There exists an abelian subgroup of maximum order whose normalizer contains every abelian subgroup it normalizes

Statement
Suppose $$p$$ is a prime number and $$P$$ is a group of prime power order with underlying prime $$p$$. Denote by $$\mathcal{A}(P)$$ the set of abelian subgroups of maximum order. Then, there exists $$A \in \mathcal{A}(P)$$ such that its normalizer in $$P$$, namely $$N_P(A)$$, contains every abelian subgroup of $$P$$ normalized by $$A$$. In particular, $$N_P(A)$$ contains every abelian normal subgroup of $$P$$ and hence contains the join of all abelian normal subgroups of $$P$$.

Similar facts

 * Thompson's replacement theorem for abelian subgroups
 * Any abelian normal subgroup normalizes an abelian subgroup of maximum order: This is a weaker version of the statement that is easier to prove, and follows directly from Thompson's replacement theorem.

Stronger facts in some cases
Note that this fact is uninteresting for small orders for the following silly reason: for small orders, it is also true that, among the abelian subgroups of maximum order, there exists a normal subgroup. The existence of a normal subgroup that is abelian of maximum order is obviously substantially stronger than this statement.

Below, we indicate which of a bunch of stronger statements is true, and why. In particular, we note from this that the smallest examples of interest where the stronger statements do not hold are $$2^9 = 512$$ for the prime 2 and $$p^{10}$$ (or possibly higher for some $$p$$, depending on other replacement results that are dependent on the value of $$p$$ -- see Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one for instance) for odd primes.