Normal not implies amalgam-characteristic

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 $$L$$.

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 contained in hypercenter is amalgam-characteristic

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 $$L$$.