# Proving that there aren't any more homomorphisms

• Dec 23rd 2011, 04:41 PM
beebe
Proving that there aren't any more homomorphisms
I was asked to find all homomorphisms $\Phi :\mathbb{Z}\to\mathbb{Z}$

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

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

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

So, how do I show now that there are no other homomorphisms?
• Dec 23rd 2011, 05:06 PM
ILikeSerena
Re: Proving that there aren't any more homomorphisms
Hi beebe! :)

Let's make an arbitrary choice for $\Phi(1)$.
Suppose $\Phi(1)=c$.
Now what does $\Phi(x)$ have to be (and why)?
• Dec 23rd 2011, 07:51 PM
Deveno
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 $\varphi(x^k) = (\varphi(x))^k$). so if $\varphi(x) = x^m$, we have $\varphi:x^k \to x^{km}$.

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

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