Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p
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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 $21$