There exist subgroups of arbitrarily large subnormal depth

Statement
Let $$k$$ be a positive integer. Then, we can find a group $$G$$ and a subgroup $$H$$ such that $$H$$ is a $$k$$-subnormal subgroup of $$G$$ but is not a $$(k-1)$$-subnormal subgroup of $$G$$. In other words, the subnormal depth of $$H$$ in $$G$$ is precisely $$k$$. Equivalently, there exists a series of subgroups:

$$H = H_0 \le H_1 \le H_2 \le \dots \le H_k = G$$

with each $$H_i$$ normal in $$H_{i+1}$$, and there exists no series of length $$k-1$$ with the property.

Related facts

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

Example of the dihedral group
Let $$G$$ be the dihedral group of order $$2^{k+1}$$. Specifically, we have:

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

Let $$H$$ be the two-element subgroup generated by $$x$$:

$$H = \langle x \rangle = \{ e, x \}$$.


 * $$H$$ is $$k$$-subnormal in $$G$$. Consider the series:

$$H = \langle x \rangle \le \langle a^{2^{k-1}}, x \rangle \le \langle a^{2^{k-2}}, x \rangle \le \dots \le \langle a^2, x \rangle \le \langle a,x \rangle = G$$.

Each subgroup has index two in its predecessor, and is thus normal. The series has length $$k$$, so $$H$$ is $$k$$-subnormal in $$G$$.


 * $$H$$ is not $$(k-1)$$-subnormal in $$G$$: To see this, note that the above subnormal series is a subnormal series of minimum length, because, starting from the right, each subgroup is the normal closure of $$H$$ in the group to its right.