7.7.5
G is abelian => every subgroup is normal => N is normal => can form the quotient G/N. It is not hard to see that N={(1,1),(2,0),(3,1),(4,0),(5,1),(0,0)}. So N has order 6. G has order 12 => G/N has order 2 => it is the cyclic group of order 2.
7.7.15
(a) Let x be in Q/Z. => There are integers a and b, b!=0, such that x=a/b+Z. Now, b(a/b+Z)=a+Z=0+Z => |x| divides b => x has finite order.
(b) Let n be an integer. Then (1/n + Z) is in Q/Z. n(1/n+Z)=1+Z=0+Z => |1/n+Z| divides n.
Let k:=|1/n+Z|. =>
0+Z=k(1/n+Z)=k/n+Z => k/n is in Z => n|k.
But we had k|n => k=n. So for every integer n, there is an element, for example 1/n+Z, in Q/Z of order n.
7.7.24
Part (a) is straightforward.
(b) Note that (a,b)+N=(c,d)+N iff (a-c,b-d) is in N iff
a-c=-(b-d) iff a+b=c+d.
So, (a,b)+N={(x,y): x+y=a+b}. It is not hard to check that the map f:RxR/N->R defined by f((a,b)+n)=a+b is an isomorphism. Therefore RxR/N is isomorphic to R.
7.8.1(a)
f(x+y)=3(x+y)=3x+3y=f(x)+f(y).
Ker(f)={x:f(x)=0 in Z(12)}={0,4,8}.
7.8.4
f(x+yi)f(z+ti)=(xx+yy)(zz+tt)=(xz-yt)(xz-yt)+(xt+yz)(xt+yz)=f(xz-yt+(xt+yz)i)=f((x+yi)(z+ti)) => f is a homomorphism.
Let r be in R**. => r>0. So sqrt(r)+0i is in C*.
f(sqrt(r)+0i)=r => Im(f)=R** i.e. f is surjective.
7.8.15
Define f:GL(2,R)->R* by 
f (a b) = ad-bc
  (c d)
It is not hard to check, that f is a surjective homomorphism [f is surjective, since for every a in R* 
f (a 0) = ad-bc=a
  (0 1)
] and that Ker(f)=SL(2,R). So by first iso. thm. GL(2,R)/SL(2,R) is isomorphic to R*.
7.8.22
1. Suppose G/N is simple. And assume there is a normal subgroup k of G s.t. N since N!=k, K/N!=G/N since K!=G => K/N is a normal proper subgroup of G/N. This is a contradiction to G/N being simple. => there is no normal K s.t. N
2. Suppose there is no normal K s.t. N there is subgroup L>=N of G s.t. L/N=H. => L/N is normal in G/N => L is normal in G, contradicting to the assumpion above => G/N is simple.