# 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)
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## 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 $H$.

## 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 $H$.