Prime Power Group has Non-Trivial Proper Normal Subgroup

Theorem
Let $$G$$ be a group, whose identity is $$e$$, such that $$\left|{G}\right| = p^n: n > 1, p \in \mathbb{P}$$.

Then $$G$$ has a proper subgroup which is non-trivial.

Proof

 * From Center is a Normal Subgroup, $$Z \left({G}\right) \triangleleft G$$ and, by Center of Group of Prime Power Order is Non-Trivial, is non-trivial.

If $$Z \left({G}\right)$$ is a proper subgroup, the job is done.


 * Otherwise, $$Z \left({G}\right) = G$$.

Then $$G$$ is abelian by Center of Abelian Group is Whole Group.

However, then any $$a \in G: a \ne e$$ generates a non-trivial normal subgroup $$\left \langle {a} \right \rangle$$, as All Subgroups of Abelian Group are Normal.

If $$\left|{a}\right| = \left|{\left \langle {a} \right \rangle}\right| < \left|{G}\right|$$, the job is done.


 * Otherwise, $$\left|{a}\right| = \left|{G}\right| = p^n$$.

Then $$\left|{a^p}\right| = p^{n-1}$$ from Subgroup of Cyclic Group whose Order Divisor, so $$a^p$$ generates that non-trivial normal subgroup.