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

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Statement

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 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 $0\leq k\leq 5$.
• 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 $0\leq k\leq 5$.

Facts used

1. 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 $N=E_{1}E_{2}$ satisfies $|N'|\leq p^{2}$ or $|N/Z(N)|\leq p^{2}$, then any maximal subgroup of $N$ containing $E_{1}\cap E_{2}$ contains an elementary abelian subgroup of order $p^{k}$.
2. Support of good lines corollary to line lemma, Jonah-Konvisser line lemma

Proof

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 $p^{k}$. $N=E_{1}E_{2}$.

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

Proof: Since $N=E_{1}E_{2}$ and both $E_{1}$ and $E_{2}$ are abelian and normal, $[N,N]=[E_{1},E_{2}]\leq E_{1}\cap E_{2}\leq Z(N)$. 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, $N=E_{1}E_{2}$ 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, $E\leq M$): 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 $p^{k}$:
   1. If $N=E_{1}E_{2}$, then either $|N'|\leq p^{2}$ or $|N/Z(N)|\leq p^{2}$: This is the lemma proved above.
   2. 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).
   3. 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 $E_{1}E_{2}$.
4. The result now follows from fact (2).

References

• Counting abelian subgroups of p-groups: a projective approach by Marc Konvisser and David Jonah, Journal of Algebra, ISSN 00218693, Volume 34, Page 309 - 330(Year 1975): ^{PDF (ScienceDirect)More info}, Theorem 2.3