Classification of nilpotent Lie rings of prime-cube order

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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:

Partition of 3 Corresponding additive group of Lie ring
3 cyclic group of prime-cube order, denoted C_{p^3} or \mathbb{Z}_{p^3}, or \mathbb{Z}/p^3\mathbb{Z}
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
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

The two non-abelian nilpotent Lie rings

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

Lie ring name Presentation Nilpotency class Additive group Corresponding group under Baer correspondence in case of odd prime p
semidirect product of cyclic Lie ring of prime-square order and cyclic Lie ring of prime order \langle a,b \mid p^2 a = pb = 0, [a,b] = a \rangle 2 direct product of cyclic group of prime-square order and cyclic group of prime order semidirect product of cyclic group of prime-square order and cyclic group of prime order
upper-triangular nilpotent matrix Lie ring:u(3,p) \langle a,b,c \mid pa = pb = pc = 0, [a,c] = b, [a,b] = [b,c] = 0 \rangle 2 elementary abelian group of prime-cube order prime-cube order group:U(3,p); see Baer correspondence between u(3,p) and U(3,p)

Proof

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.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Z is nontrivial P is nilpotent and nontrivial Given direct
2 The order of Z cannot be p^2 PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] P has order p^3
3 The order of Z is either p or p^3 Steps (1), (2)
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.
5 If Z has order p^3, P is abelian
6 We get the desired result. Steps (4), (5) Step-combination.

Second part of proof: classifying the abelian Lie rings

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:

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).