An application of Sylow theorems in p-groups!

Question

If $G$ is a finite group of order $p^{n}$ (which $p$ is a prime number) and have only one subgroup of order $p^{n-1}$ ,namely $H$ ,then $G$ is cyclic !
My "proof" is as follows:
suppose $$x\in G-H$$ then
$$|\langle x\rangle|=p^m,m=0,1,2,3,...,n$$ now suppose that $m<n$ .according to first theorem of Sylow there is a subgroup $K$ of order $p^{n-1}$ such that $x\in K$ and with the above assumption we must have $K=H$ which follows that $x\in H$ and contradics to the assumption $x\in G-H$ .this proves the result!
Help me prove that there is such a $K$ with the above properties !
Thanks !

Answer 1

Suppose $G$ is a finite $p$-group. By Burnside's basis theorem, $G$ is cyclic iff $G/\Phi(G)$ is cyclic. But if $H$ is the unique subgroup of index $p$; $H=\Phi(G)$, so $G/H=C_p$ is cyclic, and so is $G$.

(Here $\Phi(G)$ denotes the Frattini subgroup of $G$)

Answer 2

Let $G$ be a counterexample of minimal order. Since $G$ is a $p$-group, its center is non-trivial. Therefore $\overline{G}:=G/Z(G)$ has smaller order than $G$ and also has a unique maximal subgroup $\overline{H}$, and so by the minimal choice of $G$, $\overline{G}$ must be cyclic. Now there exists a very popular exercise stating that under these conditions, $G$ must be abelian. But now the classification theorem for abelian groups gives that $G$ must be cyclic, contradicting the choice of $G$ as a counterexample.