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;
群论9.4 (2)28,32@page 349; 24@page338. 28-Show f:G→G defined by f(a)=a n is a homo morphism.
proof: (数学归纳法证)
1.because ab=ba, ab*ab=aabb=a2b2,
2.(ab)n=a n b n,
ab*(ab)n=ab*a n b n=aa n bb n=a n+1b n+1 Hence, f(ab)=(ab)n=a n b n=f(a)f(b).
群论9.4 (2)28,32@page 349; 24@page338.
32-Show f:G→G by f(a)=a-1 is a isomorphism iff Abelian.
Proof: (1)Suppose f is isomorphism,
f(xy)=(xy)-1=f(x)f(y)=x-1y-1
so xy=((xy)-1)-1=(x-1y-1)-1=yx. G is Abelian. (2)Suppose G is Abelian.
f(xy)=(xy)-1=x-1y-1=f(x)f(y), homomorphism. onto: ∀x∈G, f(x-1)=(x-1)-1=x.
one-to-one: suppose f(x)=f(y), x-1=y-1,
xx-1=e=yy-1, right cancelation x=y.
群论9.4 (2)28,32@page 349; 24@page338.
24-Let A={0,1} and consider (A*, ⋅), (N,+).
(1)f(α)=length(α), show f:A*->N is homomorphism.
f(α⋅β)=f(αβ)=length(α)+length(β)=f(α)+f(β)
(2)R: f(α)=f(β),show R is congruence relation.
First show R is equivalence relation.
2. if f(α)=f(β) and f(χ)=f(δ),then f(α⋅χ) = f(β⋅δ)
(3)A*/R and N is isomorphic.
(A*/R,*) : [α]*[β]=[α⋅β].
Let g([α])=length(α), (1)show g: A*/R->N is homomorphism. g([α] *[β])=g([α⋅β])=length(α)+length(β)=g([α])+g([β])
(2) onto : ∀x∈N, let α=00…0(x factors) ,g([α])=x.
(3) one-to-one : suppose g([α])=g([β]),
length(α)=length(β),so f(α)=f(β), then αRβ, [α]=[β].
4: Show f:G 1⨯G 2→G 1by f(a,b)=a is a homomorphism. f((a,b )*‘(c,d))=f(a*c,b*d)=a*c=f(a,b)*f(c,d)
18: Prove N is normal subgroup iff a -1Na=N for all a ∈G.
If N is normal subgroup, then ∀a ∈G, aN=Na, and
e ∈N. so a -1Na=Na -1a=Ne=N.
If a -1Na=N for all a ∈G, then N=a -1aN=a -1Na,
left cancelation, aN=Na.
29: Prove H is normal subgroup if H only two left coset.
let a∉H, The left cosets of H are H and aH.
The right coset are H and Ha.
H⋂aH = ∅= H⋂Ha. and H⋃aH=G=H⋃Ha.
Thus aH=Ha.
let b∈H, bH=H=Hb. so ∀x∈G, xH=Hx.
H is a normal subgroup.
30: Prove that if N is a normal subgroup of G, then H⋂N is a normal subgroup of H.
(1)show H⋂N is a subset of H;
∀x∈H⋂N, x ∈H , H⋂N is a subset of H.
(2)show H⋂N is a closure;
∀x,y∈H⋂N, x∈H and y∈H, x*y∈H; x∈N and y∈N, x*y∈N; so x*y∈H⋂N.
(3)Since H and N are subgroup of G, e∈N and e∈H,
so e∈H⋂N.
(4)∀x∈H⋂N, x ∈H , x-1∈H, and x ∈N, x-1∈N;
so x-1∈H⋂N.
30: Prove that if N is a normal subgroup of G, then
H ⋂N is a normal subgroup of H.
(5) ∀x ∈H, x(H ⋂N) = {xn | n ∈H ⋂N}, let any n 1∈H ⋂N,
because aN=Na, x*n 1=n 2*x for some n 2∈N,
n 2= x*n 1*x -1,
since x -1 ∈H, n 1∈H, then n 2∈H.
n 2∈H ⋂N,
Thus x(H ⋂N) = (H ⋂N)x.
Hence, H ⋂N is a normal subgroup of H.
群论9.5 Ex1: Let G be a group, and let N and H be subgroups of G such that N is normal in G. Prove that (1)HN is a subgroup of G.(2)N is normal subgroup of HN.
(1) first show HN is closure;
∀x,y ∈HN, x=an 1,y=bn 2 for some a,b ∈H.
xy=an 1*bn 2, because bN=Nb => n 1*b=bn 3,
so xy=abn 3n 2∈HN.
e ∈H and e ∈N, so e*e=e ∈HN.
∀x ∈HN, x=an 1, x -1=(an 1)-1= n1-1a -1 ,
because Nb=bN,so n1-1a -1 =a -1n 4 ∈HN.
群论9.5 Ex1: Let G be a group, and let N and H be subgroups of G such that N is normal in G. Prove that
(1)HN is a subgroup of G.(2)N is normal subgroup of HN.
(2) first show N is subset of HN;
∀n∈N, because e∈H, so e*n=n∈HN.
Since N is subgroup of G, so N is closed and have identity, and inverse of all elements, thus N is subgroup of HN.
Second, show aN=Na, for ∀a∈HN;
because (1), a∈G,
N is mormal in G, so aN=Na,
Hence N is normal subgroup of HN.。