Suppose $A$ is a set of all group homomorphisms such that $f: \mathbb{Z_4} \rightarrow \mathbb{Z_6}$. To find the cardinality of $A$, can I use The First Isomorphism Theorem?
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The way I would approach this is to use the fact that $\mathbb{Z}_4$ is a cyclic group so where ever you send the generator will determine the homomorphism. Since $1$ is a generator for $\mathbb{Z}_4$ we can just send that somewhere in $\mathbb{Z}_6$, but we can't send it anywhere we must send it to an element with order $4$ (or dividing $4$). Thus, you must now count the elements in $\mathbb{Z}_6$ that when you add them to themselves $4$ times you get $0$. That will give you $0$ and $3$, so there are exactly $2$ homomorphisms determined by $\phi(1)=0$ and $\phi(1)=3$.
Steven Creech
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Thanks, I am aware of such a solution. However, I was looking for a general approach. – Fazil Safarov Aug 04 '23 at 17:29