Characteristic not implies injective endomorphism-invariant in finitely generated abelian group

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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.

Related facts

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.