Abelian-to-normal replacement fails for prime-cube index for prime equal to two
From Groupprops
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 , it is possible to have a finite -group that has an abelian subgroup of index but no abelian normal subgroup of index .
The smallest example is the case where the group has order and the abelian subgroup thus has order .
Related facts
- Abelian-to-normal replacement theorem for prime-cube index for odd prime
- Abelian-to-normal replacement theorem for prime-square index
- Jonah-Konvisser congruence condition on number of abelian subgroups of prime-square index for odd prime
- Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one
- Abelian-to-normal replacement fails for prime-sixth order for prime equal to two
References
Journal references
- Large abelian subgroups of p-groups by Jonathan Lazare Alperin, Transactions of the American Mathematical Society, Volume 117, Page 10 - 20(Year 1965): ^{Official copy}^{More info}, Page 11, shortly after Theorem 4.