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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?

Hannah_Zak
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  • Have you looked at either of these answers? They are very similar questions to yours, and knowing why they don't answer your question will help us help you ^_^ – HallaSurvivor Jan 10 '20 at 03:21
  • Thanks for the links, but I saw this question already didnt understand a thing in it, the answers given were a bit complex for me, because i dont really know how to use relations to find homomorphisms, i was looking for something simpler, like using properties of homomorphisms, and trying to understand better about homomorphic images? Like for instance i read a question that said Z6 and Z8 are homomorphic images of a group i don't remember what group but, anyways i didn't understand what to make out of it and what's the relation between homomorphic images and homomorphisms itself ? – Hannah_Zak Jan 10 '20 at 05:56

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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!

mich95
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  • Thank you for explaining that to me, i actually had a very particular question in my mind, which i remembered after reading ur answer, which is since generator maps to a generator, from Z12 to Z30, taking 1 to be a generator of z12 then shouldn't its image be a generator of z30 also,finding all the possible generators in z30 gives me different homomorphisms basically different functions, but since i am not sure if the given homomorphism is onto,i can't use this method right? Only when it says its onto,i can determine the homomorphisms just by guessing the possibilities of generators – Hannah_Zak Jan 10 '20 at 05:47
  • @HaneenHussam, the answer explains that the generator of $\mathbb{Z}{12}$ doesn't need to map to a generator of $\mathbb{Z}{30}$; it maps to an element of $\mathbb{Z}_{30}$ whose order divides $12$. – ancient mathematician Jan 10 '20 at 08:00