Prove that $G$ is cyclic group of prime order

Question

I'm at the beggining of a Group Theory course, and I'm trying to solve this problem:

Let $G = 1$ (being $1=\{e\}$, $e$ the neutral element in $G$) be a group containing no subgroup different from $1$ and $G$. Prove that $G$ is cyclic of prime order.

I've consider using that, given $g\in G$, $n\in\mathbb{N}$, $r=|g|$, I can conclude that $$g^n=e \Leftrightarrow r|n \text{ ($r$ divides $n$)},$$ but I'm not sure if it's useful without proving first that $G$ is cyclic, and I'm not sure how to do it. Any help will be appreciated.

Answer 1

1. If $G$ is non-cyclic, then any non-identity element generates a non-trivial, proper (cyclic) subgroup

2. If $G$ is cyclic with $|G| = n$ non-prime, a generator of $G$ raised to the power of a non-trivial divisor of $n$ generates a proper, non-trivial (cyclic) subgroup