Abelian-to-normal replacement fails for prime-sixth order for prime equal to two

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History

This result appeared in a paper by Alperin in 1965.

Statement

For p = 2, it is possible to have a finite p-group that has an abelian subgroup of order p^6 = 2^6 = 64 but no abelian normal subgroup of that order.

The smallest example is the case where the order of the group is p^9 = 2^9 = 512.

Related facts

Proof

Further information: Free product of class two of two Klein four-groups

Let G be the quotient of the free product of two copies of the Klein four-group by the third member of its lower central series. In other words, G is the free product of class two of two copies of the Klein four-group.

Let P be the semidirect product of G by an automorphism of order two that interchanges the two copies. Then, P is a group of order 2^9. We claim that P has exactly two abelian subgroups of order 2^6, both are elementary abelian, and neither is normal in P.

References

Journal references