I have been working through Artin's algebra book. Here's a simple exercise, but I want to make sure I am not missing anything important.
Let $(G,\cdot)$ be a group and let $x,y \in G$ with orders $m,n$ respectively. Now let $H_1 = \left<x\right>$ and $H_2 = \left<y \right>$. Now the exercise asks for the conditions on $m$, $n$, such that
$$ \phi: H_1 \to H_2, x^i \mapsto y^i $$
is a group homomorphism.
If I see it correctly, the only thing I need to ensure for this exercise, is that $\phi$ is well-defined.
In particular, in this case it holds that $\phi(x^kx^l)= \phi(x^{k+l}) = y^{k+l}$ and $\phi(x^k)\phi(x^l) = y^k y^l$. Because $y^{k+l} = y^k y^l$ it thus follows that $\phi(x^kx^l) = \phi(x^k)\phi(x^l)$.
Finally, let's ensure that $\phi$ is well defined.
It then holds that, if $x^i = x^l$, then $y^i = y^l$. But this in turn means that if $m \; \big\vert \; l-i$ then it also holds that $n \; \big\vert \; l-i$. Choosing $l-i=m$ we get that $n \big\vert m$. This is a necessary condition. Similarly we can check that it is also a sufficient condition.