Abelian-to-normal replacement theorem for prime-cube index for odd prime

History
The result appears to have been first proved in a paper by Alperin in 1965 and later proved as part of a larger array of results in a paper by David Jonah and Marc Konvisser in 1975.

Statement
Suppose $$p$$ is an odd prime. Then, if a finite $$p$$-group contains an abelian subgroup of index $$p^3$$, it contains an abelian normal subgroup (hence, an fact about::abelian normal subgroup of group of prime power order) of index $$p^3$$.

Facts used

 * 1) uses::Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime
 * 2) uses::Prime power order implies nilpotent, uses::nilpotent implies every maximal subgroup is normal

The Jonah-Konvisser proof
Given: An odd prime $$p$$, a finite $$p$$-group $$P$$, an abelian subgroup $$A$$ of $$P$$ of index $$p^3$$.

To prove: $$P$$ has an abelian normal subgroup of index $$p^3$$.

Proof:


 * 1) Let $$M$$ be a maximal subgroup of $$P$$ containing $$A$$: Such a $$M$$ exists and has index $$p$$ in $$P$$.
 * 2) By fact (1), the number of abelian subgroups of $$M$$ of index $$p^2$$ is either equal to $$2$$ or congruent to $$1$$ modulo $$p$$.
 * 3) If $$M$$ has exactly two abelian subgroups of index $$p^2$$, then both of them are normal in $$P$$: Since $$M$$ is maximal in $$P$$, it is normal (See fact (2)). Thus, any inner automorphism of $$P$$ sends subgroups of $$M$$ to subgroups of $$M$$. Since the sizes of orbits are either $$1$$ or powers of $$p$$, and there are only two abelian subgroups of index $$p^2$$, they must both be normal in $$P$$.
 * 4) If the number of abelian subgroups of $$M$$ of index $$p^2$$ is congruent to $$1$$ modulo $$p$$, then there exists a subgroup of index $$p^2$$ in $$M$$ that is normal in $$P$$: The inner automorphisms of $$P$$ permute abelian subgroups of index $$p^2$$ in $$M$$, since $$M$$ is normal in $$P$$. All orbits have size either $$1$$ or a nontrivial power (and hence, multiple) of $$p$$. Since the total number is $$1$$ modulo $$p$$, there must be at least one orbit of size one.