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 $p$.

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)).