Download PowerPoint 演示文稿

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Field (mathematics) wikipedia , lookup

Laws of Form wikipedia , lookup

Oscillator representation wikipedia , lookup

Group action wikipedia , lookup

Cayley–Hamilton theorem wikipedia , lookup

Birkhoff's representation theorem wikipedia , lookup

Complexification (Lie group) wikipedia , lookup

Polynomial ring wikipedia , lookup

Fundamental theorem of algebra wikipedia , lookup

Commutative ring wikipedia , lookup

Transcript
 Theorem 6.21: Let H be a subgroup of G.
H is a normal subgroup of G iff g-1hgH
for gG and hH.
 Proof: (1) H is a normal subgroup of G
 (2) g-1hgH for gG and hH
 For gG , Hg?=gH
 Let H be a normal subgroup of G, and let
G/H={Hg|gG}
 For Hg1 and Hg2G/H,
 Let Hg1Hg2=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=Hg4G/H,
 Hg1Hg2=H(g1*g2), Hg3Hg4=H(g3*g4),
 Hg1Hg2?=Hg3Hg4?
 H(g1*g2)=?H(g3*g4)
 g3*g4?H(g1*g2), i.e. (g3g4)(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=HG/H is identity element of G/H
 Inverse element: For HaG/H, Ha-1G/H
is inverse element of Ha, where a-1G 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 Hg1Hg2=
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  aG 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’. {xG| (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
x0
x0
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,bKer, a*b?Ker,
 i.e.(a*b)=?eG‘
 Inverse element: For aKer, a-1?Ker
 ii)For gG,aKer, 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: GG/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 KaG/K,f(Ka)=(a)
 f is an isomorphism function。
 Proof: For  KaG/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,KbG/K, f(KaKb)=?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)=e2ix
  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, cR,
(1) a · (b + c) = a · b + a · c,
(2) (b + c) · a = b · a + c · a.
We write 0R 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)nn|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, bR . A ring R is
an unitary ring if there is 1R such that
1a = a1 = a for all aR. 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|nZ+, aiR}.
 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,bR,
n
(a  b) n   C (n, i )a i b n i
i 0
 where nZ+.
 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+ab
 (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.