Characteristic not implies injective endomorphism-invariant in finitely generated 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 finitely generated abelian group. That is, it states that in a finitely generated abelian group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., injective endomorphism-invariant subgroup)
View all subgroup property non-implications | View all subgroup property implications

Statement

It is possible to have a finitely generated abelian group $G$ and a subgroup $H$ of $G$ such that $H$ is a characteristic subgroup of $G$ but is not an injective endomorphism-invariant subgroup of $G$, i.e., there exists an injective endomorphism $\sigma$ of $G$ such that $\sigma (H)$ is not contained in $H$.

Related facts

• Characteristic not implies injective endomorphism-invariant
• Center not is injective endomorphism-invariant
• Characteristic not implies fully invariant in finite abelian group

Proof

Further information: direct product of Z and Z2

Suppose $G$ is the direct product of the group of integers $\mathbb {Z}$ and the cyclic group $\mathbb {Z} /2\mathbb {Z}$. Suppose $H$ is the subgroup generated by the element $(2,1)$.

• $H$ is a characteristic subgroup of $G$: Let $A=\{x\mid \exists y,2x=4y\}$, $B=\{x\mid \exists y,x=2y\}$, , and $C=\{x\mid \exists y,2x=8y\}$. Then, the set $A\setminus (B\cup C)$ is the two-element set $\{(2,1),(-2,1)\}$. Both of these are generators of the subgroup $H$. Hence, $H$ is a characteristic subgroup of $G$.
• $H$ is not an injective endomorphism-invariant subgroup of $G$: Consider the injective endomorphism $\sigma$ defined by $\sigma ((x,y))=(2x,y)$. Then, $\sigma (H)$ is not contained in $H$, because $\sigma ((2,1))=(4,1)$, which is not in $H$.