Finite normal implies amalgam-characteristic

Verbal statement
Any finite normal subgroup (i.e., a normal subgroup that is finite as a group) is an amalgam-characteristic subgroup.

Equivalently, a finite group is an always amalgam-characteristic group: it is an amalgam-characteristic subgroup in any group in which it is a normal subgroup.

Statement with symbols
Suppose $$H$$ is a finite normal subgroup of a group $$G$$. Then, $$H$$ is a characteristic subgroup inside the amalgam $$K := G *_H G$$.

Stronger facts

 * Weaker than::Finite normal implies amalgam-normal-subhomomorph-containing

Similar facts

 * Central implies amalgam-characteristic
 * Periodic normal implies amalgam-characteristic
 * Normal subgroup contained in hypercenter is amalgam-characteristic

Opposite facts

 * Normal not implies amalgam-characteristic
 * Characteristic not implies amalgam-characteristic
 * Direct factor not implies amalgam-characteristic
 * Cocentral not implies amalgam-characteristic

Applications

 * Finite normal implies potentially characteristic
 * Finite normal implies image-potentially characteristic

Facts used

 * 1) uses::Quotient of amalgamated free product by amalgamated normal subgroup equals free product of quotient groups
 * 2) uses::Free product of nontrivial groups has no nontrivial finite normal subgroup
 * 3) uses::Normality is upper join-closed
 * 4) uses::Normality satisfies image condition

Proof
Given: A group $$G$$, a finite normal subgroup $$H$$ of $$G$$. $$K := G *_H G$$.

To prove: $$H$$ is a characteristic subgroup in $$L$$.

Proof:


 * 1) By fact (1), $$K/H \cong G/H * G/H$$.
 * 2) $$K/H$$ has no nontrivial finite normal subgroup: By fact (2), if $$H$$ is a proper subgroup of $$G$$, then $$L/H$$ has no nontrivial finite normal subgroup. If $$H = G$$, then $$L/H$$ is trivial and hence has no nontrivial finite normal subgroup.
 * 3) $$H$$ is a finite normal subgroup of $$K$$: Since $$H$$ is normal in both copies of $$G$$, it is normal in $$K$$. Also, $$H$$ is finite.
 * 4) $$H$$ is the unique largest finite normal subgroup of $$K$$: Suppose $$M$$ is a finite normal subgroup of $$K$$. Then, by fact (4), the image of $$M$$ in the quotient map $$K \to K/H$$ is a normal subgroup of $$K/H$$. Also, since $$M$$ is finite, its image is finite. In step (2), we concluded that $$K/H$$ has no nontrivial finite normal subgroup. Thus, the image of $$M$$ is trivial, so $$M \le H$$.
 * 5) $$H$$ is characteristic in $$K$$: This follows since it is the unique largest finite normal subgroup of $$K$$.