Normal not implies amalgam-characteristic

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) need not satisfy the second subgroup property (i.e., amalgam-characteristic subgroup)
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Statement

Verbal statement

A normal subgroup of a group need not be an amalgam-characteristic subgroup.

Statement with symbols

Let G be a group and H be a normal subgroup of G. Let L = G *_H G. Then, it is not necessary that H is characteristic in L.

Related facts

Converse

Similar facts

Opposite facts

Proof

Example of the free group

Let F be a free group on two generators and \mathbb{Z} be the group of integers. Let G = F \times \mathbb{Z} and H = F \times \{ 0 \} be the embedded first direct factor. We have:

L = (F \times \mathbb{Z}) *_{F \times \{ 0 \}} (F \times \mathbb{Z}) = F \times (\mathbb{Z} * \mathbb{Z}) \cong F \times F.

Thus, L is a direct product of two copies of the free group on two generators, and moreover, the embedded subgroup H in L is simply F \times \{ e \}, the first embedded direct factor. This is not a characteristic subgroup in L, because there exists an exchange automorphism swapping the two direct factors of L.