There exist subgroups of arbitrarily large subnormal depth

This is a statement of the form: there exist subnormal subgroups of arbitrarily large subnormal depth satisfying certain conditions.
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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

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

Proof

Example of the dihedral group

Further information: dihedral group

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

$G=⟨a,x\mid a^{2^k}=x^2=e,xa{x}^{-1}=a^{-1}⟩$.

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

$H=⟨x⟩=\{e,x\}$.

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

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

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.