Classification of groups of prime-square order

Statement

Let $p$ be a prime number. Then there are, up to isomorphism of groups, only two groups of order $p^2$:

• cyclic group of prime-square order, i.e., the cyclic group of order $p^2$, denoted $C_{p^2}$ or ${\mathbb{Z}}_{p^2}$.
• elementary abelian group of prime-square order, i.e., the elementary abelian group of order $p^2$, denoted $E_{p^2}$ and equal to ${\mathbb{Z}}_p\times {\mathbb{Z}}_p$. In the case $p=2$, this is more commonly called the Klein four-group.

Facts used

1. Prime power order implies not centerless
2. Cyclic over central implies abelian
3. Lagrange's theorem
4. Structure theorem for finitely generated abelian groups

Proof

The proof has two parts. First, we show that any group of order $p^2$ must be an abelian group. Then, we use the structure theorem for finitely generated abelian groups to argue that there are only two possibilities.

Proof that the group must be abelian

Given: A prime number $p$, a group $P$ of order $p^2$.

To prove: $P$ is abelian.

Proof: Let $Z$ be the center of $P$.

1. $Z$ is nontrivial: This follows from fact (1).
2. The order of $Z$ cannot be $p$: [SHOW MORE]
   If $Z$ has order $p$, then $P/Z$ also has order $p$. Thus, $P/Z$ is a group of order $p$ as well (by fact (3)) and hence a cyclic group. By fact (2), this forces $P$ to be an abelian group, in which case the order of $Z$ would have been $p^2$.
3. The order of $Z$ must be $p^2$: [SHOW MORE]
   By fact (3) (Lagrange's theorem), the order of $Z$ is one of the values $1,p,p^2$. The previous two steps rule out $1$ and $p$, leaving the only possibility that the order of $Z$ is $p^2$.
4. $Z=P$, so $P$ is abelian: This follows directly from the previous step.

Classification of abelian groups

This follows directly from fact (4) -- the group must be a direct product of cyclic groups, giving precisely the two possibilities mentioned.