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 normal subgroup which is non-trivial.

Proof
From Center is a Normal Subgroup, $Z \left({G}\right) \triangleleft G$.

By Center of Group of Prime Power Order is Non-Trivial, $Z \left({G}\right)$ is non-trivial.

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

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 proof is complete.

Otherwise:L
 * $\left|{a}\right| = \left|{G}\right| = p^n$

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