# Series-equivalent abelian-quotient abelian not implies automorphic

## Contents

## Statement

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 (and hence also and ) are both abelian groups. In other words, is an Abelian normal subgroup (?) as well as an Abelian-quotient subgroup (?) of (and so is ).
- and are
*not*Automorphic subgroups (?) in , i.e., there is no automorphism of sending to .

## Related facts

### Stronger facts

Statement | Constraint on | Smallest order of among known examples | Isomorphism class of | Isomorphism class of | Isomorphism class of quotient group |
---|---|---|---|---|---|

series-equivalent not implies automorphic in finite abelian group | is also a finite abelian group | 128 | direct product of Z8 and Z4 and V4 | direct product of Z8 and V4 | direct product of Z4 and Z2 |

series-equivalent abelian-quotient central subgroups may be distinct | and are both central subgroups of (in other words, the quotient actions are both trivial. | 64 | semidirect product of Z8 and Z8 of M-type | direct product of Z4 and Z2 | direct product of Z4 and Z2 |

### Weaker facts

## Proof

### Example of the nontrivial semidirect product of cyclic groups of order four

`Further information: nontrivial semidirect product of Z4 and Z4, subgroup structure of nontrivial semidirect product of Z4 and Z4`

We define as the nontrivial semidirect product of Z4 and Z4, explicitly by the presentation:

Define as the subgroup and as the subgroup .

Both and are isomorphic to direct product of Z4 and Z2, and they are both normal with the quotient group isomorphic to cyclic group:Z2 (because they both have index two).

To see that there is no automorphism of sending to , note that the only non-identity squares of elements in the two subgroups are (in ) and (in ) respectively. An automorphism of sending to must therefore send to . This is not possible since is in the derived subgroup of (it is the only non-identity commutator) and is not.