Abelian-to-normal replacement fails for prime-cube index for prime equal to two

From Groupprops
Jump to: navigation, search
This article discusses a failure of replacement, i.e., a situation where the analogue of a valid replacement theorem fails to hold under slightly modified conditions.
View other failures of replacement | View replacement theorems

Statement

For p = 2, it is possible to have a finite p-group that has an abelian subgroup of index p^3 but no abelian normal subgroup of index p^3.

The smallest example is the case where the group has order p^9 = 2^9 = 512 and the abelian subgroup thus has order p^6 = 2^6 = 64.

Related facts

References

Journal references