Group of prime-sixth or higher order contains abelian normal subgroup of prime-fourth order for prime equal to two

Statement
Any group of order $$2^n, n \ge 6$$contains an fact about::abelian normal subgroup (hence, fact about::abelian normal subgroup of group of prime power order) of order $$16 = 2^4$$. In other words, for the prime $$p = 2$$, any group of order $$p^n, n \ge 6$$ contains an abelian normal subgroup of order $$p^4$$.

Related facts

 * Existence of abelian normal subgroups of small prime power order: This states that if $$n \ge 1 + k(k-1)/2$$, then any finite $$p$$-group of order $$p^n$$ has an abelian normal subgroup of order $$p^k$$. Note that our result does not follow from this because the $$n$$ needed for $$k = 4$$ is $$7$$.
 * Abelian-to-normal replacement theorem for prime-cube order
 * Abelian-to-normal replacement theorem for prime-fourth order