Series-equivalent abelian-quotient abelian not implies automorphic
- 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 .
|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|
Example of the nontrivial semidirect product of cyclic groups of order four
We define as the nontrivial semidirect product of Z4 and Z4, explicitly by the presentation:
Define as the subgroup and as the subgroup .
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.