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 $$p = 2$$.

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

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)).

Related facts

 * Classification of nilpotent Lie rings of prime-cube order
 * Classification of groups of prime-square order
 * Classification of Lie rings of prime-square order

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:


 * Second cohomology group for trivial group action of group of prime order on group of prime order controls the initial branching from the root.
 * Second cohomology group for trivial group action of elementary abelian group of prime-square order on group of prime order controls the upper second level branching.
 * Second cohomology group for trivial group action of cyclic group of prime-square order on group of prime order controls the lower second level branching.

For the prime 2
This is for groups of order 8:



The cohomology groups governing the branchings are as follows:


 * Second cohomology group for trivial group action of Z2 on Z2 controls the initial branching from the root.
 * Second cohomology group for trivial group action of V4 on Z2 controls the upper second level branching.
 * Second cohomology group for trivial group action of Z4 on Z2 controls the lower second level branching.

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 $$P$$.

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.

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.

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

Here is a summary of the cases:

Finally, we note that:


 * Dihedral group:D8 and quaternion group are non-isomorphic: The latter has no non-central element of order two, for instance.
 * Unitriangular matrix group:UT(3,p) and semidirect product of cyclic group of prime-square order and cyclic group of prime order are non-isomorphic: The former has exponent $$p$$, the latter has exponent $$p^2$$.