Endomorphism kernel is not transitive

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 $$G$$.

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
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 $$G$$.