Normal not implies right-transitively fixed-depth subnormal

Statement
There exists a group $$G$$ and a normal subgroup $$K$$ of $$G$$ such that $$K$$ is not right-transitively fixed-depth subnormal in $$G$$. In other words, for any $$k \ge 1$$, there exists a $$k$$-subnormal subgroup $$H$$ of $$K$$ such that $$H$$ is not $$k$$-subnormal in $$G$$.

Related facts

 * Stronger than::Normality is not transitive
 * Stronger than::There exist subgroups of arbitrarily large subnormal depth
 * Descendant not implies subnormal, Ascendant not implies subnormal
 * Normal not implies left-transitively fixed-depth subnormal
 * Weaker than::Characteristic not implies right-transitively fixed-depth subnormal

Example of the infinite dihedral group
Let $$G$$ be the infinite dihedral group, i.e., the group given by:

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

Let $$K = \langle a^2, x \rangle$$. $$K$$ is a subgroup of index two in $$G$$, and is normal. For any $$k \ge 1$$, consider the subgroup:

$$H = \langle a^{2^{k+1}}, x \rangle$$.

Then:


 * The subnormal depth of $$H$$ in $$K$$ is $$k$$. In particular, $$H$$ is $$k$$-subnormal in $$K$$.
 * The subnormal depth of $$H$$ in $$G$$ is $$k + 1$$. In particular, $$H$$ is not $$k$$-subnormal in $$G$$.