Congruence condition on number of elementary abelian subgroups of prime-cube and prime-fourth order for odd prime

In terms of a universal congruence condition
Suppose $$p$$ is an odd prime and $$k = 3$$ or $$k = 4$$. Then, the singleton set comprising the elementary abelian group of order $$p^k$$ is a fact about::collection of groups satisfying a universal congruence condition for the prime $$p$$.

Hands-on statement
Let $$p$$ be an odd prime and $$k = 3$$ or $$k = 4$$. Then, if $$P$$ is a finite $$p$$-group, the number of elementary abelian subgroups of $$P$$ of order $$p^k$$ is either equal to zero or congruent to $$1$$ modulo $$p$$.

Related facts

 * Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime: This is a stronger version that includes all $$0 \le k \le 5$$.
 * Congruence condition on number of elementary abelian subgroups of prime-square order for odd prime: An easier version covering the case $$k = 2$$.
 * Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime: This is a stronger version that includes all $$0 \le k \le 5$$.

Facts used

 * 1) uses::Jonah-Konvisser lemma for elementary abelian-to-normal replacement for prime-cube and prime-fourth order: This states that if $$E_1, E_2$$ are two distinct elementary abelian normal subgroups of $$P$$, and $$N = E_1E_2$$ satisfies $$|N'| \le p^2$$ or $$|N/Z(N)| \le p^2$$, then any maximal subgroup of $$N$$ containing $$E_1 \cap E_2$$ contains an elementary abelian subgroup of order $$p^k$$.
 * 2) uses::Support of good lines corollary to line lemma, uses::Jonah-Konvisser line lemma

Preliminary lemma
Given: $$p$$ odd, $$k = 3$$ or $$k = 4$$, and $$E_1,E_2$$ are normal subgroups of a finite $$p$$-group both of order $$p^k$$. $$N = E_1E_2$$.

To prove: $$[N,N]$$ has order at most $$p^2$$ or $$N/Z(N)$$ has order at most $$p^2$$.

Proof: Since $$N = E_1E_2$$ and both $$E_1$$ and $$E_2$$ are abelian and normal, $$[N,N] = [E_1,E_2] \le E_1 \cap E_2 \le Z(N)$$. We consider the cases:


 * 1) $$E_1 \cap E_2$$ has order at most $$p^2$$: In this case, $$[N,N]$$ has order at most $$p^2$$.
 * 2) $$E_1 \cap E_2$$ has order $$p^3$$: In this case, $$N = E_1E_2$$ has order $$p^5$$, so $$N/Z(N)$$ has order $$p^2$$.

Main proof
We prove the statement by induction on the order of the finite $$p$$-group. Here $$k = 3$$ or $$k = 4$$.

Given: A finite $$p$$-group $$P$$ containing an elementary abelian subgroup of order $$p^k$$.

To prove: The number of elementary abelian subgroups of $$P$$ of order $$p^k$$ is congruent to $$1$$ modulo $$p$$.

Proof: If $$P$$ itself has order $$p^k$$, the number of such subgroups is $$1$$. Thus, we assume that the order of $$P$$ is greater than $$p^k$$.


 * 1) There is a maximal subgroup $$M$$ of $$P$$ containing the elementary abelian subgroup: This follows because the elementary abelian subgroup is proper.
 * 2) There is an elementary abelian subgroup $$E$$ of $$G$$ of order $$p^k$$ (in fact, $$E \le M$$): This follows from the various equivalent formulations of universal congruence condition for $$M$$. Explicitly, the number of elementary abelian subgroups of order $$p^k$$ in $$M$$ is congruent to $$1$$ modulo $$p$$. The action of $$G$$ on these subgroups by conjugation has orbits of size $$1$$ or multiples of $$p$$, so there is at least one orbit of size $$1$$, and this gives a normal subgroup.
 * 3) Any pair of distinct elementary abelian normal subgroups supports good lines for the collection of elementary abelian subgroups of order $$p^k$$. In other words, if $$E_1,E_2$$ are distinct elementary abelian normal subgroups of order $$p^k$$, any maximal subgroup containing $$E_1 \cap E_2$$ contains an elementary abelian subgroup of order $$p^k$$:
 * 4) If $$N = E_1E_2$$, then either $$|N'| \le p^2$$ or $$|N/Z(N)| \le p^2$$: This is the lemma proved above.
 * 5) Any maximal subgroup of $$N$$ containing $$E_1 \cap E_2$$ contains an elementary abelian subgroup of order $$p^k$$: This follows from the previous step and fact (1).
 * 6) Any maximal subgroup of $$P$$ containing $$E_1 \cap E_2$$ contains an elementary abelian subgroup of order $$p^k$$. In other words, the pair $$E_1,E_2$$ supports good lines: This follows from the previous step and the fact that any maximal subgroup of $$P$$ contains a maximal subgroup of $$E_1E_2$$.
 * 7) The result now follows from fact (2).