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

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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.

Related facts

Analogous facts for groups

Similar facts

Facts used

  1. Congruence condition on number of subrings of given prime power order and bounded exponent in nilpotent ring
  2. 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)).