# Series-equivalent abelian-quotient central subgroups not implies automorphic

## Contents

## Statement

### In terms of series-equivalent subgroups

It is possible to have a finite group (in fact, even a group of prime power order) and normal subgroups and of such that:

- and are Series-equivalent subgroups (?) of : and are isomorphic groups and the quotient groups and are also isomorphic groups.
- (and hence also ) is an abelian group. In other words, and are both Abelian-quotient subgroup (?)s, i.e., they both contain the derived subgroup.
- and are both Central subgroup (?)s, i.e., they are both contained in the center of .
- and are
*not*Automorphic subgroups (?) in , i.e., there is no automorphism of sending to .

### In terms of the second cohomology group

It is possible to have abelian groups and and elements of the Second cohomology group for trivial group action (?) that are not in the same orbit under the natural action of but nonetheless give isomorphic big groups.

### Equivalence of statements

The statement for the second cohomology group can be interpreted in terms of series-equivalent subgroups by identifying and .

## Proof

### Example

`Further information: semidirect product of Z8 and Z8 of M-type`

Let be the semidirect product of Z8 and Z8 of M-type (GAP ID: (64,3)) given explicitly by the presentation:

The center of is and it is isomorphic to the direct product of Z4 and Z4. The derived subgroup of is , so any subgroup between these is an abelian-quotient central subgroup.

Let be the subgroup and be the subgroup . Both are central subgroups of isomorphic to direct product of Z4 and Z2. In both cases, the quotient group is also isomorphic to direct product of Z4 and Z2. However, there is no automorphism sending to because contains an element whose square is the unique non-identity commutator of , while does not.