Is this homomorphism one-to-one?

Question

The question is: Let $\phi$:($Z_4'$) -> ($Z_8'$) be a homomorphism such that $\phi(1)$ = 6. Find $\phi(3)$ and is $\phi$ one-to-one?

Can someone please explain to me how one would go about solving this?

Thank you!

Answer 1

$\phi(3)=2$.

It's one-one because $\operatorname {ker}\phi=0$.

For $\phi(1)=6,\phi(2)=4,\phi(3)=2$.

Answer 2

I assume that "$Z_n'$" means $\mathbb{Z}/n\mathbb{Z}$.

We have $\phi(3) = \phi(1+1+1) = \phi(1) + \phi(1) + \phi(1) = 6 + 6 + 6 \equiv 2 \operatorname{Mod} 8$.

Likewise, $\phi(2) = 12 \equiv 4 \operatorname{Mod} 8$ and of course $\phi(0) = 0$.

Hence, $\phi$ is injective since $\forall x,y \in \mathbb{Z}/4\mathbb{Z},\, x\neq y \Rightarrow \phi(x) \neq \phi(y)$.