Normality is not transitive for any pair of nontrivial quotient groups

Statement
Suppose $$A$$ and $$B$$ are (possibly equal) nontrivial groups. Then, there exist groups $$H \le K \le G$$ such that all the following conditions are satisfied:


 * $$H$$ is a fact about::normal subgroup of $$K$$ and the quotient group $$K/H$$ is isomorphic to $$A$$.
 * $$K$$ is a normal subgroup of $$G$$ and the quotient group is isomorphic to $$B$$.
 * $$H$$ is not a normal subgroup of $$G$$.

Similar facts

 * Normality is not transitive
 * Conjunction of normality with any nontrivial finite-direct product-closed property of groups is not transitive
 * Normality is not transitive for any nontrivially satisfied extension-closed group property
 * There exist subgroups of arbitrarily large subnormal depth

Generalizations

 * There exist subgroups of any given subnormal depth for any given tuple of nontrivial groups as quotient groups in a subnormal series

Proof
The construction is as follows. Let $$G$$ be the wreath product of $$A$$ and $$B$$ for the regular group action of $$B$$. Let $$K$$ be the subgroup $$A^B$$, i.e., the normal subgroup that forms the base of the semidirect product, and let $$H$$ be the subgroup of $$K$$ where a particular coordinate is the identity element. (If we are thinking of $$A^B$$ as functions from $$B$$ to $$A$$, then $$H$$ can be taken as the subgroup comprising those functions that send the identity element of $$B$$ to the identity element of $$A$$ -- here, the particular coordinate becomes the coordinate corresponding to the identity element of $$B$$).

Thus, $$K$$ is isomorphic to $$H \times A$$. Then:


 * $$H$$ is normal in $$K$$ and $$K/H$$ is isomorphic to $$A$$:
 * $$K$$ is normal in $$G$$ and $$G/K$$ is isomorphic to $$B$$:
 * $$H$$ is not normal in $$G$$: The action of $$B$$ on the coordinates in $$K = A^B$$ is transitive on the coordinates, so $$H$$ is not preserved under this action.