If $G$ has no proper subgroup, then $G$ is cyclic of prime order

Solution 1:

I'll give the proof of it without using Cauchy's theorem:

1. $G$ is cyclic If not, the generating set of $G$ must have at least two elements. Let $a$ and $b$ be elements which satisfy $a\notin \langle b\rangle$ and $b\notin \langle a\rangle$, then $\langle a\rangle$ and $\langle b\rangle$ are distinct proper subgroups of $G$, a contradiction.

2. $G$ has a prime order If $G$ has infinite order, then it has proper subgroup (consider $2\Bbb{Z}$ in $\Bbb{Z}$.) so $G$ is finite. If $|G|=rs$ with $r,s>1$ and $G=\langle a\rangle$, then $\langle a^s\rangle$ is a proper subgroup of $G$.

Solution 2:

Here is a simple proof via Cauchy's theorem, proceeding by contrapositive.

Let $G$ be a group of composite order $n = pk$, $p$ prime. By Cauchy's theorem, $G$ has an element of order $p$, hence a cyclic proper subgroup of order $p$.

Solution 3:

I'm answering a 5-year old question because I had questions on this one. If G has no nontrivial subgroups then any of it's elements would generate it. If the order were composite, say $o(G)=nm \Rightarrow $ pick $a\neq e$ so that $a^{nm}=e=(\underbrace{a^{n}}_{b})^{m}=b^m\Rightarrow o(b)\leq m$.

Either $a^{n}=e$ or $b^{m}=e$. In either case, $G$ would have a nontrivial proper subgroup.