# 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 $L$.

## 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 $L$.