$\quad$ Finding the number of homomorphisms from group $G$ to a group $H$, if both are cyclic and have same order, then we have to map the generators of both the groups, this leaves us the possibility and the number of possible elements is the number of homomorphisms. What if one of them is not cyclic? or not even abelian? then of course the element $1$, and its image $a$ ($f(1)=a$) the order of $a$ should divide both the order of $H$ and order of $1$. This leaves the possible elements in $H$ one of them to be $a$, resulting in number of homomorphisms.

I have few questions

  1. At First, am I right in grasping the homomorphism concept?
  2. We all know that each homomorphism helps in creating Normal Subgroup (Kernel), so counting the number of normal subgroups of each order, can we get the number of homomorphisms?.

I read the articles in mathstack regarding the group homomorphisms but still I cant catch the monkey...


I think you are on the right track for dealing with homomorphisms. The answer to your second question is easy. In a word, no! Counting normal subgroups won't be enough. Since normal subgroups will completely be determined by the groups $G$ (the domain of your homomorphism) without even considering what the target group is! So if that were the case then the number of group homomorphisms from $G$ to any other group $H$ would be same in number and that is not the case. You can verify this using very simple examples, for example start with a cyclic group and take different abelian groups as the range.

The number of groups homomorphism from $G$ to $H$ is not that difficult when you are dealing with abelian groups. You can easily do that once you understand how group homomorphisms work (and once you now Fundamental Theorem of Finitely generated Abelian Groups). But I think it is a difficult question in general when you start dealing with non-abelian groups.


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