Endomorphism kernel is not transitive

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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., transitive subgroup property).
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Statement

It is possible to have groups H \le K \le G such that H is an endomorphism kernel in K, K is an endomorphism kernel in G, and H is not an endomorphism kernel in G.

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Proof

Further information: quaternion group, subgroup structure of quaternion group

Suppose G is the quaternion group, K is a cyclic maximal subgroup, and H is the center of quaternion group. Explicitly:

\! G = \{ 1,-1,i,-i,j,-j,k,-k \}

\! K = \{ 1,-1,i,-i \}

\! H = \{ 1,-1 \}

Then:

  1. K is an endomorphism kernel in G: In fact, G/K is isomorphic to H.
  2. H is an endomorphism kernel in K: In fact, K/H is isomorphic to H.
  3. H is not an endomorphism kernel in G: In fact, G/H is isomorphic to the Klein four-group which does not occur as a subgroup of G.