Simple not implies co-Hopfian

Statement
A simple group need not be a co-Hopfian group: it is possible to have a simple group that is isomorphic to a proper subgroup of itself.

Facts used

 * 1) uses::Finitary alternating groups are simple

Example of the finitary alternating group
Let $$A$$ be an infinite set and $$B$$ a proper subset of $$A$$ of the same cardinality. Let $$\varphi:A \to B$$ be a bijection. Let $$G$$ be the finitary alternating group on $$A$$ and $$H$$ be the subgroup of $$G$$ comprising those permutations that fix the complement of $$B$$. Note that $$H$$ can be naturally identified with the finitary alternating group on $$B$$.

Then, $$\varphi$$ induces an isomorphism from $$G$$ to $$H$$. Thus, $$G$$ is isomorphic to the proper subgroup $$H$$. by fact (1), we thus have a simple group that is isomorphic to a proper subgroup of itself.