Abelian-to-normal replacement theorem for prime-square index

Statement
Let $$p$$ be any prime number (possibly $$p = 2$$). Then, if $$P$$ is a finite $$p$$-group (i.e., a group of prime power order where the underlying prime is $$p$$) and $$A$$ is an abelian subgroup of index $$p^2$$ in $$P$$, then there is an abelian normal subgroup of $$P$$ of index $$p^2$$. Moreover, we can choose this abelian normal subgroup so that it is contained in the normal closure of $$A$$ in $$P$$.

Stronger facts

 * Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime

Facts used

 * 1) uses::Congruence condition on number of abelian subgroups of prime index

Proof
Given: A finite $$p$$-group $$P$$ for some prime $$p$$, an abelian subgroup $$A$$ of $$P$$ of index $$p^2$$.

To prove: There exists an abelian normal subgroup $$B$$ of $$P$$ of index $$p^2$$, contained inside the normal closure of $$A$$ in $$P$$.

Proof:


 * 1) If $$A$$ is normal in $$P$$, we are done. Otherwise, $$A$$ is a 2-subnormal subgroup and its normal closure is a maximal subgroup $$M$$ of $$P$$ containing $$A$$. $$M$$ is normal and has index $$p$$ in $$P$$, and $$A$$ is normal and has index $$p$$ in $$M$$.
 * 2) By fact (1), the number of abelian subgroups of $$M$$ is congruent to $$1$$ modulo $$p$$.
 * 3) Since $$M$$ is normal in $$P$$, $$P$$ acts on $$M$$ by conjugation and the orbits of abelian maximal subgroups are therefore of size equal to a power of $$p$$. Since the total number of abelian maximal subgroups is $$1$$ modulo $$p$$, there exists an orbit of size $$1$$, and the member of this orbit is thus an abelian normal subgroup of index $$p$$.

Journal references

 * : This seems to be the first paper where a proof appeared, though the result may have been known or observed by others earlier.
 * : Here, they prove stronger results including abelian-to-normal replacement theorem for prime-cube index for odd prime and congruence condition on number of abelian subgroups of prime-square index for odd prime.