Normal not implies amalgam-characteristic

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

Characteristic not implies amalgam-characteristic: This is a stronger fact, and examples of this also serve as examples of normal not implying amalgam-characteristic.
Direct factor not implies amalgam-characteristic
Cocentral not implies amalgam-characteristic

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 $L$ .