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 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 $G$.

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 $A$: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
• $K$ is normal in $G$ and $G/K$ is isomorphic to $B$: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
• $H$ 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.