北京邮电大学 离散数学下 群论 作业错题讲解

群论9.1-9.2 （1）20,28 @323-324

a. not commutative, a*b ≠b*a.

b. a*(b*c)=a*c=b, (a*b)*c=c*c=b.

c. not associative, (c*c)*c=b*c=c ≠a=c*b=c*(c*c).

群论9.1-9.2 （1）20,28 @323-324

1. ≤is reflexive

Since a= a*a, a≤a for all a in A.

2. ≤is antisymmetric

suppose that a≤b and b≤a.

Then, b = a*b= b*a= a, so a= b.

3. ≤is transitive

If a≤b and b≤c, then c= b*c= (a*b)*c= a*(b*c)= a*c, so a≤c.

群论9.1-9.2 （1）28 @323-324

show a*b= a∨b, for all a and b in A

(1)a*b= a*(b*b) =(a*b)*b=b*(a*b), so b ≤a*b.

In a similar way, a*b= (a*a)*b=a*(a*b), so a≤a*b.

so a*b is a upper bound for a and b.

(2)if a≤c and b≤c.then c= a*c and c= b*c by definition. Thus c= a*(b*c)= (a*b)*c. so a*b≤c.

This shows that a*b is the leastest upper bound of a and b.

群论9.1-9.2 （2）12,16@page 348;

︒f1 f2 f3 f4 f5 f6closure;

f1 f1 f2 f3 f4 f5 f6identity=f1;

f2 f2 f1 f5 f6 f3 f4reverse:f2-1=f2,f3-1=f3, f3 f3 f4 f1 f2 f6 f5f4-1=f5,f6-1=f6.

f4 f4 f3 f6 f5 f1 f2associate: ︒is associate.

f5 f5 f6 f2 f1 f4 f3

f6 f6 f5 f4 f3 f2 f1

群论9.1-9.2 （2）12,16@page 348; If a*a=e, ∀a∈G. show commutative.

(1)group propertise (a*b)-1=b-1*a-1 because a*a=e=a*a-1 so, b-1*a-1=b*a, (2) (a*b)*(a*b)=e => (a*b)-1=a*b, reverse be only one in Group.

so b*a=a*b.

群论9.1-9.2 Ex1: Let G be a group. For

a,b∈G,we say that b is conjugate to a,written by b~a, if there exist g∈G such that b=gag-1.show that ~ is a equivalence relation on G. The equivalence classes of are called the conjugacy classes of G.

proof: (1)reflexive a=eae, so a~a;

(2)symmetric If b~a, then b=gag-1,

a=eae=g-1gag-1g=g-1bg.so a~b.

(3)transitive suppose a~b and b~c,

a=gbg-1 ,b=h ch-1; so a=ghch-1g-1=(gh)c(gh)-1

群论9.1-9.2 Ex2: Let G be a group, and suppose that a and b are any elements of G.

Show that if (ab)2=a2b2,,then ba=ab.

proof:

(ab)2=ab*ab=a*(b*a)*b

a2b2=aa*bb=a*(a*b)*b,

based left/right cancellation, b*a=a*b.

群论9.1-9.2 Ex3: Let G ={x∈R|x>1} be the set of all real numbers greater than 1. For x,y∈G, define

x*y=xy-x-y+2. Show that (G,*) is a group.

proof: (1)closure; because x>1, xy>x+y,

so x*y=xy-x-y+2>2, x*y∈G.

(2)associate; (x*y)*z=(xy-x-y+2)*z=xyz-xz-yz+2z-(xy-x-y+2)-z+2=xyz-xy-xz-yz+x+y+z .

x*(y*z)=x*(yz-y-z+2)=xyz-xy-xz+2x-x-yz+y+z-

2+2=xyz-xy-xz-yz+x+y+z .

(3)identity: e=2, 2*x=2x-x-2+2=x.

(4)reverse: x*x-1=xx-1-x-x-1+2=e=2, x-1=x/(x-1).

群论9.4 (1)22,28@page 331;

22-Show S 1⋂S 2is a subsemigroup of S.

proof: closure; ∀a,b ∈S 1⋂S 2, then a*b ∈S 1, a*b ∈S 2, so a*b ∈S 1⋂S 2.

if S 1⋂S 2 =∅, still be subsemigroup.

28-If f:S 1→S 2and g :S 2→S 3 be isomorphisms, show g ︒f :S 1→S 3is an isomorphism.

proof: (g ︒f)(x *1y)= g(f(x *1y))=g(f(x) *2f(y))

=g(f(x))*3g(f(y))=(g ︒f)(x) *3(g ︒f)(y).

one-to-one: suppose (g ︒f)(a)=(g ︒f)(b), g(f(a))=g(f(b)), f(a)=f(b), a=b.

onto: ∀z ∈S 3 有g(y)=z; ∀y ∈S 2 有f(x)=y;