Center of Group of Order Prime Cubed
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Theorem
Let $G$ be a group of order $p^3$, where $p$ is a prime.
Let $\map Z G$ be the center of $G$.
Then $\order {\map Z G} \ne p^2$.
Proof
Aiming for a contradiction, suppose $\order {\map Z G} = p^2$.
Then $\order {G / \map Z G} = p$.
From Prime Group is Cyclic it follows that $G / \map Z G$ is cyclic.
From Quotient of Group by Center Cyclic implies Abelian it follows that $G$ is abelian.
by definition of abelian group it follows that $\order {\map Z G} = p^3$.
The result follows by Proof by Contradiction.
$\blacksquare$
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