Congruence condition fails for abelian subgroups of prime-sixth order

Statement
Let $$p$$ be any prime number. Then, there exists a finite $$p$$-group of order $$p^8$$ that contains exactly two abelian subgroups of order $$p^6$$, both of which are elementary abelian normal subgroups. Thus:


 * The collection of abelian groups of order $$p^6$$ fails to satisfy a universal congruence condition for any prime $$p$$.
 * The singleton collection of the elementary abelian group of order $$p^6$$ fails to satisfy a universal congruence condition for any prime $$p$$.

Related facts
See also collection of groups satisfying a universal congruence condition.


 * Abelian-to-normal replacement fails for prime-sixth order for prime equal to two
 * Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one: This in particular shows that for $$p \ge 11$$, the existence of an abelian subgroup of order $$p^6$$ implies the existence of an abelian normal subgroup of order $$p^6$$.
 * Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime: For $$p$$ odd, the universal congruence condition does hold for abelian subgroups of order $$p^k$$, for any fixed $$k$$ between $$0$$ and $$5$$.

Proof
Let $$G$$ be the quotient of the free product of two elementary abelian groups of order $$p^2$$ by the third member of its derived series. Thus, $$G$$ is the free product of class two of two elementary abelian groups of order $$p^2$$. Then, $$G$$ is a group of order $$p^8$$. We claim that $$G$$ has exactly two abelian subgroups of order $$p^6$$, and both of these are elementary abelian and normal.

Journal references

 * : This describes the construction of counterexamples.
 * : This is a companion paper that proves the opposite results for orders up to $$p^5$$.
 * : A paper published in 1965 that explores similar questions.