Abelian-to-normal replacement fails for prime-cube index for prime equal to two

Statement
For $$p = 2$$, it is possible to have a finite $$p$$-group that has an abelian subgroup of index $$p^3$$ but no abelian normal subgroup of index $$p^3$$.

The smallest example is the case where the group has order $$p^9 = 2^9 = 512$$ and the abelian subgroup thus has order $$p^6 = 2^6 = 64$$.

Related facts

 * Abelian-to-normal replacement theorem for prime-cube index for odd prime
 * 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
 * Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one
 * Abelian-to-normal replacement fails for prime-sixth order for prime equal to two

Journal references

 * , Page 11, shortly after Theorem 4.