Free product of nontrivial groups has no nontrivial periodic normal subgroup

Statement
Suppose $$G_1$$ and $$G_2$$ are nontrivial groups, and $$L = G_1 * G_2$$. Then, $$L$$ has no nontrivial periodic normal subgroup.

Related facts

 * Free product of nontrivial groups has no nontrivial finite normal subgroup
 * Free product of nontrivial groups is centerless

Applications

 * Periodic normal implies amalgam-characteristic
 * Periodic normal implies potentially characteristic

Proof
Given: Nontrivial groups $$G_1$$ and $$G_2$$. $$L$$ is the free product of $$G_1$$ and $$G_2$$.

To prove: If $$M$$ is a periodic (i.e., every element has finite order) normal subgroup of $$L$$, then $$M$$ is trivial.

Proof: Suppose $$g \in M$$ is a nontrivial word.


 * 1) If $$g$$ has even length, then $$g$$ has infinite order, since any finite positive power of $$g$$ is obtained simply by concatenating the word for $$g$$ with itself, and this is a non-identity word.
 * 2) If $$g$$ has odd length, then its first and last letter are both in the same free factor. Suppose both the first and last letter are in $$G_1$$. Then, pick any non-identity element $$a \in G_2$$. The commutator of $$g$$ and $$a$$ is a non-identity word of even length. Since $$M$$ is normal, this element is also in $$M$$. Step (1) yields that it has infinite order. A similar argument follows if the first and last letter of $$g$$ are in $$G_2$$.

Thus, $$M$$ has an element of infinite order, hence cannot be a periodic normal subgroup.