Endomorphism kernel is not transitive

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 $G$.

Related facts

Similar facts about similar properties

• Normality is not transitive
• Complemented normal is not transitive
• Permutably complemented is not transitive

Opposite facts about similar properties

• Direct factor is transitive
• Endomorphism image is transitive

Combined facts with other properties

• Endomorphism kernel of direct factor implies endomorphism kernel

Related facts about endomorphism kernel

• Endomorphism kernel is quotient-transitive
• Endomorphism kernel is not finite-intersection-closed

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 $G$.