Characteristic not implies fully invariant

Statement
A characteristic subgroup of a group need not be a fully invariant subgroup.

For particular kinds of groups

 * Every nontrivial characteristic subgroup is potentially characteristic-and-not-fully invariant
 * Every nontrivial normal subgroup is potentially characteristic-and-not-fully invariant

Some kinds of characteristic subgroups that are not fully invariant

 * Center not is fully invariant
 * Characteristic direct factor not implies fully invariant
 * Characteristic not implies fully invariant in finite abelian group
 * Characteristic not implies fully invariant in odd-order class two p-group
 * Characteristic not implies fully invariant in class three maximal class p-group

Opposite facts

 * Characteristic equals fully invariant in odd-order abelian group

Some subgroup-defining functions yield fully invariant subgroups
All members of the derived series and the lower central series of a group are fully characteristic. This follows from the fact that they are all verbal subgroups (i.e., can be described as being generated by words of a certain form.

The example of the dihedral group
In the dihedral group of order eight, there exists a cyclic characteristic subgroup of order four that is not fully characteristic. Specifically, if we have:

$$G := \langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle$$

then the subgroup generated by $$a$$ is characteristic. On the other hand, it is not fully characteristic: consider the homomorphism that sends both $$a$$ and $$x$$ to $$x$$. This is a retraction to the two-element subgroup $$\{ x, e \}$$, and under this retraction, the subgroup generated by $$a$$ is not invariant (since $$a$$ gets mapped to $$x$$).

The example of the center, or a characteristic direct factor
If $$G$$ is a nontrivial centerless group and $$A$$ is an Abelian group isomorphic to a nontrivial Abelian subgroup $$B$$ of $$G$$, then we have:


 * $$A$$ is characteristic in $$G$$: In fact, $$A$$ is the center of $$G \times A$$.
 * $$A$$ is not fully characteristic in $$G$$: Consider the endomorphism with kernel $$G$$, mapping $$A$$ isomorphically to $$B$$. $$A$$ is not invariant under this endomorphism.