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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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 $G$ be the direct sum of the infinite cyclic group and the cyclic group of order two:

$G := \mathbb{Z}/8\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.

Let $H$ be the cyclic subgroup of $G$ generated by $(2,1)$.

The subgroup is characteristic

Set:

$A := \{g \in G \mid \exists x \in G, 4x = 2g \} = \{ (2r,s) \mid r \in \mathbb{Z}/8\mathbb{Z}, s \in \mathbb{Z}/2\mathbb{Z} \}$

and

$B := \{g \in G \mid \exists x \in G, g = 2x \} = \{ (2r,0) \mid r \in \mathbb{Z}/8\mathbb{Z} \}$.

and:

$C := \{ g \in G \mid \exists x \in G, 8x = 2g \} = \{ (4r,s) \mid r \in \mathbb{Z}/8\mathbb{Z}, s \in \mathbb{Z}/2\mathbb{Z} \}$.

Thus, we have:

$D := A \setminus (B \cup C) = \{ (2r,1) \mid r \in \mathbb{Z}/8\mathbb{Z}, r \notin 2\mathbb{Z}/8\mathbb{Z} \}$.

Clearly, any automorphism of $G$ sends $A$ to itself, sends $B$ to itself, and sends $C$ to itself. Thus, any automorphism of $G$ sends $A \setminus (B \cup C)$ to itself. Thus, any automorphism of $G$ sends $D$ to itself. Note that $D$ comprises precisely those elements of $H$ that have the second coordinate equal to $1$: in particular, $(2,1) \in D \subseteq H$, so the subgroup generated by $D$ equals $H$. Thus, any automorphism of $G$ preserves $H$.

The subgroup is not fully invariant

Consider the map:

$\pi_1:G \to G, \qquad \pi_1(a,b) = (a,0)$.

This map is an endomorphism of $G$, but the image of $(2,1)$ under this map is $(2,0)$, which is not an element of $H$. Thus, $H$ is not fully invariant in $G$.