Retract not implies every permutable complement is normal

Statement
Suppose $$G$$ is a group and $$H$$ is a retract of $$G$$. In other words, $$H$$ has a normal complement in $$G$$: there exists a normal subgroup $$N$$ of $$G$$ such that $$N \cap H$$ is trivial and $$NH = G$$.

There may exist non-normal permutable complements to $$H$$ in $$G$$. In other words, there may exist a non-normal subgroup $$K$$ of $$G$$ such that $$HK = G$$ and $$H \cap K$$ is trivial.

Related facts

 * Retract not implies normal complements are isomorphic
 * Every group of given order is a permutable complement for symmetric groups

Example
Let $$G$$ be the symmetric group on the set $$\{ 1,2,3,4 \}$$, $$H$$ be the subgroup comprising permutations on $$\{ 1,2,3 \}$$. Then:


 * $$H$$ has a normal complement, namely the subgroup $$N = \{, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$$.
 * $$H$$ has a permutable complement that is not normal, namely the subgroup $$K = \{, (1,2,3,4), (1,3)(2,4), (1,4,3,2)\}$$.