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Theorem 6.21: Let H be a subgroup of G.
H is a normal subgroup of G iff g-1hgH
for gG and hH.
Proof: (1) H is a normal subgroup of G
(2) g-1hgH for gG and hH
For gG , Hg?=gH
Let H be a normal subgroup of G, and let
G/H={Hg|gG}
For Hg1 and Hg2G/H,
Let Hg1Hg2=H(g1*g2)
Lemma 3: Let H be a normal subgroup of
G. Then [G/H; ] is a algebraic system.
Proof: is a binary operation on G/H.
For Hg1=Hg3 and Hg2=Hg4G/H,
Hg1Hg2=H(g1*g2), Hg3Hg4=H(g3*g4),
Hg1Hg2?=Hg3Hg4?
H(g1*g2)=?H(g3*g4)
g3*g4?H(g1*g2), i.e. (g3g4)(g1*g2)-1?H.
Theorem 6.22: Let [H;] be a normal
subgroup of the group [G;]. Then [G/H;]
is a group.
Proof: associative
Identity element: Let e be identity element
of G.
He=HG/H is identity element of G/H
Inverse element: For HaG/H, Ha-1G/H
is inverse element of Ha, where a-1G is
inverse element of a.
Definition 19: Let [H;*] be a normal
subgroup of the group [G;*]. [G/H;] is
called quotient group, where the operation
is defined on G/H by Hg1Hg2=
H(g1*g2).
If G is a finite group, then G/H is also a
finite group, and |G/H|=|G|/|H|
6.5 The fundamental theorem of
homomorphism for groups
6.5.1.Homomorphism
kernel
and
homomorphism image
Lemma 4: Let [G;*] and [G';] be groups,
and be a homomorphism function from
G to G'. Then (e) is identity element of
[G';].
Proof: Let x(G)G'. Then aG such
that x=(a).
Definition 20: Let be a
homomorphism function from group
G with identity element e to group G'
with identity element e’. {xG| (x)=
e'} is called the kernel of
homomorphism function . We
denoted by Ker( K(),or K).
Example: [R-{0};*] and [{-1,1};*] are
groups.
1
( x)
1
x0
x0
Ker {x | x 0, x R}
Theorem 6.23:Let be a homomorphism
function from group G to group G'. Then
following results hold.
(1)[Ker;*] is a normal subgroup of [G;*].
(2) is one-to-one iff K={eG}
(3)[(G); ] is a subgroup of [G';].
proof:(1)i) Ker is a subgroup of G
For a,bKer, a*b?Ker,
i.e.(a*b)=?eG‘
Inverse element: For aKer, a-1?Ker
ii)For gG,aKer, g-1*a*g?Ker
6.5.2 The fundamental theorem
of homomorphism for groups
Theorem 6.24 Let H be a normal
subgroup of group G, and let [G/H;] be
quotient group. Then f: GG/H defined
by f(g)=Hg is an onto homomorphism,
called the natural homomorphism.
Proof: homomorphism
Onto
Theorem 6.25 : Let be a homomorphism
function from group [G;*] to group [G';].
Then [G/Ker();][(G);]
isomorphism function f:G/ Ker()(G).
Let K= Ker(). For KaG/K,f(Ka)=(a)
f is an isomorphism function。
Proof: For KaG/K,let f(Ka)=(a)
(1)f is a function from G/K to (G)
For Ka=Kb,(a)=?(b)
(2)f is a homomorphism function
For Ka,KbG/K, f(KaKb)=?f(Ka)f(Kb)
(3) f is a bijection
One-to-one
Onto
Corollary 6.2: If is a homomorphism
function from group [G;*] to group
[G';], and it is onto, then
[G/K;][G';]
Example: Let W={ei|R}. Then
[R/Z;][W;*].
Let (x)=e2ix
is a homomorphism function from
[R;+] to [W;*],
is onto
Ker={x|(x)=1}=Z
6.6 Rings and fields
6.6.1 Rings
Definition 21: A ring is an Abelian group [R, +] with
an additional associative binary
operation(denoted · such that for all a, b, cR,
(1) a · (b + c) = a · b + a · c,
(2) (b + c) · a = b · a + c · a.
We write 0R for the identity element of the group
[R, +].
For a R, we write -a for the additive inverse of a.
Remark: Observe that the addition operation is
always commutative while the multiplication need
not be.
Observe that there need not be inverses for
multiplication.
Example: The sets Z,Q, with the usual
operations of multiplication and addition
form rings,
[Z;+,],[Q;+,] are rings
Let M={(aij)nn|aij is real number}, Then
[M;+,]is a ring
Example: S,[P(S);,∩],
Commutative ring
Definition 23: A ring R is a commutative
ring if ab = ba for all a, bR . A ring R is
an unitary ring if there is 1R such that
1a = a1 = a for all aR. Such an
element is called a multiplicative identity.
Example: If R is a ring, then R[x] denotes the
set of polynomials with coefficients in R. We
shall not give a formal definition of this set, but
it can be thought of as: R[x] = {a0 + a1x + a2x2
+ …+ anxn|nZ+, aiR}.
Multiplication and addition are defined in the
usual manner; if
n
m
f ( x) ai x and g ( x) bi x i
i
i 0
f ( x) g ( x)
then
i 0
max{ n , m}
(a b ) x
i 0
i
i
i
n m
f ( x) g ( x)
k
(
a
b
)
x
i j
k 0 i j k
One then has to check that these operations define a ring.
The ring is called polynomial ring.
Theorem 6.26: Let R be a commutative
ring. Then for all a,bR,
n
(a b) n C (n, i )a i b n i
i 0
where nZ+.
Quiz:1.Let G be a cyclic group generated by the
element g, where |G|=18. Then g6 is not a
generator of G
2.Let Q be the set of all rational numbers.
Define & on Q by
a&b=a+b+ab
(1)Prove[Q;&]is a monoid.
(2)Is [Q;&] a group?Why?
Exercise:P367 6
1.Prove Theorem 6.23(2)(3)
2.Let W={ei|R}. Then [C*/W;][R+;*].
3.Let X be any non-empty set. Show that
[P(X); ∪, ∩] is not a ring.