# Classification of groups of prime-cube order

## Statement

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

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 
3 cyclic group of prime-cube order, denoted  or , or  1
2 + 1 direct product of cyclic group of prime-square order and cyclic group of prime order, denoted  or  2
1 + 1 + 1 elementary abelian group of prime-cube order, denoted , or , or  5

### The two non-abelian groups

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

For the case of odd , these are unitriangular matrix group:UT(3,p) (GAP ID: (,3)) and semidirect product of cyclic group of prime-square order and cyclic group of prime order (GAP ID: (,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 , helps determine nature of quotient group if center has order  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 , helps determine isomorphism class of center of order  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  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
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

## 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 , a group  of order .

To prove: Either  is abelian, or we have:  is a cyclic group of order  and  is an elementary abelian group of order 

Proof: Let  be the center of .

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
CD1  is nontrivial Fact (1)  has order , specifically, a power of a prime Fact+Given direct
CD2 The order of  cannot be  Facts (2), (3), (4), (5)  has order  [SHOW MORE]
CD3 The order of  is either  or  Fact (4)  has order  Steps (CD1), (CD2) [SHOW MORE]
CD4 If  has order , then  is cyclic of order  and the quotient  is elementary abelian of order  Facts (2), (3), (4), (5), (6)  has order . [SHOW MORE]
CD5 If  has order ,  is abelian.  has order . [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  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  of order . Let  be the center of .

Previous steps:  is cyclic of order , and  is elementary abelian of order .

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)  equals , so  has class two.  is non-abelian of order .  is abelian,  has order . [SHOW MORE]
CN2 We can find elements  such that the images of  in  are non-identity elements of  that generate it.  is elementary abelian of order  [SHOW MORE]
CN3  together generate . Step (CN2)
CN4  and  do not commute. Steps (CN2), (CN3) [SHOW MORE]
CN5 Let . Then,  is a non-identity element of  and . Steps (CN1), (CN4) [SHOW MORE]
CN6 The elements  both have order either  or . Also, the elements  and  are both in . [SHOW MORE]

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

Case A:  and  both have order .

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



These relations already restrict us to order at most , because we can use the commutation relations to express every element in the form , where  are integers mod . To show that there is no further reduction, we note that there is a group of order  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  elements.

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

Case B:  has order ,  has order 

In this case, we first note that . Since  is a non-identity element, there exists nonzero  (taken mod ) such that . Consider the element  Then, by Fact (8), and the observation that  has class two (Step (1) in the above table), we obtain:



Consider the presentation:



We see that all these relations are forced by the above, and further, that this presentation defines a group of order , 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  and for odd primes. The nature of the group, though, is different for , we get dihedral group:D8, which is the same isomorphism class as Case A.

Case B2:  has order ,  has order .

Interchange the roles of  and replace  by  and we are back in Case B.

Case C:  and  both have order .

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 , working out the presentation yields quaternion group.

Here is a summary of the cases:

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

Finally, we note that: