Monolith is fully invariant in co-Hopfian group

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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.

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Facts used


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.