Show that Hom$_\mathbb{Z}(\mathbb{Z}/n\mathbb{Z}) \cong A_n$.
Where $A_n:=${$a\in A\ |\ na = 0$} (The annihilator of ($n$) in $A$)
This question has previously been answered.
Prove Hom$_{\mathbb{Z}}(\mathbb{Z}/n \mathbb{Z},A) \cong A_n$
But I have still have some doubts that I want to ask.
Previous part of the question I proved the following:
$\phi_a:\mathbb{Z}/n\mathbb{Z} \rightarrow A$ where $\phi_a (\bar{k}) = ka, a\in A$
$\phi_a$ is well-defined homomorphism iff $na = 0$
So naturally for the second part, I would do something like $\psi(\phi_a) = a$
(exactly like what he did in his original post)
My question is:
If thats the case, ain't we assuming that all homomorphism from $\mathbb{Z}/n\mathbb{Z} \rightarrow A$ "looks" like $\phi$ ?
If yes I would be grateful if some explanation can be provided, if not what should be the correct isomorphism ?
If by the "look like" you mean that $\Phi$ is uniquely determined by $\Phi(1)$, then yes - every homomorphism from a cyclic group/module is fully determined by the image of $\overline{1}$.