Composition of homomorphisms is a homomorphism

[futurebird]

Homework Statement

Prove that if f: G \to H and g: H \to K are homomorphisms, then so is g \circ f: G \to K.

2. The attempt at a solution

Since f is a homomorphism (G, * ) and (H, \circ) are groups and f(a*b)= f(a) \circ f(b), \forall a,b \in G. Likewise, (K, +) is a group and g(f(a) \circ f(b)) = g(f(a)) + g(f(b)), \forall f(a), f(b) \in H with a, b \in G. Hence, g \circ f = g(f(x)) \forall x\in G. g \circ f is a homomorphism.Is this anything like correct?

 

[futurebird]

*bump*

 

[mistermath]

Someone will correct me on this, but I'm pretty sure I'm not incorrect.

First of all, you shouldn't say the homomorphism implies G and H are groups. The problem should start out with

Let <G,*>, <H,o>, and <K,+> be groups and let f: G->H be a homomorphism and g: H->K be a homomorphism prove that (g o f): G->K is a homomorphism.

If I was in trouble doing this problem, I'd first state what I need to show:

NTS: (g o f) (x*y) = (g o f)(x) + (g o f)(y) for all x,y in G.

This is the homomorphism property.

So if you pick any 2 random elements in G and "combine" them, then pop them into (g o f) you should get the same result as if you first popped each one into (g o f) then "combined" them. This is english for homomorphism.

I guess the only thing left is take 2 elements in G and show that what we said in two different ways is true.. ie:

Proof:
Let x, y \element of G, then { ... } done!

I think what you've done above is good.. just need to structure it better =)