Group of Order Prime Squared is Abelian

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 $Z \left({G}\right)$ be the center of $G$.

By Lagrange's Theorem, $\left\vert{Z \left({G}\right)}\right\vert \backslash \left\vert{G}\right\vert$, so $\left\vert{Z \left({G}\right)}\right\vert = 1, p$ or $p^2$.

By Center of Group of Prime Power Order is Non-Trivial, $\left\vert{Z \left({G}\right)}\right\vert \ne 1$.

Now suppose $\left\vert{Z \left({G}\right)}\right\vert = p$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{G / Z \left({G}\right)}\right\vert\)	\(=\)	\(\displaystyle \left\vert{G}\right\vert / \left\vert{Z \left({G}\right)}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of quotient group
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle p^2 / p\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle p\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $G / Z \left({G}\right)$ is non-trivial, and of prime order and therefore cyclic.

But from Cyclic Quotient Group of Center, it can not be.

Therefore $\left\vert{Z \left({G}\right)}\right\vert = p^2$ and therefore $Z \left({G}\right) = G$.

Therefore $G$ is abelian.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 51 \beta$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 51.2$
• John F. Humphreys: A Course in Group Theory (1996): $\S 10$: Corollary $10.22$