Center of Group of Order Prime Cubed

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.

$\blacksquare$

This article is complete as far as it goes, but it could do with expansion. In particular: This results generalizes naturally to a group of order $p^n$: its center is not of order $p^{n-1}$. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Expand}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.