Endomorphism kernel does not satisfy intermediate subgroup condition

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This article gives the statement, and possibly proof, of a subgroup property (i.e., endomorphism kernel) not satisfying a subgroup metaproperty (i.e., intermediate subgroup condition).
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It is possible to have a group G and subgroups H \le K \le G such that:

  1. H is an endomorphism kernel in G, i.e., H is a normal subgroup of G and there is a subgroup M of G such that G/H \cong M.
  2. H is not an endomorphism kernel in K.


Finite example

Take the following:


Infinite example that always works

We combine the fact that normal not implies endomorphism kernel with the fact that normal iff potential endomorphism kernel (note that the roles of the letters K and G as used on that page are the reverse of the roles here).