Congruence condition on number of subrings of given prime power order in nilpotent ring

Statement
Suppose $$L$$ is a finite nilpotent ring (not necessarily a Lie ring or an associative ring or a commutative ring) and $$p^r$$ is a prime power dividing the order of $$L$$. Then, the number of subrings of $$L$$ of order $$p^r$$ is congruent to 1 mod $$p$$.

Similar facts

 * Congruence condition on number of ideals of given prime power order in nilpotent Lie ring
 * Congruence condition on number of subrings of given prime power order in nilpotent Lie ring
 * Congruence condition on number of subgroups of given prime power order

Opposite facts

 * Congruence condition fails for number of subrings of given prime power order

Facts used

 * 1) uses::Congruence condition relating number of subrings in maximal subrings and number of subrings in the whole ring
 * 2) uses::Formula for number of maximal subrings in nilpotent ring of prime power order

Reduction to the case of prime power order
Any finite ring is a direct sum of rings of prime power order (Sylow subrings) and any ring of prime power order is contained in the corresponding Sylow subring. Thus, it suffices to prove the result for the case of nilpotent rings of prime power order.

Proof for the prime power order case
Given: A nilpotent ring $$L$$ of order $$p^k$$, $$p$$ a prime number.

To prove: For any $$r \le k$$, the number of subrings in $$L$$ of order $$p^r$$ is congruent to $$1$$ modulo $$p$$.

Proof: In this proof, we induct on $$k$$, i.e., we assume the statement is true inside nilpotent rings of order $$p^l, l \le k$$.

Base case for induction: The case $$k = 0$$ is obvious.

Inductive step: If $$r = k$$, the number of subrings is 1, so the statement is true. So we consider $$r < k$$.

For a subring $$S$$ of $$L$$, denote by $$n(S)$$ the number of subrings of $$S$$ of order $$p^r$$.