Find all homomorphisms from $(\mathbb{Z}_8,+)$ into $(\mathbb{Z}_6,+)$.
The solution given is as follows :
We have , $\mathbb{Z}_8=\langle [1]\rangle$ . Let $f:\mathbb{Z}_8\longrightarrow \mathbb{Z}_6$ be a homomorphism . For any $[a]\in\mathbb{Z}_8$ ,$f([a])=af([1])$ , shows that that f is completely known if $f([1])$ is known . Now , $o(f[1])$ divides $o([1])$ and $|\mathbb {Z}_6|$ , i.e , $o(f[1])$ divides $8$ and $6$. Hence, $o(f[1])=1$ or $2$ . Thus, $f([1])=[0]$ or $[3]$ . If $f([1])=[0]$, then $f$ is the trivial homomorphism which maps every element to $[0]$. On the other hand, $f([1])=3$ implies $f([a])=[3a]$ for all $[a]\in\mathbb{Z}_8$. Thus , $f([a]+[b])=[3a+3b]=[3a]+[3b]=f([a])+f([b])$, proving the mapping $f:\mathbb{Z}_8\longrightarrow \mathbb{Z}_6$ defined by $f([a])=[3a]$ for all $[a]\in\mathbb{Z}_8$ is a homomorphism. Hence, there are two homomorphism from $\mathbb{Z}_8$ to $\mathbb{Z}_6$.
However, I am not getting how are they concluding the fact "$o(f[1])$ divides $o([1])$ and $|\mathbb {Z}_6|$" ? Also, how is the fact " $f([1])=[0]$ or $[3]$ " as mentioned in the solution true?...