Periodic normal implies amalgam-characteristic

Statement
Suppose $$G$$ is a group and $$H$$ is a periodic normal subgroup of $$G$$: in other words, $$H$$ is a normal subgroup of $$G$$ as well as a periodic group (every element of $$H$$ is of finite order). Then, $$H$$ is an amalgam-characteristic subgroup of $$G$$: $$H$$ is characteristic in $$G *_H G$$.

Related facts

 * Finite normal implies amalgam-characteristic
 * Central implies amalgam-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 periodic normal subgroup
 * 3) uses::Normality is upper join-closed
 * 4) uses::Normality satisfies image condition

Proof
Given: A group $$G$$, a periodic 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 periodic normal subgroup: By fact (2), if $$H$$ is a proper subgroup of $$G$$, then $$L/H$$ has no nontrivial periodic normal subgroup. If $$H = G$$, then $$L/H$$ is trivial and hence has no nontrivial periodic 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 periodic 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 periodic, its image is periodic. In step (2), we concluded that $$K/H$$ has no nontrivial periodic 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 periodic normal subgroup of $$K$$.