Homomorph-containing not implies no nontrivial homomorphism to quotient group

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., homomorph-containing subgroup) need not satisfy the second subgroup property (i.e., normal subgroup having no nontrivial homomorphism to its quotient group)
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Statement

It is possible to have a group G and a subgroup H such that:

  1. H is a homomorph-containing subgroup in G, i.e., for any homomorphism in \operatorname{Hom}(H,G) the image of H is contained in H.
  2. H is not a normal subgroup having no nontrivial homomorphism to its quotient group. In other words, there exists a nontrivial homomorphism from H to G/H.

Proof

Take the following:

Then:

  • H is homomorph-containing in G: This is because the image of H under any homomorphism must also have exponent at most two, hence must be inside H.
  • There is a nontrivial homomorphism from H to G/H: In fact, both are isomorphic to cyclic group:Z2.