I wanted to find all the homomorphisms $\theta : \mathbb{Z}/12\mathbb{Z} \ \rightarrow \ \mathbb{Z}/42\mathbb{Z} $.
I thought that it would be enough to describe the map by $1 \mapsto a$ for some $a$, since $1$ generates the ring $\mathbb{Z}/12\mathbb{Z}$. The order of $\theta(1)$ has to devide $12$. That means that its order must belong to $\{0,2,3,4,6,12\}$.
- For $ord(a) = 2$, we can only choose $a=21$
- If $ord(a) = 3$, we can choose $a=14,28$
- The number $4$ does not divide $42$.
- For $ord(a)=6$ we have $7,35$
- And $12$ does not divide $42$
Those numbers are all the possible values for $a$.
Is this the right way to find those maps, or is there something I forgot?
There is no ring homomorphism $\theta:\Bbb Z/(12\Bbb Z)\to \Bbb Z/(42\Bbb Z)$ because $42$ not divide $12$ ,indeed ; if $\theta $ is such homomorphism, since $\theta(1)=1$, we have $12=12\theta(1)=\theta(12)=0 \mod [42]$, $i.e$ $42$ divide $12$.