# Composition of homomorphisms is a homomorphism

1. Aug 2, 2008

### futurebird

1. The problem statement, all variables and given/known data

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?

Last edited: Aug 2, 2008
2. Aug 3, 2008

*bump*

3. Aug 4, 2008

### 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 =)