Task
Let $G$ be a group and $g \in G$ an element with a finite order. Show that the >order of $g$ is identical with the minimal $d \in \Bbb N_+$ for which there is >a group homomorphism
$$\Bbb Z / (d) \rightarrow G,$$
where $g$ is a part of its image.
Solution
Assume $g$ is an element of the image of a group homomorphism
$$\Bbb Z / (d) \rightarrow G.$$
Then, $g$ is an element of a sub group with an order $\le d$. Applying the theorem of Lagrange, we can conclude that the order of $g$ is also $\le d$. Therefore, the order is at most as big as the minimum of the natural numbers $d$ for which there is such a group homomorphism.
Now, on the other hand, assume $d$ is the order of $g$. The canonical group homomorphism $\Bbb Z \rightarrow G$, $n \mapsto g^n$, has the kernel $\Bbb Z d$. Applying the theorem of the canonical homomorphism, this group homomorphism induces a group homomorphism $\Bbb Z / (d) \rightarrow G$, and g is part of its image.
Questions
I have my issues with understanding the proof in general. What is he actually showing here, what is the underlying structure of this proof? It reads like the author proves an iff-statement, but that is not the structure of the task.
Let $g\in G$, and let $n$ be the order of $g$. Let $D_g$ be the set of positive integers $d$ such that there is a group homomorphism $\Bbb Z/(d)\to G$ with $g$ in its range.
The first part of the proof shows that $n\le d$ for each $d\in D_g$, so $n\le\min D_g$.
The second part of the proof shows that $n\in D_g$.
These two statements together say precisely that $n=\min D_g$: in words, the order of $g$ is the smallest positive integer $k$ such that $g$ is in the range of some homomorphism from $\Bbb Z/(d)$ to $G$.