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.
View other such facts for p-groups|View other such facts for finite groups

This article defines a replacement theorem
View a complete list of replacement theorems| View a complete list of failures of replacement

This article is about a congruence condition.
View other congruence conditions

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 $p$.

Related facts

Breakdown at the prime two

• Elementary abelian-to-normal replacement fails for Klein four-group

Similar replacement theorems

• Jonah-Konvisser elementary abelian-to-normal replacement theorem does the analogue for elementary abelian subgroups of order $p^k$, for $k\le 5$ and $p$ odd.
• Jonah-Konvisser abelian-to-normal replacement theorem

Facts used

1. Prime power order implies nilpotent, Nilpotent implies every maximal subgroup is normal
2. Local origin corollary to line lemma
3. Maximal subgroup of join of two elementary abelian normal subgroups of prime-square order contains elementary abelian subgroup of prime-square order for odd prime

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 $N_1{N}_2$. 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 $N_1{N}_2$ with any maximal subgroup of $G$ contains an elementary abelian group of order $p^2$.
6. Thus, $N_1{N}_2$ 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 $p$.

References

Journal 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}, Application 1.7, Page 313