Normal not implies left-transitively fixed-depth subnormal

Statement
It is possible to have a group $$K$$ and a normal subgroup $$H$$ such that for every $$k \ge 1$$, there exists a group $$G$$ containing $$K$$ as a $$k$$-subnormal subgroup, but in which $$H$$ is not a $$k$$-subnormal subgroup.

Note that this also gives an example where $$K$$ is a subnormal subgroup of $$H$$, and for every $$k \ge 1$$, there exists a group $$G$$ containing $$K$$ as a $$k$$-subnormal subgroup, but in which $$H$$ is not a $$k$$-subnormal subgroup.

Related facts

 * Left transiter of normal is characteristic
 * Cofactorial automorphism-invariant implies left-transitively 2-subnormal
 * Stronger than::Normality is not transitive
 * Stronger than::There exist subgroups of arbitrarily large subnormal depth
 * Ascendant not implies subnormal, descendant not implies subnormal
 * Normal not implies right-transitively fixed-depth subnormal

Example of the dihedral group
Let $$K$$ be the dihedral group of order eight, given by:

$$K := \langle a,x \mid a^4 = x^2 = 1, xax = a^{-1} \rangle$$.

Let $$H$$ be the subgroup of $$K$$ generated by $$a^2$$ and $$x$$. $$H$$ has index two in $$K$$ and is hence normal. (subgroup of index two is normal).

For any $$k \ge 1$$ define $$G$$ as the group:

$$G := \langle b,x \mid b^{2^{k+2}} = x^2 = 1, xbx^{-1} = b^{-1} \rangle$$.

In other words, $$G$$ is a dihedral group of order $$2^{k+3}$$. Consider $$K$$ as a subgroup of $$G$$ by identifying $$a = b^{2^k}$$ and $$x = x$$. Then:


 * The subnormal depth of $$K$$ in $$G$$ is $$k$$.
 * The subnormal depth of $$H$$ in $$G$$ is $$k + 1$$.

This completes the proof.