Group of Order Prime Squared is Abelian

Theorem

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

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 $p^2$.

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

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

Then:

$\ds \order {\map Z G}$

$=$

$\ds \index G {\map Z G}$

Definition of Index of Subgroup

$\ds $

$=$

$\ds \order G / \order {\map Z G}$

$\ds $

$=$

$\ds p^2 / p$

$\ds $

$=$

$\ds p$

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.

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

Therefore $G$ is abelian.

$\blacksquare$

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 $10.22$