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#Examples/facts.

• 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 $p\geq 11$, 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=

Further information: free product of class two of two elementary abelian groups of prime-square order

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.

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References

Journal references

• Abelian subgroups of p-groups, an algebraic approach by David Jonah and Marc Konvisser, Journal of Algebra, ISSN 00218693, Volume 34, Page 386 - 402(Year 1975): ^{Official copyMore info}: This describes the construction of counterexamples.
• Counting abelian subgroups of p-groups: a projective approach by Marc Konvisser and David Jonah, Journal of Algebra, ISSN 00218693, Volume 34, Page 309 - 330(Year 1975): ^{PDF (ScienceDirect)More info}: This is a companion paper that proves the opposite results for orders up to $p^{5}$.
• Large abelian subgroups of p-groups by Jonathan Lazare Alperin, Transactions of the American Mathematical Society, Volume 117, Page 10 - 20(Year 1965): ^{Official copyMore info}: A paper published in 1965 that explores similar questions.