Prime Power Group has Non-Trivial Proper Normal Subgroup

Theorem

Let $G$ be a group, whose identity is $e$, such that $\order G = p^n: n > 1, p \in \mathbb P$.

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

Proof

From Center of Group is Normal Subgroup, $\map Z G \lhd G$.

By Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial.

If $\map Z G$ is a proper subgroup, the proof is finished.

Otherwise, $\map Z G = G$.

Then $G$ is abelian by definition.

However, then any $a \in G: a \ne e$ generates a non-trivial normal subgroup $\gen a$, as Subgroup of Abelian Group is Normal.

If $\order a = \order {\gen a} < \order G$, the proof is complete.

Otherwise:

$\order a = \order G = p^n$

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

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Conjugacy, Normal Subgroups, and Quotient Groups: $\S 52$ Lemma $2$