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

Statement
Suppose $$L$$ is a nilpotent Lie ring of order $$p^k$$ for some prime number $$p$$. Let $$\mathcal{S}$$ be the collection of subrings of $$L$$ of order $$p^2$$ that are abelian (i.e., the Lie bracket is trivial) with additive group the elementary abelian group of prime-square order. Then, either $$\mathcal{S}$$ is empty or the size of $$\mathcal{S}$$ is congruent to 1 modulo $$p$$.

Analogous facts for groups

 * Congruence condition on number of elementary abelian subgroups of prime-square order for odd prime
 * On the other hand, the analogous statement for groups fails for the prime two, i.e., for the Klein four-group.

Similar facts

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

Facts used

 * 1) uses::Congruence condition on number of subrings of given prime power order and bounded exponent in nilpotent ring
 * 2) uses::Classification of Lie rings of prime-square order

Proof
Given: A nilpotent Lie ring $$L$$ of order $$p^k$$. $$\mathcal{S}$$ is the collection of abelian subrings of $$L$$ that have order $$p^2$$ and whose additive group is elementary abelian.

To prove: Either $$\mathcal{S}$$ is empty or the size of $$\mathcal{S}$$ is congruent to 1 mod $$p$$.

Proof: The proof follows directly from Fact (1), setting the bound on exponent as $$p$$ and noting that the only nilpotent Lie ring of order $$p^2$$ and exponent $$p$$ is the abelian Lie ring whose additive group is elementary abelian of order $$p^2$$ (by Fact (2)).