Abelian-to-normal replacement fails for prime-sixth order for prime equal to two

History
This result appeared in a paper by Alperin in 1965.

Statement
For $$p = 2$$, it is possible to have a finite $$p$$-group that has an abelian subgroup of order $$p^6 = 2^6 = 64$$ but no abelian normal subgroup of that order.

The smallest example is the case where the order of the group is $$p^9 = 2^9 = 512$$.

Related facts

 * Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one: This states that if $$k \le (p + 1)/2$$, the existence of an abelian subgroup of order $$p^k$$ implies the existence of an abelian normal subgroup of order $$p^k$$. In particular, for $$p \ge 11$$, it shows that the existence of an abelian subgroup of order $$p^6$$ implies the existence of an abelian normal subgroup of order $$p^6$$.
 * Abelian-to-normal replacement theorem for prime-cube order
 * Abelian-to-normal replacement theorem for prime-fourth order
 * Congruence condition on number of abelian subgroups of prime-cube order
 * Congruence condition on number of abelian subgroups of prime-fourth order
 * Abelian-to-normal replacement fails for prime-cube index for prime equal to two: The same example works.
 * Abelian-to-normal replacement theorem for prime-cube index for odd prime: This was proved by Alperin and later in a paper by Jonah and Konvisser.

Proof
Let $$G$$ be the quotient of the free product of two copies of the Klein four-group by the third member of its lower central series. In other words, $$G$$ is the free product of class two of two copies of the Klein four-group.

Let $$P$$ be the semidirect product of $$G$$ by an automorphism of order two that interchanges the two copies. Then, $$P$$ is a group of order $$2^9$$. We claim that $$P$$ has exactly two abelian subgroups of order $$2^6$$, both are elementary abelian, and neither is normal in $$P$$.

Journal references

 * , Page 11, shortly after Theorem 4.