# 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

• 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=

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.