# 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 $p$.

## 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 $N = E_1E_2$ 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. 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 $p^k$. $N = E_1E_2$.

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 $p^k$:
1. If $N = E_1E_2$, then either $|N'| \le p^2$ or $|N/Z(N)| \le p^2$: 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 $E_1E_2$.
4. The result now follows from fact (2).