Prime Group has no Proper Subgroups

Theorem
A nontrivial group $G$ has no proper subgroups except the trivial group iff $G$ is finite and its order is prime.

Sufficient Condition
Suppose $G$ is finite and of prime order $p$.

Then from Lagrange's Theorem the order of any subgroup of $G$ must divide the order $p$ of $G$.

From the definition of prime, any subgroups of $p$ can therefore only have order $1$ or $p$.

Hence $G$ can have only itself and the trivial group as subgroups.

Necessary Condition
Now suppose $G$ is not finite and prime.

Take any element $h \in G, h \ne e$. Then $H = \left \langle {h} \right \rangle$ is a cyclic subgroup of $G$.

If $H \ne G$ then $H$ is a non-trivial proper subgroup of $G$, and the proof is done.

Otherwise, $H = G$ is a cyclic group and there are two possibilities:
 * $(1): \quad G$ is infinite
 * $(2): \quad G$ is finite (and of non-prime order).

First, suppose $G$ is infinite.

Then $G$ is isomorphic to $\left({\Z, +}\right)$ from Infinite Cyclic Group Isomorphic to Integers.

Now, from Subgroups of Additive Group of Integers, $\left({\Z, +}\right)$ has proper subgroups, for example $\left \langle {2} \right \rangle$.

Because $G \cong \left({\Z, +}\right)$, then so does $G$ have proper subgroups, and the proof is complete.

Now suppose $G$ is finite, and of an order $n$ where $n$ is not prime.

Then $\exists d \in \N: d \mathop \backslash n, 1 < d < n$.

From Subgroup of Cyclic Group Determined by Order, $G$ has a proper subgroup of order $d$ and again, the proof is complete.