$p$ prime, Group of order $p^n$ is cyclic iff it is an abelian group having a unique subgroup of order $p$ I've just read from An introduction to the theory of groups from Rotman's the following theorem:

Theorem 2.19. Let $p$ be a prime. A group $G$ of order $p^n$ is cyclic
  if and only if it is an abelian group having a unique subgroup of order $p$.
Proof Necessity follows at once from Lemma 2.15. For the converse, let 
  $a \in G$ have largest order, say $p^k$ (it follows that $g^{p^k} = 1$ for
  all $g \in G$). Of course, the unique subgroup $H$ of order $p$ is a subgroup
  of $\langle a\rangle$. If $\langle a\rangle$ is a proper subgroup of $G$,
  then there is $x \in G$ with $x \not \in \langle a\rangle$ but with
  $x^p \in \langle a\rangle$; let $x^p = a^l$. If $k = 1$, then $x^p = 1$ and
  $x \in H \subset \langle a\rangle$, a contradiction; we may, therefore,
  assume that $k>1$. Now $$1 = x^{p^k}=(x^p)^{p^{k-1}}=a^{lp^{k-1}},$$
  so that $l = pm$ for some integer $m$, by Exercise 2.13. Hence,
  $x^p = a^{mp}$, and so $1 = x^{-p}a^{mp}$. Since $G$ is abelian,
  $x^{-p}a^{mp} = (x^{-1}a^m)^p$, and so
  $x^{-1}a^m \in H \subset \langle a\rangle$.
  This gives $x \in \langle a\rangle$, a contradiction. Therefore,
  $G = \langle a\rangle$ and hence is cyclic.

As usual, there is some trivial part of the proof which I don't understand:
I don't see why "If $\langle a\rangle$ is a proper subgroup of $G$,
then there is $x \in G$ with $x \not \in \langle a\rangle$ but with
$x^p \in \langle a\rangle$", I understand the rest of the proof.
I would appreciate if someone could explain to me that statement.
 A: For someone who knows about quotient groups and Cauchy's theorem, the existence of $x$ can be explained in a simple way that bypasses mysterious and delicate group theory calculations.
Let's express the conditions $x \not\in \langle a\rangle$ and $x^p \in \langle a\rangle$ in terms of the group $G/\langle a\rangle$ (which makes sense since $G$ is abelian). Those conditions are saying in $G/\langle a\rangle$ that $\overline{x}$ is nontrivial and the $p$th power of $\overline{x}$ is trivial, i.e., $\overline{x}$ has order $p$ in $G/\langle a\rangle$. So all we need to do is appeal to Cauchy's theorem for the group $G/\langle a\rangle$: assuming $\langle a\rangle$ is not $G$, the group $G/\langle a\rangle$ is a nontrivial $p$-group, so it has an element $\overline{x}$ of order $p$ and that gives you the desired element: $x \in G$, $x \not\in \langle a\rangle$, and $x^p \in \langle a\rangle$.
The theorem you cite from Rotman is in Chapter 2, which comes before quotient groups and probably Cauchy's theorem as well.  I still think the argument above is a much simpler way to understand why there is an $x$ of the desired kind.
A: Clearly, $\langle a\rangle$ being proper means there is $x\in G\setminus \langle a\rangle$. Now, there is $p^n$ such that $x^{p^n}=1\in \langle a \rangle$. In particular, there is $n$ such that $x^{p^{n-1}}\notin \langle a\rangle $ but $x^{p^n}\in \langle a\rangle$. Indeed, if $x^p\in \langle a\rangle$, we're done. Else $x^p \notin \langle a \rangle$. Now take $x'=x^p$. If $x'^p= x^{p^2}\in \langle a\rangle $ we're done. Eventually we reach $n$ such that $x^{p^n}=1$, so the process furnishes some "good" $x^{(n)}$ for some $n\geqslant 0$. 
