Composition of homomorphisms is a homomorphism

  • Thread starter futurebird
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  • #1
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Homework Statement



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

2. The attempt at a solution

Since [tex]f[/tex] is a homomorphism [tex](G, * )[/tex] and [tex](H, \circ)[/tex] are groups and [tex]f(a*b)= f(a) \circ f(b), \forall a,b \in G[/tex]. Likewise, [tex](K, +)[/tex] is a group and [tex]g(f(a) \circ f(b)) = g(f(a)) + g(f(b)), \forall f(a), f(b) \in H[/tex] with [tex]a, b \in G[/tex]. Hence, [tex]g \circ f = g(f(x))[/tex] [tex]\forall x\in G[/tex]. [tex]g \circ f[/tex] is a homomorphism.


Is this anything like correct?
 
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Answers and Replies

  • #3
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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 =)
 

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