Exercise: How many actions are there from $ \mathbb {Z}_n $ to $ \mathbb {Z}_2 $ with $ n \geq 2$ ?
My idea is as follows: Let $\psi:\mathbb{Z}_n \to S(\mathbb{Z}_2)\cong S_2$ be a homomorphism. Then since $ \mathbb {Z} _n $ is abelian, it follows that $\ker(\psi)=K/\langle n\rangle $ with $K\leq \mathbb{Z}$ such that $\langle n\rangle \; \subseteq K.$ Then $\ker(\psi)=\langle m\rangle/\langle n\rangle$ con $m \mid n.$ Note that, $$|\mathbb{Z}_n/\ker(\psi)|=|\mathbb{Z}/\langle n\rangle/\ker(\psi)|=|\mathbb{Z}/\langle m\rangle|=|\mathbb{Z}_m|=m.$$
Hence, since $\mathbb{Z}_n/\ker(\psi)\leq S_2$ then $m=1$ or $m=2.$ In this way, if $ n $ is odd there is only one action of $\mathbb{Z}_n$ in $\mathbb{Z}_2$ and if $ n $ is even there are $2$ actions.
Is this correct? Thanks so much for reading.