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 $\mathbb {Z} _{p^{3}}$, or $\mathbb {Z} /p^{3}\mathbb {Z}$	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 $\mathbb {Z} _{p^{2}}\times \mathbb {Z} _{p}$	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 $\mathbb {Z} _{p}\times \mathbb {Z} _{p}\times \mathbb {Z} _{p}$	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

1. Prime power order implies not centerless
2. Center is normal
3. Cyclic over central implies abelian
4. Lagrange's theorem
5. Equivalence of definitions of group of prime order: This basically states that any group of prime order must be cyclic.
6. Classification of groups of prime-square order
7. 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	Facts used	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 $\langle z\rangle =Z$.			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:

$\langle a,b,z\mid a^{p}=b^{p}=z^{p}=e,az=za,bz=zb,[a,b]=z\rangle$

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 $\alpha ,\beta ,\gamma$ 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.