Characteristic not implies fully invariant in finite abelian group
This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finite abelian group. That is, it states that in a finite abelian group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., fully invariant subgroup)
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Contents
Statement
In a finite abelian group, a characteristic subgroup need not be a fully invariant subgroup.
Related facts
- Characteristic equals fully invariant in odd-order abelian group
- Characteristic not implies fully invariant in finitely generated abelian group
- Characteristic equals verbal in free abelian group
Proof
Let be the direct sum of the infinite cyclic group and the cyclic group of order two:
.
Let be the cyclic subgroup of
generated by
.
The subgroup is characteristic
Set:
and
.
and:
.
Thus, we have:
.
Clearly, any automorphism of sends
to itself, sends
to itself, and sends
to itself. Thus, any automorphism of
sends
to itself. Thus, any automorphism of
sends
to itself. Note that
comprises precisely those elements of
that have the second coordinate equal to
: in particular,
, so the subgroup generated by
equals
. Thus, any automorphism of
preserves
.
The subgroup is not fully invariant
Consider the map:
.
This map is an endomorphism of , but the image of
under this map is
, which is not an element of
. Thus,
is not fully invariant in
.