Monolith is fully invariant in co-Hopfian group

Statement

If a Co-Hopfian group (?) (for instance, a Finite group (?)) has a 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

• 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: $\sigma (N)\leq N$.

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 $\sigma (N)\leq N$.

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 $N\leq \sigma ^{-1}(N)$, so $\sigma (N)\leq N$, completing the proof.