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.

Note that both of these are abelian groups, so in particular we see that any prime-square is an abelianness-forcing number.

Similar classifications

 * Classification of Lie rings of prime-square order
 * Classification of connected unipotent two-dimensional algebraic groups over an algebraically closed field
 * Classification of groups of prime-cube order
 * Classification of nilpotent Lie rings of prime-cube order

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 $$P$$. Our goal is to show that $$Z = P$$.

Classification of abelian groups
This follows directly from fact (6) -- the group must be a direct product of cyclic groups, giving precisely the two possibilities mentioned.