Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one

Hands-on statement
Suppose $$p$$ is a prime number and $$P$$ is a finite $$p$$-group. Suppose $$A$$ is an abelian subgroup of $$P$$ of order $$p^k$$, where $$k \le (p + 1)/2$$ and exponent dividing $$p^d$$. Then, there exists an abelian normal subgroup $$B$$ of $$P$$ contained in the normal closure of $$A$$ such that $$B$$ has order $$p^k$$ and exponent dividing $$p^d$$.

Statement in terms of normal replacement condition
Suppose $$p$$ is a prime number and $$0 \le d \le k \le (p + 1)/2$$. Then, the collection of abelian groups of order $$p^k$$ and exponent dividing $$p^d$$ is a fact about::collection of groups satisfying a strong normal replacement condition.

Related facts

 * Congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime
 * Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime
 * Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime
 * Elementary abelian-to-normal replacement theorem for large primes (this is a weaker version that is superseded by the result on this page and the results of Jonah and Konvisser)
 * Mann's replacement theorem for subgroups of prime exponent