please help me prove this.... abstract algebra..homomorphisms?

Question

1.) show that if G, G', and G'' are groups and if Ø: G -> G' and ¥: G' -> G'' are homomorphisms, then the composite map ¥Ø:G-> G'' is a homomorphism.

2.) Let Ø:G-> H be a group homomorphism. show that Ø[G] is abelian if and only if for all x, y element of G, we have xyx^-1y^-1 element of Ker(Ø).

3.) Let G be any group and let a be any element of G.
Let Ø: Z-> G be defined by Ø(n)= a^n. show that Ø is a homomorphism. Describe the image and the possibilities for the kernel of Ø.

4.) let Ø: G-> G' be a homomorphism with kernel H and let a element of G. Prove the set equality {x є G l Ø(x) = Ø(a) } = Ha.

Answer 1

#1

Since each is a homomorphism, we know:
φ(ab) = φ(a)φ(b)
y(pq) = y(p)y(q)

So yφ(ab) = y(φ(ab))=y(φ(a)φ(b))
= y(φ(a))y(φ(b)) = yφ(a)yφ(b)

Thus yφ is also a homomorphism. Note that if you want to follow things explicitly, p = φ(a), q = φ(b). QED.

#2

I will use ' to denote inverses, for simplicity's sake.

φ(xyx'y') = φ(x)φ(y)φ(x')φ(y')

We know that φ(x') = φ(x)' and likewise for φ(y').

That means:

φ(xyx'y') = φ(x)φ(y)&phi(x)'φ(y)'

φ(xyx'y') is the identity e, if and only if:

φ(x)φ(y)&phi(x)'φ(y)' = e

Which is equivalent to:

φ(x)φ(y) = (φ(x)'φ(y)' ) ' = φ(y)φ(x)

This is the definition of commutativity. So φ(xyx'y') = e if and only if φ(G) is commutative (Abelian).


#3

To show that it's a homomorphism, we just need to show a few things. First of all, φ(0) = a^0 = e. Second, we know:

φ(m+n) = a^(m+n) = a^m * a^n = φ(m)φ(n)

Don't forget - Z is an additive group. That's why it's m+n.

That's all it takes to be to be a homomorphism. Now, the kernel is the set of n values such that a^n = 0. If you take m = the order of a, then those n values must be multiples of n. So you'd have the kernel of φ to be:

{ ... -3 o(a) , -2 o(a), - o(a) , 0 , o(a) , 2 o(a) , 3 o(a) ... }

If a does not have a finite order, however, there will be no such elements, and the kernel will be simply {0}.

#4

Consider the set in question to be called X. Take x∈X, so that φ(x) = φ(a). There is SOME element g such that x=ga. However, take φ of both sides:

φ(x) = φ(ga) = φ(g)φ(a)

This means that φ(g) = e, thus g∈H, the kernel. Thus x∈Ha after all. And so X ⊂ Ha.

On the other hand, take ha ∈ Ha. Then

φ(ha) = φ(h)φ(a) = e φ(a) = φ(a)

And thus Ha ⊂ X, and thus Ha = X.

Answer 2

You should do your own homework. You already asked a question about group homomorphisms.