Order of a cyclic group with a single proper subgroup of order 7

Question

Let $G$ be a cyclic group with its only proper subgroup of order 7. Find out the order of the group.

let the subgroup of order 7 be denoted by H. since 7 is prime H is cyclic. Now if G = then H= for some r in Z. There can be two cases 1. x belongs to H 2. x dose not belong to H.Now if x belongs to H then x = (x$^r$)$^k$ where k<7 ....I need some help here.

Answer 1

If $p$ is a prime divisor of $|G|$ then by Cauchy theorem there is a subgroup $H$ of order $p$.

So in your case $|G| = 7^{n}$, $n \geq 1.$

But $G$ is cyclic, so for every divisor d of $|G|$ there is an unique subgroup of order d $\Rightarrow |G| = 7^{2}$.

Answer 2

$G$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ and the subgroups of $G$ are in bijection with the divisors of $n$. Therefore $n$ has only $7$ as a proper divisor. What is the only value of $n$ for this to happen?