Characteristic 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., characteristic subgroup) need not satisfy the second subgroup property (i.e., amalgam-characteristic subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about characteristic subgroup|Get more facts about amalgam-characteristic subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property characteristic subgroup but not amalgam-characteristic subgroup|View examples of subgroups satisfying property characteristic subgroup and amalgam-characteristic subgroup

Statement

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

Related facts

Proof

Example of the infinite dihedral group

Let G = D_\infty be the infinite dihedral group given by:

G = \mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}.

Let H be the subgroup of G given as the normal subgroup \mathbb{Z}. Note that H is characteristic in G, since it is the centralizer of derived subgroup. We have:

L = G *_H G = \mathbb{Z} \rtimes (\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z}) = \mathbb{Z} \rtimes D_\infty = (\mathbb{Z} \times \mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}

Note here that we're using the fact that the infinite dihedral group can be identified with the free product of two copies of the cyclic group of order two, whence either of these takes the role of the acting \mathbb{Z}/2\mathbb{Z}. The upshot is that we have:

L = (\mathbb{Z} \times \mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}

where the action is by the inverse map. Further, H is the first embedded direct factor \mathbb{Z} \times \{ 0 \} in \mathbb{Z} \times \mathbb{Z}.

Finally, observe that the coordinate exchange automorphism of \mathbb{Z} \times \mathbb{Z} extends to an automorphism of L, since it commutes with the inverse map, and can therefore be taken to fix the complementary \mathbb{Z}/2\mathbb{Z}.

Thus, H is not a characteristic subgroup of L.