How do I show that $G$ of order $p^n$ is cyclic if and only if it is abelian with a unique subgroup of order $p$? 
I spent quite some time thinking about it but I don't see why must be $x^p\in\left\langle a\right\rangle$ in the given conditions. Can someone please explain?
 A: Assume, by contradiction, that for every $x \in G$ with $x \notin \langle a\rangle$, we have $x^p\notin\langle a\rangle$. Then $x^{p^2}=(x^p)^p \notin \langle a\rangle$ because $x^p \in G$ and $x^p\notin\langle a\rangle$. And also $x^{p^3}\notin \langle a\rangle$, for the same reason, and so on and so on. In particular, $1 = x^{p^k} \notin \langle a \rangle$, which is a contradiction. Therefore, there should exist at least one $x \in G\setminus\langle a\rangle$ such that $x^p\in\langle a\rangle$.
A: Since $H$ is unique and $H \subset \left< a \right>$, any $z \notin \left< a \right> $ has order greater than $p$ (otherwise $ \left< z \right>$ would be another subgroup of order $p$). So order of $z = p^m$ where $1 \lt m \le k$. Let $x = z^{p^{m - 1}}$
A: Notice that the proof does not say any $x$ will satisfy this, just that there is some. Because all $x$ satisfy $x^{p^n}=1$, you can see what you want by iteration: if $x\not \in\langle a \rangle$ and $x^p\in \langle a\rangle$, you are done, else substitute $x$ by $x^p$. At some point $x^p$ will be $1$ and hence inside $\langle a \rangle$.
