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

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 fact about::collection of groups satisfying a universal congruence condition for the prime $$p$$.

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) uses::Prime power order implies nilpotent, uses::Nilpotent implies every maximal subgroup is normal
 * 2) uses::Local origin corollary to line lemma
 * 3) uses::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 $$N_1N_2$$. 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 $$p$$.

Journal references

 * , Application 1.7, Page 313