# Congruence condition on number of elementary abelian subgroups of prime-square order for odd prime

This article states and (possibly) proves a fact that is true for odd-order p-groups: groups of prime power order where the underlying prime is odd. The statement is false, in general, for groups whose order is a power of two.
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## Statement

### Hands-on statement

Suppose $p$ is an odd prime and $G$ is a finite $p$-group. Suppose $G$ has an elementary abelian subgroup of order $p^2$. Then, the following are true:

1. The number of elementary abelian subgroups of $G$ of order $p^2$ is congruent to $1$ modulo $p$.
2. The number of elementary abelian normal subgroups of $G$ of order $p^2$ is congruent to $1$ modulo $p$. In particular, there is an elementary abelian normal subgroup of order $p^2$.
3. If $G$ is a normal subgroup of a bigger finite $p$-group $L$ and $G$ contains an elementary abelian subgroup of order $p^2$, then the number of elementary abelian subgroups of $G$ that are normal in $L$ is congruent to $1$ modulo $p$. In particular, there is an elementary abelian subgroup of $G$ that is normal in $L$.

### Statement in terms of a universal congruence condition

Let $p$ be an odd prime.

Let $\mathcal{S}$ be a singleton set comprising the elementary abelian subgroup of order $p^2$.

Then, $\mathcal{S}$ is a Collection of groups satisfying a universal congruence condition (?) for the prime $p$.

## Proof

We prove here the stronger version.

### Equivalence of conditions (1)-(3)

For the equivalence of (1) and (2), consider the action of $G$ on itself by conjugation. Under this action, the non-normal elementary abelian subgroups of order $p^2$ form orbits whose size is a multiple of $p$. Thus, the number of elementary abelian subgroups of order $p^2$ is congruent modulo $p$ to the number of elementary abelian normal subgroups of order $p^2$.

For the equivalence of definitions (1) and (3), consider the action of $L$ on $G$ by conjugation. The subgroups of $G$ that are not normal in $L$ have orbits whose sizes are multiples of $p$, so the number of elementary abelian subgroups of order $p^2$ in $G$ is congruent modulo $p$ to the number of such subgroups that are normal in $L$. In particular, (1) implies (3). (3) clearly implies (2), and (2) is equivalent to (1).

We will freely use this equivalence in the proof below.

### Main proof

Given: A $p$-group $G$ (odd $p$). $G$ has an elementary abelian subgroup $K$ of order $p^2$.

To prove: The number of elementary abelian subgroups of $G$ of order $p^2$ is $1$ modulo $p$.

Proof: We prove the claim by induction on $G$, assuming the result is true for smaller orders.

If $K = G$, the number of subgroups is $1$, and we are done. We consider the other case:

1. There exists a maximal subgroup $M$ of $G$ containing $K$: This follows since $K$ is proper.
2. $M$ is normal in $G$: This follows from fact (1).
3. $G$ contains an elementary abelian normal subgroup $N$ of order $p^2$ (in fact, $N \le M$): We apply the induction hypothesis in form (3) with $M$ in place of $G$ and $G$ in place of $L$.
4. If $G$ contains only one elementary abelian normal subgroup of order $p^2$, we've proved the statement for $G$ in form (2). So, we assume that $G$ contains two distinct elementary abelian normal subgroups $N_1,N_2$.
5. Consider the product $N_1N_2$. This is either elementary abelian of order $p^4$, or is isomorphic to prime-cube order group:U(3,p), i.e., is a non-abelian group of order $p^3$ and exponent $p$. In either case, every maximal subgroup of it contains an elementary abelian subgroup of order $p^2$ (for more details, see fact (3)). In particular, the intersection of $N_1N_2$ with any maximal subgroup of $G$ contains an elementary abelian group of order $p^2$.
6. Thus, $N_1N_2$ is a local origin in the terminology of fact (2), and the induction hypothesis yields that the number of elementary abelian subgroups of order $p^2$ in $G$ is congruent to $1$ modulo $p$.