Congruence condition fails for abelian subgroups of prime-sixth order

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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.

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