Let $G$ be a finite abelian $p$-group with a unique subgroup $H$ of index $p$. It is a fact that $G$ is cyclic. This can be deduced from the classification theorem for finite abelian groups by writing $G$ as a product of cyclic groups and noting that $G$ does not have a unique subgroup of index $p$ unless there is only one such group. I am wondering if there is a nice proof of this result which does not use the classification theorem. Does anyone know of one?
If $G$ is finite then every proper subgroup is contained in a maximal subgroup, so every proper subgroup is contained in $H$. I claim therefore that $\langle x \rangle = G$ for any $x\in G\setminus H$ - why?
If $\langle x \rangle$ were proper in $G$, then we would have $\langle x \rangle \leqslant H$, a contradiction. So $\langle x \rangle=G$.
Note that this argument does not require that $G$ is abelian. In fact, we could even extend your hypothesis to "$G$ is a finite group with a unique maximal subgroup."
Hint: It suffices to show that $G$ has precisely one subgroup of order $m$ for each $m$ dividing $|G|$. Show by induction that if we get more than one subgroup of order $p^k$ with $p^k\mid |G|$ then we can lift this to two distinct index $p$ subgroups. Thus, $G$ will have precisely one subgroup of order $p^k$ for each $p^k\mid|G|$ and thus $G$ will be cyclic.
In fact the assumptions can be weakened and the conclusion will remain (almost) the same:
Let $G$ be a group of order $p^n$ for some prime $p$ and assume that $G$ contains a unique subgroup of order $p^k$ for some $1\leq k \leq n-1$.
Then either $G$ is cyclic or $p = 2$, $k = 1$ and $G$ is generalized quaternion (so $n\geq 3$).
For a proof of this, one can see Berkovich's Groups of Prime Power Order (it is Proposition 1.3).