Prime Group has no Proper Subgroups

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

Proof

 * Suppose $$G$$ is finite and of prime order $$p$$.

Then from Lagrange's Theorem any subgroups must divide $$p$$.

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.


 * 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) $$G$$ is infinite;
 * 2) $$G$$ is finite (and of non-prime order).

First, suppose $$G$$ is infinite.

Then $$G$$ is isomorphic to $$\left({\Z, +}\right)$$ from Cyclic Groups Same Order Isomorphic.

Now from [work in progress] $$\left({\Z, +}\right)$$ 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 done.

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

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

From Subgroup of Cyclic Group whose Order Divisor, $$G$$ has a proper subgroup of order $$d$$ and again, the proof is done.