Homomorphisms from $\mathbb{Z}_p$ to $\mathbb{Z}_3$

Question

1. For which odd values of $p$ can we find a non-trivial homomorphism from $\mathbb{Z}_p$ to $\mathbb{Z}_3$ ?

2. Is there any method to find those homeomorphisms explicitly?

I have no any idea to handle this problem. Thank you.

Answer 1

If $f:\ \Bbb{Z}_p\ \longrightarrow\ \Bbb{Z}_3$ is a homomorphism, then for every $k\in\Bbb{Z}_p$ we have $$f(k)=f(1+1+\ldots+1)=f(1)+f(1)+\ldots+f(1)=k\cdot f(1),$$ which shows that $f$ is entirely determined by the value of $f(1)$. Now in $\Bbb{Z}_3$ we have $3=0$, so $$f(3)=3\cdot f(1)=0\cdot f(1)=0,$$ which shows that $3\in\ker(f)$. If $f$ is nontrivial then $\ker(f)\neq\Bbb{Z}_p$, so $3$ must not generate all of $\Bbb{Z}_p$. If $p$ is prime, this is only the case if $p=3$. In this case there are precisely two nontrivial homomorphisms.

Answer 2

For a group homomorphism $\phi$ which maps into $\mathbb{Z}/3\mathbb{Z}$ to be non trivial it must be surjective. This means something in the image of $\phi$ must have order $3$.

Since the order of $\phi(x)$ divides the order of $x$ for any homomorphism $\phi$ and element $x$, the order of $x$ must be a multiple of $3$. By Lagranges theorem, $|\mathbb{Z}/p\mathbb{Z}|$ must be divisible by $3$, hence $3|p$ is a necessary condition.

This is also a sufficient condition though as you can then construct a nontrivial surjective homomorphism, which I will leave to you to do.