Classification of nilpotent Lie rings of prime-cube order

Statement
Let $$p$$ be a prime number. There are, up to isomorphism, five possibilities for a nilpotent Lie ring of order $$p^3$$. Three of them are abelian and two are non-abelian.

For odd primes, we can use the Baer correspondence (a special case of the Lazard correspondence) to achieve a one-to-one correspondence between the nilpotent Lie rings of order $$p^3$$ and the groups of order $$p^3$$.

For the prime $$p = 2$$, although the number of nilpotent Lie rings equals the number of groups, we cannot use the Baer correspondence. More specifically, there is no natural way of matching the two non-abelian groups of order $$p^3$$ and the two non-abelian nilpotent Lie rings of order $$p^3$$.

The three abelian Lie rings
The three abelian Lie rings are the Lie rings with trivial Lie bracket; these thus correspond to the abelian groups of order $$p^3$$. They are classified by the partitions of 3:

The two non-abelian nilpotent Lie rings
The two non-abelian nilpotent Lie rings are given as follows:

First part of proof: crude description of center and quotient by center
Given: A prime number $$p$$, a nilpotent Lie ring $$P$$ of order $$p^3$$.

To prove: Either $$P$$ is abelian, or $$Z(P)$$ is cyclic of order $$p$$ and the quotient is an abelian Lie ring of order $$p^2$$ whose additive group is elementary abelian of order $$p^2$$.

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

Note: Some proof details need to be clarified, but the outline is complete and good enough for people thorough with group theory.

Third part of proof: classifying the non-abelian nilpotent Lie rings
Since we already know $$Z(P)$$ and $$P/Z(P)$$, we need to specify two things:


 * The alternating bilinear map $$P/Z(P) \times P/Z(P) \to Z(P)$$ given by the Lie bracket: There is only one option for this (up to isomorphism) because $$\bigwedge^2(P/Z(P))$$ is isomorphic to $$Z(P)$$, and, up to conjugacy, there is a unique isomorphism.
 * The nature of the additive group extension of $$P/Z(P)$$ on top of $$Z(P)$$: There are two possibilities for this: elementary abelian group of prime-square order (corresponding to the trivial or split extension)and direct product of cyclic group of prime-square order and cyclic group of prime order (corresponding to the nontrivial extensions).

Working out the corresponding Lie ring structures, we get upper-triangular nilpotent matrix Lie ring:u(3,p) (corresponding to the trivial extension) and semidirect product of cyclic Lie ring of prime-square order and cyclic Lie ring of prime order (corresponding to the nontrivial extension).