Normal not implies potentially fully invariant
This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) need not satisfy the second subgroup property (i.e., potentially fully invariant subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about potentially fully invariant subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not potentially fully invariant subgroup|View examples of subgroups satisfying property normal subgroup and potentially fully invariant subgroup
Contents
Statement
It is possible to have a normal subgroup of a group
that is not a potentially fully invariant subgroup of
-- in other words, there is no group
containing
such that
is a fully invariant subgroup of
.
Related facts
- Normal not implies image-potentially fully invariant
- Normal not implies potentially verbal
- NPC theorem: Normal equals potentially characteristic.
- Normal equals potentially normal-subhomomorph-containing
- Complete and potentially fully invariant implies homomorph-containing
- Fully normalized and potentially fully invariant implies centralizer-annihilating endomorphism-invariant
Facts used
- Equivalence of definitions of complete direct factor: This states that for a complete subgroup, being normal is equivalent to being a direct factor.
- Equivalence of definitions of fully invariant direct factor: This states that for a direct factor, being a fully invariant subgroup is equivalent to being a homomorph-containing subgroup.
- Homomorph-containment satisfies intermediate subgroup condition
Proof
Example involving a complete group
Further information: Complete and potentially fully invariant implies homomorph-containing
Let be a nontrivial complete group. Define
and
. Clearly,
is a normal subgroup of
.
Suppose is a group containing
, such that
is fully invariant in
. In particular,
is normal in
. Since
is complete, it is a direct factor, so there exists a group
that is a complement to
, so
as an internal direct product. Further, since
is a subgroup of
,
has a subgroup, say
, isomorphic to
.
Then, consider the endomorphism of
that sends
to the trivial subgroup and
isomorphically to the subgroup
. This endomorphism does not send
to within itself.
More general example
Further information: Fully normalized and potentially fully invariant implies centralizer-annihilating endomorphism-invariant
More generally, suppose is a fully normalized subgroup of
that is normal in
, but such that there is a homomorphism
whose kernel contains
such that
is not contained in
(in other words,
is not a centralizer-annihilating endomorphism-invariant subgroup).
Then,
is not a potentially fully invariant subgroup of
.
Examples include:
-
is the dihedral group of order 16, say
, and
. Then,
and we have a homomorphism
from
to
such that
with kernel containing
and with
not contained in
. Thus,
is not a potentially fully invariant subgroup of
. Further information: D8 is not potentially fully invariant in D16