# 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 unitriangular matrix group:UT(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

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 (CD1) Eliminates the case of trivial center class equation of a group 3 click here
2 Center is normal Steps (CD2), (CD4) Used to note that the quotient by the center exists as a group 0 click here
3 Cyclic over central implies abelian Steps (CD2) (CD4) Rules out the possibility of center of order $p^2$, helps determine nature of quotient group if center has order $p$ click here
4 Lagrange's theorem Step (CD2) 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: This basically states that any group of prime order must be cyclic. Steps (CD2), (CD4) Helps rule out center of order $p^2$, helps determine isomorphism class of center of order $p$ click here
6 Classification of groups of prime-square order Step (CD4) Narrows down possibilities for quotient by the center once we have determined that the center has order $p$ 3 click here
7 Structure theorem for finitely generated abelian groups Classification of abelian groups case click here
8 Class two implies commutator map is endomorphism Case B analysis Helps reduce general case to a specific presentation by a change of variable click here
9 Formula for powers of product in group of class two Case C analysis Helps reduce Case C to Case B by a change of variable click here

## Cohomology tree

This can be formulated as an alternative proof if we prove assertions for each of the cohomology groups.

### For odd primes

The cohomology groups governing the branchings are as follows:

### For the prime 2

This is for groups of order 8:

The cohomology groups governing the branchings are as follows:

## Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

### 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
CD1 $Z$ is nontrivial Fact (1) $P$ has order $p^3$, specifically, a power of a prime Fact+Given direct
CD2 The order of $Z$ cannot be $p^2$ Facts (2), (3), (4), (5) $P$ has order $p^3$ [SHOW MORE]
CD3 The order of $Z$ is either $p$ or $p^3$ Fact (4) $P$ has order $p^3$ Steps (CD1), (CD2) [SHOW MORE]
CD4 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 (2), (3), (4), (5), (6) $P$ has order $p^3$. [SHOW MORE]
CD5 If $Z$ has order $p^3$, $P$ is abelian. $P$ has order $p^3$. [SHOW MORE]
CD6 We get the desired result. Steps (CD3), (CD4), (CD5) 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
CN1 The derived subgroup (commutator subgroup) $P'$ equals $Z$, so $P$ has class two. $P$ is non-abelian of order $p^3$. $P/Z$ is abelian, $Z$ has order $p$. [SHOW MORE]
CN2 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]
CN3 $Z,a,b$ together generate $P$. Step (CN2)
CN4 $a$ and $b$ do not commute. Steps (CN2), (CN3) [SHOW MORE]
CN5 Let $z = [a,b]$. Then, $z$ is a non-identity element of $Z$ and $\langle z \rangle = Z$. Steps (CN1), (CN4) [SHOW MORE]
CN6 The elements $a,b$ both have order either $p$ or $p^2$. Also, the elements $a^p$ and $b^p$ are both in $Z$. [SHOW MORE]

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 $a^\alpha b^\beta z^\gamma$, 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 unitriangular matrix group:UT(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 so far 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.

Case B: $a$ has order $p^2$, $b$ has order $p$

In this case, we first note that $a^p \in Z = \langle z \rangle$. Since $a^p$ is a non-identity element, there exists nonzero $r$ (taken mod $p$) such that $a^p = z^r$. Consider the element $c = b^r$ Then, by Fact (8), and the observation that $P$ has class two (Step (1) in the above table), we obtain:

$\! [a,c] = [a,b^r] = [a,b]^r = z^r = a^p$

Consider the presentation:

$\langle a,c \mid a^{p^2} = c^p = e, [a,c] = a^p \rangle$

We see that all these relations are forced by the above, and further, that this presentation defines a group of order $p^3$, namely semidirect product of cyclic group of prime-square order and cyclic group of prime order.

Thus, there is a unique isomorphism class in Case B. Note that the analysis so far works both for $p = 2$ and for odd primes. The nature of the group, though, is different for $p = 2$, we get dihedral group:D8, which is the same isomorphism class as Case A.

Case B2: $a$ has order $p$, $b$ has order $p^2$.

Interchange the roles of $a,b$ and replace $z$ by $z^{-1}$ and we are back in Case B.

Case C: $a$ and $b$ both have order $p^2$.

By Fact (9), we can show that for odd prime, it is possible to make a substitution and get into Case B. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

For $p = 2$, working out the presentation yields quaternion group.

Here is a summary of the cases:

Case letter What it means Isomorphism class of group for $p = 2$ Isomorphism class of group for odd prime
A Both $a,b$ have order $p$ dihedral group:D8 unitriangular matrix group:UT(3,p)
B, B2 One of the elements has order $p^2$, the other has order $p$ dihedral group:D8 semidirect product of cyclic group of prime-square order and cyclic group of prime order
C Both elements have order $p^2$ quaternion group semidirect product of cyclic group of prime-square order and cyclic group of prime order

Finally, we note that: