Chapter 1
1、proof Let A,B,C be sets .Suppose that x∈B,we get x∈A∩B or
x?A?B?A,and x?A?C or x?A?C?A since A?B?A?C and A?B?A?C.so x∈C and B?C.Similarly ,we have C?Band so B=C .
2、proof ① First,consider x?A?(B?A).Then x?A or x?B,but x?A.This
implies if x is not an element of A ,then x?B.Hence x?A?B and
A?(B?A)?A?B.
Conversely, if x?A?B,then by definition , x?A or x?B.
This generates two cases:
(a1) If x?A,clearly x?A?(B?A);
(b2) If x?B,then either x?A or not . i.e.,either x?B and x?A or
x?B but x?A, in either case, we have x?A?(B?A).
Hence A?B?A?(B?A).
Therefore A?(B?A)=A?B.
② Suppose that x?A?(B?C).Then x?A but x?B and x?C. So
x?A-B and x?A?C and x?(A?B)?(A?C) by definition .
Hence A?(B?C)?(A?B)?(A?C).
Converssely, Assume that x?(A?B)?(A?C),then x?A-B and
x?A?C,and we have x?A,but x?B and x?C.Hence
x?B?C, x?A, i.e., x?A?(B?C).
Therefore(A?B)?(A?C)?A?(B?C)and,soA?(B?C)=
(A?B)?(A?C)
3.(a) surjective (b) bijective (c) bijective
4.proof if f: X?Y and g: Y?Z are functions,then their composite denoted by
g?f, is the function X?Z given by g?f: X?g(f(x)) (i)
suppose that (g?f)(a)= (g?f)(b), where a,b ?X. we have g(f(a))=g(f(b)) by definition, and f(a)=f(b) since g is injective, similarly, a=b for f is injective. Therefore, g?f is injective.
(ii)
For each Z?Z, there is y ?Y with g(y)=z since g is surjective, and for each y ? Y, there exists a ? x with f(a)=y since f is surjective. So for ?z ?Z, there is a ? x with (g?f)(a)=g(f(a))=g(y)=z. which implies g?f is surgective.
5. proof clearly, ?:R?R is a function. Suppose that ?(a)= ?(b) where a, b
?R are distinct. Then
a23a?1?b23b?1, cross multiplying yields
33332323a?ab?ba?b, which simplifies to a?b and hence a?b,so ? is
injective. for ?given y ? R,
32x23x?1?yfrom,we get equation
x?yx?y?0, which can be solved for x, i.e .for each y ? R,there is at least
x ?x such that bijective.
x23x?1?y.whic implies ? is surjective. Therefore ? is
6、(a) R is reflexive, symmetric, transitive. (b) R is reflexive, not symmetric, transitive. (c ) R is reflexive, symmetric, transitive.
(d) R is reflexive, symmetric, transitive.
7、proof (1) For every a∈R-{0}，we have aa?a?0, and so aRa, (2) If
2
aRb, where a,b?R?{0}, i.e.,ab?0, then ba?0,
i.e., bRa,
(3) If aRb,bRc, where a,b,c?R?{0}, i.e. ab?0,bc?0,
then ac?0.i.e.,aRc.Therefore, the relation ～ is an equivalence relation .
8、 There are 1,3,5,15 equivalence relations on a set S with 1,2,3 or 4 elements,
separately.
9、 We can list the elements of the residue classes of modulo 3: [0]={…,-9,-6,-3,0,6,9,…} [1]={…,-8,-5,-2,1,4,7,10,…} [2]={…，-7,-4,-1,2,5,8,11,…}
Chapter 2
1、proof i)
ii）For every x,y,z∈G,
(x*y)*z=(xy-x-y+2)*z=(xy-x-y+2)z-z-(xy-x-y+2)+2=xyz-xz-yz+z-xy+x+y x*(y*z)=x*(yz-y-z+2)=x(yz-y-z+2)-x-(yz-y-z+2)+2=xyz-xy-xz+x-yz+y+z we have (x*y)*z=x*(y*z). And so the associative law holds.
3、Solution Straightforward calculation shows that A?I?B. (AB)?I,
46n?1since (AB)???0n?n?)??I (n?0.
1?2224、proof Suppose (ab)?ab for all
2a,b?G，then
(ab)?(ab)(ab)?a2b2=(aa)(bb),
i.e., abab?aabb. Applying left cancellation , this becomes bab?abb,and by right cancellation, this reduces to ba?ab. 5、proof For every a?G, there is a,a(identity)
?1?G such that a?1a?e
So (aba?1n)?aba?1aba?1…aba?1aba?1=abna?1。
6、proof Firstly ,the identity element belongs to N since it is equal to
e?e,
n (e is an identity of G ). The product of any two elements of N is an element of N , since G is abelian And abnn?(ab).Also the inverse of an element of N is an element of
?1nN .Since (a)n?(a)
?1n For some positive integer. Thus N is a subgroup of G .
?17、Proof Clearly，e???00??cos0???1??sin0?sin0???H. cos0??sin?2???H,we have cos?2??cos?1For every ??sin?1?cos?1??sin?1?sin?1??cos?2?,
cos?1???sin?2?sin?2??= cos?2??sin?1??cos?2?cos?1???sin?2?cos(?1??2)??sin(?1??2)?sin(?1??2)??cos(?1??2)??cos??H ,and the inverse of ??sin??sin??? is cos???cos(??)??sin(??)?sin(??)??cos(??)??H
Thus H is a subgroup of G .
?1e?8、Proof Firstly ,the identity element ??00???K 1??aFor every ???2bb??c?,?a???2dd??c??k,
ad?bc???k, And the inverse of
ac?2bd??a???2bb??c??a???2dd??ac?2bd???c???2(ad?bc)?a??2b?a???2bb?1? is 22a?a?2bb???k, so k is a subgroup of GL2(R). a?9. Hint: Theorem 2.4.6.
10 .Proof For every a?G, we have ax=xa since G is abelian. For a subgroup H,
naturally, Ha=aH and H is normal.
11.o(0)=1, o(1)?6,o(2)?3,o(3)?2,o(4)?3,o(5)?6.
12. Solution AU positive divisors of 6 are 1,2,3,6, so the subgroup of Z6 are : <1>=Z6={0,1,2,3,4.5}, <2>=2Z6={0,2,4}, <3>=3Z6={0,3}, <6>=6Z6={0}.
13. Yes. 14. need not. 15. H={mz/z?Z},where m is integer . 16.Answers may vary.
17. proof suppose that o(ab)=r, then arn?ab?(ab)?e since o(b)=n,
rnrnrnand we have m/r n. From gcd(m,n)=1, so m/r. Similary n/r, and mn/r since gcd(m,n)=1. On the other hand, (ab)r/mn ,Hence r=mn.
19、(14)(265),odd. 20、5！ 21、30
23、Proof firstly, we will prove Ha=H iff a?H. If Ha=H, then a=ea?Ha=H .i.e.
a?H. Conversely , If a?H,H is a subgroup of the group G, then a?1mn?abmnmn?e?e?e and so
?H and
Ha?H?H=H. Also H=Haiff a?H.
a?1?1=(Ha)a?Ha. Suppose .So Ha=H. Hence Ha=H
?1?1?1Suppose that Ha=Ha, then Hab=Hbb=H. and ab?H. from the above
proof .If ab?1?H, then Hab?1=H and so Ha=Hb.
?124、Proof First, aa=e?H, then a～a.
?1?1?1If a～b ,then ab?H and Ha= Hb by 23.So Hba=H and ba?H. we get b～
a.
?1?1?1?1?1
If a～b, b～c, then ab?H, bc?H, and ac=(ab)bc?H, so a～c.