# Abelian group of order $p^m$ having only one subgroup of order $p^{m-1}$ [duplicate]

Let $G$ be an abelian group such that $|G|=p^m$ for some $m\geq 1$ where $p$ is a prime. Assume that $G$ has only one subgroup of order $p^{m-1}$.

How to prove that $G$ is cyclic ?

## marked as duplicate by Watson, Derek Holt, Claude Leibovici, user91500, meselNov 8 '16 at 9:53

I think if we take some $x\in G-H$ where $|H|=p^{n-1}$ then nesserily $G=<x>$. Lagrange Theo. admits that $|x|=p^{m}$ for some $m$. Of coures $1\leq m\leq n$. If $m<n$ then there is a subgroup say $K$ such that $|K|=p^{n-1}$ and $x\in K$. As we assumed alreay this $K$ should be the same as $H$. So $x\in H$. A nice Contradiction! So $|x|=p^{n}$. I assumed $|G|=p^n$ than your notation.
• May I know why "If $m<n$ then there is a subgroup say $K$ such that $|K|=p^{n-1}$ and $x\in K$." – Alan Wang Nov 8 '16 at 7:31
• Where do you use the assumption that $G$ is abelian? – Chas Brown Nov 8 '16 at 9:00
• @AlanWang Take $K = \langle x \rangle$, by assumption there is only one subgroup of $G$ and has it has order $p^{n-1}$, therefore $|K|=p^{n-1}$ – Bysshed Nov 8 '16 at 9:48