Complemented characteristic not implies left-transitively complemented normal

Statement with symbols
It is possible to have the following situation: groups $$H \le K \le G$$ such that $$H$$ is a complemented characteristic subgroup of $$K$$ and $$K$$ is a complemented normal subgroup of $$G$$, but $$H$$ is not a complemented normal subgroup of $$G$$.

Related facts

 * Complemented normal is not transitive

Opposite facts

 * Characteristically complemented characteristic implies left-transitively complemented normal

Converse

 * Left-transitively complemented normal implies complemented characteristic

Facts used

 * 1) uses::Complemented normal satisfies intermediate subgroup condition

Example of the dihedral group of order sixteen
Let $$G$$ be a dihedral group of order $$16$$. In other words, $$G$$ is given by the presentation:

$$G = \langle a,x \mid a^8 = x^2 = e, xax = a^{-1} \rangle$$.

Consider the subgroups:

$$H = \langle a^2 \rangle, \qquad K = \langle a^2, x \rangle$$.


 * $$H$$ is complemented characteristic in $$K$$: $$K$$ is a dihedral group of order eight, with $$H$$ as the unique cyclic normal subgroup of order four, hence $$H$$ is characteristic in $$K$$. Further, $$\langle x \rangle$$ is a complement to $$H$$ in $$K$$.
 * $$K$$ is complemented normal in $$G$$: Since $$K$$ has index two, it is normal in $$G$$ (subgroup of index two is normal, or alternatively, nilpotent implies every maximal subgroup is normal). Further, the two-element subgroup $$\langle ax \rangle$$ is a permutable complement to $$K$$ in $$G$$.
 * $$H$$ is not complemented normal in $$G$$: If $$H$$ were complemented normal in $$G$$, it would also (by fact (1)) be complemented normal in the intermediate subgroup $$L = \langle a \rangle$$, which is a cyclic group of order eight. But we know that $$H$$ is not complemented in $$L$$, since any element of $$L$$ not in $$H$$ generates $$L$$.