Introduction to group theory - proof of equivalency of a group $G$ being cyclic and some property of order of $G$ The following is from Czesław Bagiński, "Wstęp do teorii grup". The original is in Polish, the translation is mine.
Remark: "the previous claim" in the last sentence of the quoted proof refers to the claim before theorem 5.10. I don't quote it because it is not relevant for the question.

Theorem 5.10. A finite abelian group of order $n$ is a cyclic group if and only if for every divisor $d$ of the number $n$ there exists exactly $d$ elements satisfying the condition $x^d = e$. 
Proof. The implication "=>" is the the theorem 4.4. Let us assume that for every $d$ dividing $n$ there exists exactly $d$ elements whose orders are divisors of $d$. Let $G_d$ be a set of this elements, that is $G_d = \{x \in G: x^d = e\}$. The commutativity of $G$ implies that the subset $G_d$ is a subgroup with $d$ elements. If $G = |p|$ is a prime number then the theorem is obvious. Let us assume that $|G|$ is not a prime and that the theorem is true for groups of the order lesser than $|G|$. If $|G| = p^k$ for some prime $p$ then by inductive hypothesis $h = G_{p^{k-1}}$ is the only cyclic subgroup of the order $p^{k-1}$. Additionally for any $x \in G \setminus H$, $x^{p^{k-1}} \neq e$. At the same time $o(x)|p^k$. Thus $o(x) = p^k$ and by this $G$ is cyclic. If $|G|$ is not a power of a prime then by the fundamental theorem of arithmetic $|G| = rs$ where $r$ and $s$ are relatively prime and $r, s > 1$. By induction hypothesis $G_s$ and $G_r$ are cyclic subgroups of the group $G$ of orders equivalently $s$ and $r$. If $ x = G_s$ and $y = G_r$ then by commutitivity of $G$ and the previous claim $xy$ is an element of the order $rs$ which proves that $G$ is cyclic.

How does the fact that $o(x) = p^k$ imply that $G$ is cyclic?
How does the fact that $xy$ is an elmeent of the order $rs$ proves that $G$ is cyclic?
 A: Thanks for all who commented, I write the full (the full from my point of view) answer for reference in case it is needed in the future.
If we have a finite group $G$ of order $n$ and $o(x) = n$ then $G$ is cyclic because this element raised to 1-th, 2-nd, 3-rd, ..., n-th power must yield all the elements contained in $G$. For if it was not the case, we would have the situation in which at least two elements in such genenerated sequence would be duplicates - but it would mean that generating some initial fragment of this sequence we would obtain $\{x, x^2, ..., x^k, u\}$ where the elements from the set $\{x, x^2, ..., x^k\}$ are different, $k < n$, and $u$ would be equal to one of the previous values and the said sequence (that is $\{x, x^2, ..., x^k, u\}$) is not $\{x, x^2, ..., x^{n-1}, e\}$. Now, there are following possibilities:

*

*$u \neq e$. In this case there is no $e$ in the whole sequence of the length $n$ (because raising $x$ to subsequent powers would just yield the already obtained elements). Contradiction because $o(x) = n$ implies that $x^n = e$ (can also be justified in other ways).

*$u = e$. If $k < n - 1$ it would mean that $o(x) < n$ which would contradict the assumption. The case $k = n - 1$ is excluded by the disclaimer.

