# Thread: Proving that there aren't any more homomorphisms

1. ## Proving that there aren't any more homomorphisms

I was asked to find all homomorphisms $\displaystyle \Phi :\mathbb{Z}\to\mathbb{Z}$

I'm fairly convinced in my own mind that all of them are $\displaystyle \Phi (x)=cx \ \forall \ c \in \mathbb{Z}$

I can show that these are homomorphisms by claiming that $\displaystyle \Phi$ is defined as above and:

1. $\displaystyle \Phi (a+b)=\Phi(a)+\Phi(b)$ (definition of homomorphism)
2. $\displaystyle c(a+b)=c(a)+c(b)$ (1, my def. of $\displaystyle \Phi$)
3. $\displaystyle ca+cb=ca+cb$ (2, distributive prop.)

So, how do I show now that there are no other homomorphisms?

2. ## Re: Proving that there aren't any more homomorphisms

Hi beebe!

Let's make an arbitrary choice for $\displaystyle \Phi(1)$.
Suppose $\displaystyle \Phi(1)=c$.
Now what does $\displaystyle \Phi(x)$ have to be (and why)?

3. ## Re: Proving that there aren't any more homomorphisms

more generally, if G is any cyclic group, with generator x, and φ:G→G is a homomorphism, you can show that φ is completely determined by φ(x).

(because $\displaystyle \varphi(x^k) = (\varphi(x))^k$). so if $\displaystyle \varphi(x) = x^m$, we have $\displaystyle \varphi:x^k \to x^{km}$.

now $\displaystyle \mathbb{Z}$ is cyclic, with generator 1, so....

(recall that in $\displaystyle \mathbb{Z}$, $\displaystyle x^k = x+x+\dots+x\ \text{(k times)}= kx$).