Congruence condition on number of abelian subrings of prime-cube order in nilpotent Lie ring

Statement
Suppose $$L$$ is a nilpotent Lie ring of order $$p^n$$ for some prime number $$p$$. Let $$\mathcal{S}$$ be the collection of Lie subrings of $$L$$ that are abelian Lie rings of order $$p^3$$. Then, either $$\mathcal{S}$$ is empty or the size of $$\mathcal{S}$$ is congruent to 1 mod $$p$$.

Analogous facts for groups

 * Congruence condition on number of abelian subgroups of prime-cube order

Similar facts

 * Congruence condition on number of abelian ideals of prime-cube order in nilpotent Lie ring

Facts used

 * 1) uses::Existence of abelian ideals of small prime power order in nilpotent Lie ring
 * 2) uses::Congruence condition on number of abelian subrings of prime index in nilpotent ring
 * 3) uses::Congruence condition relating number of subrings in maximal subrings and number of subrings in the whole ring
 * 4) uses::Congruence condition on number of subrings of given prime power order in nilpotent ring

Proof
Given: A nilpotent Lie ring $$L$$ of order $$p^n$$, $$n \ge 3$$, that contains an abelian subgroup of order $$p^3$$.

To prove: The number of abelian subrings of $$L$$ of order $$p^3$$ is congruent to $$1$$ modulo $$p$$.

Proof: We consider three cases:


 * 1) $$n = 3$$: In this case, there is exactly one abelian subgroup of order $$p^3$$, namely $$G$$ itself.
 * 2) $$n = 4$$: In this case, fact (2) completes the proof.
 * 3) $$n \ge 5$$: In this case, we use induction on $$n$$. By fact (1), every maximal subring of $$L$$ contains an abelian ideal of order $$p^3$$, which is still a subring in the whole ring, so by the inductive hypothesis, the number of abelian subrings of order $$p^3$$ contained in each maximal subring is congruent to $$1$$ modulo $$p$$. Since the number of maximal subrings is itself congruent to $$1$$ modulo $$p$$ by fact (4), fact (3) tells us that the number of abelian subrings of $$L$$ of order $$p^3$$ is congruent to $$1$$ modulo $$p$$.