Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p

Theorem

Let $p$ be a prime number.

Let $G$ be a non-abelian group of order $p^3$.

Then $G$ contains exactly one normal subgroup of order $p$.

Proof

From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is not the trivial subgroup.

From Quotient of Group by Center Cyclic implies Abelian, $G / \map G Z$ cannot be cyclic and non-trivial.

Thus $\order {G / \map G Z}$ cannot be $p$ and so must be $p^2$.

Thus $\order {\map G Z} = p$.

Let $N$ be a normal subgroup of $G$ of order $p$.

Then from Normal Subgroup of p-Group of Order p is Subset of Center:

$N \subseteq \map G Z$

Thus there is no normal subgroup of order $p$ different from $\map G Z$.

Hence the result.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$