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

• Amalgam-characteristic implies potentially characteristic
• Amalgam-characteristic implies normal

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

• Finite normal implies amalgam-characteristic
• Central implies amalgam-characteristic
• Normal subgroup in hypercenter is amalgam-characteristic

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 $L$.