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 ![]() |
Isomorphism class of ![]() |
Isomorphism class of ![]() |
Isomorphism class of quotient group ![]() |
---|---|---|---|---|---|
series-equivalent not implies automorphic in 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 | ![]() ![]() ![]() |
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.