# 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 $p$.

## 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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