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.

For detailed information comparing the two groups side by side, see groups of prime-square order

Related facts

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

Facts used

The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.

Fact no.	Statement	Steps in the proof where it is used	Qualitative description of how it is used	What does it rely on?	Difficulty level	Other applications
1	Prime power order implies not centerless	Step (1)	Eliminates the case of trivial center	class equation of a group	3	click here
2	Center is normal	Step (2)	Justifies taking the quotient by the center		0	click here
3	Cyclic over central implies abelian	Step (2)	Rules out the possibility of a center of order $p$, since the quotient would then be forced to be cyclic			click here
4	Lagrange's theorem	Steps (2), (3)	Narrows down possibilities for order of center, and allows for computation of order of quotient group		2	click here
5	Equivalence of definitions of group of prime order	Step (2)	Show cyclicity of $P/Z$ if $Z$ has order $p$			click here
6	Structure theorem for finitely generated abelian groups	Classifying the abelian groups				click here

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

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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 $Z = P$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$Z$ is nontrivial	Fact (1)	$P$ is nontrivial and its order is a power of a prime.	--	--
2	The order of $Z$ cannot be $p$	Facts (2), (3), (4), (5)	$P$ has order $p^2$	--	[SHOW MORE] $P/Z$ exists by Fact (2). $P/Z$ has order $|P|/|Z|$ by fact (4). If $Z$ has order $p$, the order of $P/Z$ becomes $p^2/p = p$. Hence, $P/Z$ has order $p$. By fact (5), it must therefore be a cyclic group. By fact (3), 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 $p^2$	Fact (4)	$P$ has order $p^2$	Steps (1), (2)	[SHOW MORE] By fact (4) (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			Step (3)	Follows directly.

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.