How to determine all homomorphisms regarding two groups

Question

I am confused of what actually it means to find all the homomorphisms between lets say Zn to itself or any other group. I keep confusing it with the idea of finding all the automorphisms ( i know that any automorphism is determined by the image of 1, basically it maps a generator to a generator so the number of auto morphisms in Zn is gonna be phi(n) ) but for homomorphisms I need to know whats the process in order to find any homomorphism between two groups, what do i really need to know ? I have seen different cases while trying to find them, one where the map is onto, and the other where it's just "to", what's the difference and what properties i can use which can make it easier to find them, will properties of homomorphism differ if the groups were not cyclic?

Answer 1

Determining homomorphisms between two groups is not usually an easy thing: one very simple answer for finite groups is to map the generators "coherently". Easier said than done! But, luckily, to determine a homomorphism from $\mathbb{Z}_{n}$ to any other group $G$, we do not have to do much! We just have to map a generator coherently! $\mathbb{Z}_{n}$ is a cyclic group generated by $1_{n}$. Hence any homomorphism $\phi:\mathbb{Z}_{n} \to G$ is determined by $\phi(1_{n})$ (if you know where $1_{n}$ goes to, then you know where $1_{n}+1_{n}$ goes to, and so on!). Now, we know that $n \phi(1_{n})=\phi(0_{n})=e_{G}$. Hence $\phi(1_n)$ must have order dividing $n$. Pick any element of $G$ whose order divides $n$ (such an element exists, $e_{G}$) and call it $h$. Show that the map $\phi : \mathbb{Z}_{n} \to G$, $\phi([k])=g^k$ is a WELL DEFINED homomorphism of groups. If $G$ is finite, you can look for elements of order dividing gcd(n,|G|), but I will leave that for you to think about!