Download Normal Subgroups The following definition applies. Definition B.2: A

Normal Subgroups
The following definition applies.
Definition B.2:
A subgroup H of G is a normal subgroup if a−1H a ⊆
H, for every a ∈ G, or, equivalently,
if aH = H a, i.e., if the right and left cosets coincide.
Note that every subgroup of an abelian group is normal.
The importance of normal subgroups comes from the following result (proved in Problem B.17).
Theorem B.8: Let H be a normal subgroup of a group G. Then the cosets of H
form a group under coset
multiplication:
(aH)(bH ) = abH
This group is called the quotient group and is denoted by G/H.
Suppose the operation in G is addition or, in other words,
G is written additively. Then the cosets of a
subgroup H of G are of the form a + H. Moreover, if H is a normal subgroup of G, then the cosets form a group
under coset addition, that is,
(a + H ) + (b + H ) = (a + b) + H
EXAMPLE B.11
(a) Consider the permutation group S3of degree 3 which is investigated above. The set H = {ε, σ1} is a
subgroup of S3. Its right and left cosets follow:
Right Cosets
H = {ε, σ1}
H φ1= {φ1, σ2}
H φ2= {φ2, σ3}
Left Cosets
H = {ε, σ1}
φ1H = {φ1, σ3}
φ2H = {φ2, σ2}
Observe that the right cosets and the left cosets are distinct; hence H is not a normal subgroup of S3.
(b) Consider the group G of 2 × 2 matrices with rational entries and nonzero determinants. (See Example A.10.)
Let H be the subset of G consisting of matrices whose upper-right entry is zero; that is, matrices of the form
&
'
a 0
c d
Then H is a subgroup of G since H is closed under multiplication and inverses and I ∈ H . However, H is
not a normal subgroup since, for example, the following product does not belong to H:
&
1 2
1 3
'−1&
1 0
1 1
'&
1 2
1 3
'
'
=
& −1 −4
1
3
On the other hand, let K be the subset of G consisting of matrices with determinant 1. One can show
that K is also a subgroup of G. Moreover, for any matrix X in G and any matrix A in K, we have
det(X−1AX)= 1
Hence X−1AX belongs to K, so K is a normal subgroup of G.
Integers Modulo m
Consider the group Z of integers under addition. Let H denote the multiples of 5, that is,
H = {. . . , −10, −5, 0, 5, 10, . . .}
Then H is a subgroup (necessarily normal) of Z. The cosets of H in Z appear in Fig. B-5(a). By the above
Theorem B.8, Z/H = {0, 1, 2, 3, 4} is a group under coset addition; its addition table appears in Fig. B-5(b).
This quotient group Z/H is referred to as the integers modulo 5 and it is frequently denoted by Z5. Analogously,
for any positive integer n, there exists the quotient group Zncalled the integers modulo n.
Fig. B-5
Cyclic Subgroups
Let G be any group and let a be any element of G. As usual, we define a0
= e and an+1 = an
· a.
Clearly,
aman = am+nand(am)n = amn, for any integers m and n. Let S denote the set of all the powers of a; that is
S = {· · · , a−3, a−2, a−1, e, a, a2, a3, · · · }
Then S is a subgroup of G called the cyclic group generated by a. We denote this group by gp(a).
Furthermore, suppose that the powers of a are not distinct, say ar= aswith, say, r > s. Then ar−s = e
where r, s > 0. The smallest positive integer m such that am
= e is called the order of a and it will be denoted
by |a|. If |a| = m, then the cyclic subgroup gp(a) has m elements as follows:
2
3
m
gp(a) = {e, a, a , a , . . . , a −1}
Consider, for example, the element φ1in the symmetric group S3discussed above. Then:
φ11=φ1,
//
φ12=φ2,
φ13=φ2· φ1= e
//
//=3 and gp(φ1) = {e, φ1, φ2}. Observe that φ1//
Hence φ1
divides the order of S3. This is true in general;
that is, for any element a in a group G, |a| equals the order of gp(a) and hence |a| divides |G| by Lagrange’s
Theorem B.7. We also remark that a group G is said to be cyclic if it has an element a such that G = gp(a).
Generating Sets, Generators
Consider any subset A of a group G. Let gp(A) denote the set of all elements x in G such that x is equal to a
product of elements where each element comes from the set A ∪ A−1
(where A−1 denotes the set of inverses of
elements of A). That is,
//
gp(A) = {x ∈ G x = b1b2. . . bmwhere each bi∈ A ∪ A−1}
Then gp(A) is a subgroup of G withgenerating set A. In particular, A is said to generate the group G if G = gp(A),
that is, if every g in G is a product of elements from A ∪ A−1. We say A is a minimal set of generators of G if
A generates G and if no set with fewer elements than A generates G. For example, the permutations a
= σ1and
b = φ1form a minimal set of generators of the symmetric group S3(Fig. B-4). Specifically,
e = a 2,
σ1= a,
σ2= ab,
σ3= ab2,
φ1= b,
φ2= b2
and S3is not cyclic so it cannot be generated by one element.
Homomorphisms
A mapping f from a group G into a group G is called a homomorphism if, for every a, b ∈ G,
f (ab) = f (a)f (b)
In addition, if f
is one-to-one and onto, then f
is called an isomorphism; and G and G
are said to be isomorphic,
written G ∼= G .
If f : G → G is a homomorphism, then the kernel of f , written Ker f , is the set of elements whose image
is the identity element e of G ; that is,
Ker f = {a ∈ G | f (a) = e }
Recall that the image of f , written f (G) or Im f , consists of the images of the elements under f ; that is,
Im f = {b ∈ G | there exists a ∈ G for which f (a) = b} .
The following theorem (proved in Problem B.19) is fundamental to group theory.
Theorem B.9:
Suppose f : G → G
is a homomorphism with
kernel K. Then K is a normal subgroup of G, and
the quotient group G/K is isomorphic to f (G).
EXAMPLE B.12
(a) Let G be the group of real numbers under addition, and let G be the group of
positive real numbers under
multiplication. The mapping f : G → G
defined by f (a) = 2ais a
homomorphism because
f (a + b) = 2a+b=2a2b= f (a)f (b)
In fact, f is also one-to-one and onto; hence G and G are isomorphic.
(b) Let a be any element in a group G. The function f : Z → G defined by f (n) = an is
a homomorphism since
f (m + n) = am+n = am· an= f (m) · f (n)
The image of f is gp(a), the cyclic subgroup generated by a. By Theorem B.9,
gp(a) ∼= Z/K
where K is the kernel of f. If K = {0}, then gp(a) = Z. On the other hand, if m is
the order of a, then
K ={multiples of m}, and so gp(a) ∼= Zm. In other words, any cyclic group is
isomorphic to either the
integers Z under addition, or to Zm, the integers under addition modulo m.