⇒ g =
e =⇒ ker φ = {e}.
Conversely, assume that ker φ = {e}. Let φ(a) = φ(b). Then φ(a)φ(b)−1 = e0 =⇒
φ(ab−1 ) = e0 =⇒ ab−1 ∈ ker φ =⇒ ab−1 = e =⇒ a = b. Thus φ is one-to-one.
The First Isomorphism Theorem
Theorem 10.5. Let φ : G → G0 be a group homomorphism. Then the factor group
G/ ker φ is isomorphic to φ(G). The isomorphism is given by g ker φ →
7 φ(g).
CHAPTER 10.
MORE ON GROUP HOMOMORPHISMS
Homework
1. Let φ : G → G0 be a group homomorphism and a ∈ G.
a. Prove that φ(an ) = (φ(a))n .
b. If |a| is finite, prove that |φ(a)| divides |a|.
c. If H is cyclic, prove that φ(H) is cyclic.
d. Prove that ker φ is a normal subgroup of G.
2. Let φ : Z15 → Z5 be a homomorphism defined by φ(k) = k mod 5.
a. Find ker φ.
b. Find all the left cosets of ker φ.
c. Find φ−1 (0), φ−1 (1), φ−1 (2), φ−1 (3) and φ−1 (4).
d. Compare your answers in b. and c.
33
Chapter 11
Introduction to Rings
Homework: # 5, 8, 12, 18.
Definition 11.1. A ring R is a set with two binary operations addition and multiplication such that for all a, b, c ∈ R the following properties are satisfied:
1. a + b = b + a.
2. (a + b) + c = a + (b + c)
3. There is a 0 ∈ R such that a + 0 = a.
4. There is −a ∈ R such that a + (−a) = 0.
5. a(bc) = (ab)c.
6. a(b + c) = ab + ac and (b + c)a = ba + ca.
So (R, +) is an abelian group. A ring R is called commutative if ab = ba for all
a, b ∈ R. A unity or identity is a nonzero element e such that ea = ae = a for all
a ∈ R. A nonzero element a ∈ R is called unit if it has a multiplicative inverse.
Let R be a commutative ring and a, b ∈ R, a 6= 0. We say that a divides b and
write a | b if there is an element c ∈ R such that b = ac. Otherwise we say a does not
divide b and write a - b.
Example 11.1. The set of integers Z under the usual addition and multiplication is
a ring. It is commutative. 1 is the identity (unity). The units of Z are 1 and −1.
Example 11.2. The set Zn is a ring under addition and multiplication modulo n. It
is commutative, 1 is unity. The units are the elements in U (n).
Example 11.3. The set 2Z of even integers under ordinary addition and multiplication is a ring. It is a commutative ring without unity.
Example 11.4. The set M
2 (Z) of
2 × 2 matrices of with integer entries is a noncom1 0
mutative ring with unity
.
0 1
34
CHAPTER 11. INTRODUCTION TO RINGS
35
Properties of Rings
Theorem 11.1. Let R be a ring and a, b, c ∈ R. Then
1. a0 = 0a = 0.
2. a(−b) = (−a)b = −(ab).
3. (−a)(−b) = ab.
4. a(b − c) = ab − ac and (b − c)a = ba − ca.
If R has a unity element 1, then
1. (−1)a = −a
2. (−1)(−1) = 1.
Proof.
1. a0 + 0 = a0 = a(0 + 0) = a0 + a0 =⇒ a0 = 0.
2. a(−b) + ab = a(−b + b) = a0 = 0 =⇒ a(−b) = −(ab). Similarly (−a)b =
−(ab).
3. (−a)(−b) = −(a(−b)) is inverse of a(−b). But a(−b) = −(ab) is inverse
of ab. Thus (−a)(−b) and ab are inverses of a(−b). By the uniqueness
(−a)(−b) = ab.
4. a(b − c) = a(b + (−c)) = ab + a(−c) = ab − ac.
Theorem 11.2. If a ring R has unity , it is unique. If a ring element has a multiplicative inverse, it is unique.
Subrings
Definition 11.2. A subset S of a ring R is a subring of R if S is itself a ring with
the operations of R.
Theorem 11.3. A nonempty subset S of a ring R is a subring if a − b ∈ S and
ab ∈ S whenever a, b ∈ S.
Proof. For every a, b ∈ S, a − b ∈ S implies S is a subgroup of R under addition. S is
commutative under addition because R is commutative under addition. Thus S is an
abelian group under addition. For every a, b, c ∈ S, a(bc) = (ab)c and a(b+c) = ab+ac
because a, bc ∈ R.
Example 11.5. For each positive integer n,
nZ = {0, ±n, ±2n, ±3n, · · · }
is a subring of a the ring Z.
Chapter 12
Integral Domains
Homework: # 4, 8, 18, 22, 41.
Definition 12.1. A zero-divisor is a nonzero element a of a commutative ring R such
that there is a nonzero element b ∈ R with ab = 0.
Definition 12.2. An integral domain is a commutative ring R with unity and no
zero-divisors, that is, if ab = 0 then either a = 0 or b = 0.
Example 12.1. The ring of integers Z is an integral domain.
Example 12.2. The ring of Guassian integers Z[i] = {a + bi : a, b ∈ Z} and the ring
Z[x] of polynomials with integer coefficients are integral domains.
Example 12.3. The ring Zp of integers modulo p is an integral domain if p is prime
otherwise it is not an integral domain. For instance Z10 is not an integral domain
because 5 · 2 = 0.
Example 12.4. The ring M2 (Z) of 2 × 2 matrices over integers is not an integral
domain.
The cancellation property holds in integral domain.
Theorem 12.1. Let R be an integral domain and a, b, c ∈ R. If a 6= 0 and ab = ac,
then b = c.
Proof. ab = ac =⇒ ab − ac = 0 =⇒ a(b − c) = 0. Since R is an integral domain
and a 6= 0 we must have b − c = 0 =⇒ b = c.
Definition 12.3. A field is a commutative ring with unity in which every nonzero
element is a unit.
Theorem 12.2. Every field is an integral domain.
36
CHAPTER 12. INTEGRAL DOMAINS
37
Proof. Let F be a field. Let a, b ∈ F such that ab = 0. We want to show that
either a = 0 or b = 0. If a 6= 0 then there a is unit. There exist a−1 ∈ F such that
aa−1 = a−1 a = 1. Thus a−1 (ab) = 0 =⇒ b = 0. Hence F is an integral domain.
Theorem 12.3. A finite integral domain is a field.
Proof. Let D be a finite integral domain. Let a ∈ D, a 6= 0. We want to show that a is
a unit. If a = 1 then it is unit. So assume that a 6= 1. Since D is finite there must be
repetition in a, a2 , a3 , · · · . There exist integers i > j such that ai = aj =⇒ ai−j = 1.
Since a 6= 1, i − j > 1. Hence aai−j−1 = 1 =⇒ ai−j−1 is the inverse of a and a is
unit.
Corollary 12.4. For p prime, Zp is a field.
Proof. We just have to show that Zp , p prime, is an integral domain. Suppose that
a, b ∈ Zp and ab = 0 =⇒ ab = pk for some integer k. Since p is prime either p | a or
p | b =⇒ either a = 0 or b = 0.
Characteristic of a Ring
Definition 12.4. The characteristic of a ring R, denoted by char R, is the least
positive integer n such that nx = 0 for all x ∈ R, that is, x+x+· · ·+x(n times ) = 0.
If no such integer exists we say that the characteristic of R is zero.
The ring Zn of integers modulo n has characteristic n.
Theorem 12.5. Let R be a ring with unity 1. If 1 has order n under addition, then
the characteristic of R is n. If 1 has infinite order under addition, then char R = 0.
Proof. If 1 has order n, n is the smallest positive integer such that
1 + 1 + · · · + 1(n times ) = 0.
For any x ∈ R,
x + x + · · · + x = 1x + 1x + · · · + 1x = (1 + 1 + · · · + 1)x = 0x = 0.
Therefore char R = n.
If the order of 1 is infinity, then there is no n such that n1 = 0. Hence char R =
0.
Theorem 12.6. The characteristic of an integral domain is 0 or prime.
CHAPTER 12. INTEGRAL DOMAINS
38
Proof. Let 1 be the unity of the integral domain. If 1 has infinite order, then we are
done. So suppose that the order of 1 is finite, say n. We want to show that n is
prime. Let n = st where 1 ≤ s, t ≤ n. Then
n·1=0
(st) · 1 = 0
(s · 1)(t · 1) = 0
=⇒ s · 1 = 0 or t · 1 = 0.
Since n is the smallest positive integer with this property either s = n or t = n.
Thus n is prime.
Chapter 13
Ideals and Factor Rings
Homework: # 8, 10, 12, 14
Definition 13.1. A subring I of a ring R is called
(i) a left ideal of R if for every r ∈ R, rI = {ra : a ∈ I} ⊆ I, that is, ra ∈ I for
every a ∈ I.
(ii) a right ideal of R if for every r ∈ R, Ir = {ar : a ∈ I} ⊆ I, that is, ar ∈ I
for every a ∈ I.
If I is both a left ideal and a right ideal, we simply call it an ideal of R or two-sided
ideal of R for obvious reason. If an ideal I is a proper subset of R then it is called
proper ideal.
If a ring R is commutative then all the ideals are two-sided ideals. In this course
we will be talking about the two-sided ideals only.
Examples
1. For any ring R, I = {0} is an ideal of R. It is called the trivial ideal. I = R
is also an ideals of R.
2. 2Z = {2n : n ∈ Z} is an ideal of Z. Similarly, nZ for any positive integer n
is an ideal of Z.
3. Let R be a commutative ring with unity and a ∈ R. The set
hai = {ra : r ∈ R}
is an ideal of R called the principal ideal generated by a. The assumption
that R is commutative is necessary otherwise there is nothing that guarantees
that r1 as1 + r2 as2 ∈ hai where r1 , r2 , s1 , s2 ∈ R.
39
CHAPTER 13. IDEALS AND FACTOR RINGS
40
4. Let R be a commutative ring with unity and let a1 , a2 , · · · , an ∈ R. Then
I = ha1 , a2 , · · · , an i = {r1 a1 + · · · + rn an : ri ∈ R}
is an ideal of R called the ideal generated by a1 , a2 , · · · , an . Verify that I is
an ideal.
Factor Rings
We have seen that if H is a normal subgroup of a group G, we can form a factor
group G/H of left cosets of H in G. In the similar fashion we can form the factor
rings.
Theorem 13.1. Let R be a ring and let I be a subring of R. The set of cosets
{r + I : r ∈ R} is a ring under the operations
Addition: (r + I) + (s + I) = (r + s) + I
Multiplication: (r + I)(s + I) = rs + I
if and only if I is an ideal of R.
Example 13.1. Let R = Z and I = 6Z. Then
Z/6Z = {0 + 6Z, 1 + 6Z, 2 + 6Z, 3 + 6Z, 4 + 6Z, 5 + 6Z}
is a factor ring. Discuss the addition and multiplication on Z/6Z.
a b
Example 13.2. Let R = M2 (Z) =
: a, b, c, d ∈ Z and I be the subset of
c d
R consisting of matrices with even entries. It is easy to see that I is an ideal of R.
What is in the factor ring R/I ?
s t
R/I =
: s, t, u, v ∈ {0, 1}
u v
. How many elements does R/I have? It has only 16 elements count carefully.
Prime Ideals and Maximal Ideals
Definition 13.2. Let R be a commutative ring and P be a proper ideal of R. Then
P is called prime ideal if whenever ab ∈ P where a, b ∈ R, then a ∈ P or b ∈ P . A
maximal ideal M of a commutative ring R is a proper ideal of R such that if there is
an ideal I of R containing M , that is, M ⊂ I then I = M or I = R.
Example 13.3. Let p > 1 be an integer. Then pZ is a prime ideal of Z, the ring of
integers, if and only if p is prime.
CHAPTER 13. IDEALS AND FACTOR RINGS
41
Solution: mn ∈ pZ if and only if mn = pk for some k ∈ Z if and only if p|m or p|n
if and only if m ∈ pZ or n ∈ pZ.
Example 13.4. h2i and h3i are the only maximal ideals of Z36 . We have the following
chains of ideals
Z36 > h2i > h4i > h12i > h0i
Insert h3i, h6i, h9i, h18i.
Theorem 13.2. Let R be a commutative ring with unity and let I be an ideal of R.
Then R/I is an integral domain if and only if I is prime.
Proof. Let R/I be an integral domain. Let a, b ∈ R such that ab ∈ I. Now
(a + I)(b + I) = ab + I = I
Since I is the zero element of R/I and R/I is an integral domain either a + I = I or
b + I = I =⇒ a ∈ I or b ∈ I =⇒ I is prime.
Conversely, assume that I is prime and (a + I)(b + I) = 0 + I = I =⇒ ab + I =
I =⇒ ab ∈ I =⇒ a ∈ I or b ∈ I =⇒ a + I = I or b + I = I. Thus R/I has
no zero divisors. R/I is commutative with unity 1 + I. Therefore R/I is an integral
domain.
Theorem 13.3. Let R be a commutative ring with unity and let I be an ideal of R.
Then R/I is a field if and only if I is maximal.
Proof. The proof is excluded to this level. Interested students can read it from the
book.
Chapter 14
Ring Homomorphisms
Definition 14.1. A ring homomorphism φ from a ring R to a ring R0 is a mapping
φ : R → R0 such that for all a, b ∈ R we have
1. φ(a + b) = φ(a) + φ(b)
2. φ(ab) = φ(a)φ(b)
A bijective ring homomorphism is called a ring isomorphism.
Example 14.1. For any positive integer n, the mapping φ : Z → Zn defined by
φ(k) = k mod n is a ring homomorphism. It is called the natural homomorphism
from Z to Zn .
Example 14.2. Let R be a commutative ring of characteristic 2. Then the mapping
φ : R → R given by φ(a) = a2 is a ring homomorphism from R → R.
Solution: φ(a + b) = (a + b)2 = a2 + 2ab + b2 = a2 + b2 = φ(a) + φ(b) and
φ(ab) = (ab)2 = a2 b2 = φ(a)φ(b).
Example 14.3. The group of integers Z and the group of even integers 2Z are
group-isomorphic under addition. For instance φ : Z → 2Z, k 7→ 2k is the groupisomorphism under addition. However, Z and 2Z are not ring-isomorphic. Z has
unity 1, 2Z does not have a unity.
Example 14.4. An integer is divisible by 9 if and only if sum of its digits is divisible
by 9.
Proof. Define the natural ring-homomorphism φ : Z → Z9 , φ(n) = n mod 9. Observe
that any integer divisible by 9 in Z goes to 0 in Z9 . Let n ∈ Z be an integer divisible
by 9 and ak , ak−1 , · · · , a1 , a0 are the digits. So we want to prove that 9 | n if and
only if 9 | (ak + ak−1 + · · · + a1 + a0 ). Which is equivalent to φ(n) = 0 if and only if
φ(ak + ak−1 + · · · + a1 + a0 ) = 0. Write
n = ak 10k + ak−1 10k−1 + · · · + a1 10 + a0
42
CHAPTER 14. RING HOMOMORPHISMS
43
Now
φ(n) = 0 ⇔ φ(ak 10k + ak−1 10k−1 + · · · + a1 10 + a0 ) = 0
⇔ φ(ak )φ(10)k + φ(ak−1 )φ(10)k−1 + · · · + φ(a1 )φ(10) + φ(a0 )) = 0
⇔ φ(ak ) + φ(ak−1 ) + · · · + φ(a1 ) + φ(a0 ) = 0
⇔ φ(ak + ak−1 + · · · + a1 + a0 ) = 0
Example 14.5. Determine all the ring homomorphisms from Z12 to Z30 .
Solution: The ring homomorphism φ : Z12 → Z30 is completely determined if we
know φ(1). Let φ(1) = a. Then |a| divides both 12 and 30. This implies |a| = 1, 2, 3
or 6. Therefor the possibilities for a are 0, 15, 10, 20, 5 or 25. Since a2 = a, 20 and 5
are not the possibilities. Therefore φ(1) = 0 or φ(1) = 15 or φ(1) = 10 or φ(1) = 25.
Properties of Ring Homomorphisms
Theorem 14.1. Let φ : R → R0 be a ring homomorphism. Let A be a subring of R
and let B be an ideal of R0 .
1. φ(nr) = nφ(r) = nφ(r) and φ(rn ) = φ(r)n for any positive integer n and any
r ∈ R.
2. φ(A) = {φ(a) : a ∈ A} is a subring of R0 .
3. If A is an ideal of R and φ is onto, then φ(A) is an ideal of R0 .
4. φ−1 (B) = {r ∈ R : φ(r) ∈ B} is an ideal of R.
5. If R is commutative then φ(R) is commutative.
6. If R has unity 1, R0 6= 0, and φ is onto, then φ(1) is the unity of R0 .
7. φ is an isomorphism if and only if φ is onto and ker φ = {r ∈ R : φ(r) =
0} = {0}.
8. If φ : R → R0 is an isomorphism, then φ−1 : R0 → R is an isomorphism.
Proof. Some of these proofs will be asked in the final exam.
Theorem 14.2. Let φ : R → R0 be a ring homomorphism. Then ker φ = {r ∈ R :
φ(r) = 0} is an ideal of R.
CHAPTER 14. RING HOMOMORPHISMS
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Proof. We want to show that r ker φ ⊆ ker φ for every r ∈ R. For any r ∈ R, let ra ∈
r ker φ. Then φ(ra) = φ(r)φ(a) = φ(r)0 = 0 =⇒ ra ∈ ker φ =⇒ r ker φ ⊆ ker φ.
Similarly ker φr ⊆ ker φ for every r ∈ R. Thus ker φ is an ideal of R.
Theorem 14.3 (First Isomorphism Theorem for Rings). Let φ : R → R0 be a ring homomorphism. Then the factor ring R/ ker φ is isomorphic to φ(R). The isomorphism
is given by
r + ker φ 7→ φ(r)
Proof. No proof please it’s too much for them.
Theorem 14.4. Every ideal of a ring R is the kernel of a ring homomorphism of R.
In particular, an ideal I of R is the kernel of the homomorphism φ : R → R/I given
by r 7→ r + I.
Proof. φ(r + s) = (r + s) + I = (r + I) + (s + I) = φ(r) + φ(s) and φ(rs) = rs + I =
(r + I)(s + I) = φ(r)φ(s) implies φ : R → R/I is, indeed, a ring homomorphism.
ker φ = {r ∈ R : φ(r) = I} = {r ∈ R : r + I = I} = I.
Theorem 14.5. Let R be a ring with unity 1. The mapping φ : Z → R given by
n 7→ n · 1 is a ring homomorphism.
Proof. Recall that n · 1 = 1 + 1 + · · · + 1 (n 1’s).
1. φ(m + n) = (m + n) · 1 = m · 1 + n · 1 = φ(m) + φ(n).
2. φ(mn) = (mn) · 1 = (m · 1)(n · 1) = φ(m)φ(n)
This shows that φ : Z → R is a ring homomorphism.
The Field of Quotients
We construct a quotient field from a given integral domain in a similar fashion the
rationals are constructed from integers.