Classification of groups of prime-cube order

Statement

Let $p$ be a prime number. Then there are, up to isomorphism, five groups of order $p^{3}$ . These include three abelian groups and two non-abelian groups. The nature of the two non-abelian groups is somewhat different for the case $p = 2$ .

For more information on side-by-side comparison of the groups for odd primes, see groups of prime-cube order. For information for the prime 2, see groups of order 8

The three abelian groups

The three abelian groups correspond to the three partitions of 3:

Partition of 3	Corresponding abelian group	GAP ID among groups of order $p^{3}$
3	cyclic group of prime-cube order, denoted $C_{p^{3}}$ or $Z_{p^{3}}$ , or $Z / p^{3} Z$	1
2 + 1	direct product of cyclic group of prime-square order and cyclic group of prime order, denoted $C_{p^{2}} \times C_{p}$ or $Z_{p^{2}} \times Z_{p}$	2
1 + 1 + 1	elementary abelian group of prime-cube order, denoted $E_{p^{3}}$ , or $C_{p} \times C_{p} \times C_{p}$ , or $Z_{p} \times Z_{p} \times Z_{p}$	5

The two non-abelian groups

For the case $p = 2$ , these are dihedral group:D8 (GAP ID: (8,3)) and quaternion group (GAP ID: (8,4)).

For the case of odd $p$ , these are prime-cube order group:U(3,p) (GAP ID: ( $p^{3}$ ,3)) and semidirect product of cyclic group of prime-square order and cyclic group of prime order (GAP ID: ( $p^{3}$ ,4)).

Facts used

Prime power order implies not centerless
Center is normal
Cyclic over central implies abelian
Lagrange's theorem
Equivalence of definitions of group of prime order: This basically states that any group of prime order must be cyclic.
Classification of groups of prime-square order
Structure theorem for finitely generated abelian groups

Proof

First part of proof: crude descriptions of center and quotient by center

Given: A prime number $p$ , a group $P$ of order $p^{3}$ .

To prove: Either $P$ is abelian, or we have: $Z (P)$ is a cyclic group of order $p$ and $P / Z (P)$ is an elementary abelian group of order $p^{2}$

Proof: Let $Z = Z (P)$ be the center of $P$ .

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$Z$ is nontrivial	Fact (1)	$P$ has order $p^{3}$ , specifically, a power of a prime		Fact+Given direct
2	The order of $Z$ cannot be $p^{2}$	Facts (2), (3), (4), (5)	$P$ has order $p^{3}$		[SHOW MORE] If $Z$ has order $p^{2}$ , then $P / Z$ (which exists by fact (2)) has order $p$ (by fact (4)) hence must be cyclic (by fact (5)). Then, by fact (3), $P$ would be abelian, but this would imply that $Z = P$ , in which case the order of $Z$ would be $p^{3}$ .
3	The order of $Z$ is either $p$ or $p^{3}$	Fact (4)	$P$ has order $p^{3}$	Steps (1), (2)	[SHOW MORE] By fact (4), the order of $Z$ must divide the order of $P$ . The only possibilities are $1, p, p^{2}, p^{3}$ . Step (1) eliminates the possibility of $1$ , and step (3) eliminates the possibility of $p^{2}$ . This leaves only $p$ or $p^{3}$ .
4	If $Z$ has order $p$ , then $Z$ is cyclic of order $p$ and the quotient $P / Z$ is elementary abelian of order $p^{2}$	Facts (3), (4), (5), (6)	$P$ has order $p^{3}$ .		[SHOW MORE] If $Z$ has order $p$ , then by fact (5), $Z$ must be cyclic. By Fact (4), $P / Z$ has order $p^{2}$ . Further, by fact (3), $P / Z$ cannot be cyclic, because if it were cyclic, then $P$ would be abelian, which would mean that $Z = P$ has order $p^{3}$ . Thus, $P / Z$ is a non-cyclic group of order $p^{2}$ . By fact (6), it must be the elementary abelian group of order $p^{2}$ .
5	If $Z$ has order $p^{3}$ , $P$ is abelian.		$P$ has order $p^{3}$ .		[SHOW MORE] We get $P = Z$ , so $P$ is abelian.
6	We get the desired result.			Steps (3), (4), (5)	Step-combination.

Second part of proof: classifying the abelian groups

This classification follows from fact (7): the abelian groups of order $p^{3}$ correspond to partitions of 3, as indicated in the original statement of the classification.

Third part of proof: classifying the non-abelian groups

Given: A non-abelian group $P$ of order $p^{3}$ . Let $Z$ be the center of $P$ .

Previous steps: $Z$ is cyclic of order $p$ , and $P / Z$ is elementary abelian of order $p^{2}$ .

We first make some additional observations.

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation
1	The derived subgroup (commutator subgroup) $P^{'}$ equals $Z$ .	$P$ is non-abelian of order $p^{3}$ .	$P / Z$ is abelian, $Z$ has order $p$ .	[SHOW MORE] Since $P / Z$ is abelian, $P^{'}$ is contained in $Z$ . Since $Z$ has prime order, the only possible subgroups are the trivial subgroup and $Z$ itself. The derived subgroup cannot be trivial since $P$ is non-abelian, hence the derived subgroup must be $Z$ .
2	We can find elements $a, b \in P$ such that the images of $a, b$ in $P / Z$ are non-identity elements of $P / Z$ that generate it.		$P / Z$ is elementary abelian of order $p^{2}$	[SHOW MORE] $P / Z$ is elementary abelian of order $p^{2}$ , hence is generated by a set of size two comprising non-identity elements. Choose $a, b$ to be arbitrary inverse images of these elements.
3	$Z, a, b$ together generate $P$ .
4	$a$ and $b$ do not commute.		Steps (2), (3)	[SHOW MORE] If $a$ and $b$ commute, then $C_{P} (a)$ contains $Z, a, b$ , hence equals all of $P$ . Thus, $a \in Z$ , so its image in $P / Z$ is the identity element, a contradiction to what we assumed.
5	Let $z = [a, b]$ . Then, $z$ is a non-identity element of $Z$ and $⟨ z ⟩ = Z$ .		Steps (1), (4)	[SHOW MORE] We must have $z \in P^{'}$ , which equals $Z$ by Step (1). By Step (4), $[a, b]$ is not the identity element. So $z$ is a non-identity element of $Z$ .
6	The elements $a, b$ both have order either $p$ or $p^{2}$ .			[SHOW MORE] The order of $a$ must divide the order of the group, which is $p^{3}$ . It cannot equal $p^{3}$ , because that would make the group cyclic and hence abelian. It cannot equal $1$ , because $a$ is a non-identity element. The only possibilities are $p$ and $p^{2}$ . The same goes for $b$ .

We now make cases based on the orders of $a$ and $b$ . Note that these cases may turn out to yield isomorphic groups, because the cases are made based on $a$ and $b$ , and there is some freedom in selecting these.

Case A: $a$ and $b$ both have order $p$ .

In this case, the relations so far give the presentation:

$⟨ a, b, z ∣ a^{p} = b^{p} = z^{p} = e, a z = z a, b z = z b, [a, b] = z ⟩$

These relations already restrict us to order at most $p^{3}$ , because we can use the commutation relations to express every element in the form Failed to parse (unknown function "\alphab"): {\displaystyle a^\alphab^\betaz^\gamma} , where $α, β, γ$ are integers mod $p$ . To show that there is no further reduction, we note that there is a group of order $p^{3}$ satisfying all these relations, namely prime-cube order group:U(3,p). This is the multiplicative group of unipotent upper-triangular matrices with entries from the field of $p$ elements.

Thus, Case A gives a unique isomorphism class of groups. Note that the analysis here works both for $p = 2$ and for odd primes. The nature of the group obtained, though, is different for $p = 2$ , where we get dihedral group:D8 which has exponent $p^{2}$ . For odd primes, we get a group of prime exponent.