# Group of Order Prime Squared is Abelian

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## Theorem

A group whose order is the square of a prime is abelian.

## Proof

Let $G$ be a group of order $p^2$, where $p$ is prime.

Let $\map Z G$ be the center of $G$.

By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.

- $\order {\map Z G} \divides \order G$

It follows that $\order {\map Z G} = 1, p$ or $p^2$.

By Center of Group of Prime Power Order is Non-Trivial:

- $\order {\map Z G} \ne 1$

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

Then:

\(\displaystyle \order {\map Z G}\) | \(=\) | \(\displaystyle \index G {\map Z G}\) | Definition of Index of Subgroup | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \order G / \order {\map Z G}\) | Lagrange's Theorem | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle p^2 / p\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle p\) |

So $G / \map Z G$ is non-trivial, and of prime order.

From Prime Group is Cyclic, $G / \map Z G$ is a cyclic group.

But by Quotient of Group by Center Cyclic implies Abelian, that cannot be the case.

Therefore $\order {\map Z G} = p^2$ and therefore $\map Z G = G$.

Therefore $G$ is abelian.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51 \beta$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 51.2$ The quotient group $G / Z$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Corollary $10.22$