Berkovich's theorem on failure of abelian-to-normal replacement for subgroups of small index

Statement
Suppose $$p$$ is a prime number such that $$p \ge 5$$. Then, there exists a group of order $$p^{2p + 1}$$ that contains an abelian subgroup of order $$p^{(3p - 1)/2}$$ but does not contain any abelian normal subgroup of order $$p^{(3p - 1)/2}$$. In particular:


 * 1) If $$k = (3p - 1)/2$$, there exists a finite $$p$$-group that has an abelian subgroup of order $$p^k$$ but no abelian normal subgroup of order $$p^k$$. Then, the collection of abelian groups of order $$p^{(3p - 1)/2}$$ is not a fact about::collection of groups satisfying a weak normal replacement condition.
 * 2) If $$k  =(p + 3)/2$$, there exists a finite $$p$$-group that has an abelian subgroup of index $$p^k$$ but no abelian normal subgroup of index $$p^k$$.

Related facts

 * Abelian-to-normal replacement theorem for prime-square index
 * Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime
 * Abelian-to-normal replacement theorem for prime-cube index for odd prime
 * Alperin's conjecture on abelian-to-normal replacement for small index

Journal references

 * , Page 540 (Page 69 within the document), Section 14: On Alperin's conjecture on abelian subgroups of small index