Monolith is fully invariant in co-Hopfian group

Statement
If a fact about::co-Hopfian group (for instance, a fact about::finite group) has a fact about::monolith (a minimal normal subgroup contained in every nontrivial normal subgroup) then the monolith is a fully characteristic subgroup.

Related facts

 * Monolith is strictly characteristic

Applications

 * Self-centralizing and minimal normal implies fully invariant in co-Hopfian group

Facts used

 * uses::Normality satisfies inverse image condition

Proof
Given: A co-Hopfian group $$G$$, a minimal normal subgroup $$N$$ contained in every nontrivial normal subgroup of $$G$$, an endomorphism $$\sigma$$ of $$G$$.

To prove: $$\sigma(N) \le N$$.

Proof: If $$\sigma$$ is not injective, it has a kernel. The kernel is a nontrivial normal subgroup, so it contains $$N$$, so $$\sigma(N)$$ is trivial, and hence $$\sigma(N) \le N$$.

If $$\sigma$$ is injective, then its image is a subgroup of $$G$$ isomorphic to $$G$$. Since we assumed $$G$$ to be co-Hopfian, $$\sigma(G) = G$$, so $$\sigma$$ is surjective. But then, by fact (1), $$\sigma^{-1}(N)$$ is nontrivial and normal, so $$N \le \sigma^{-1}(N)$$, so $$\sigma(N) \le N$$, completing the proof.