Normalizer of 2-subnormal subgroup may have arbitrarily large subnormal depth

Statement
Suppose $$k$$ is a positive integer. Then, we can find a 2-subnormal subgroup of a finite group whose normalizer is $$(k+1)$$-subnormal but not $$k$$-subnormal.

Related facts

 * Abnormal normalizer and 2-subnormal not implies normal: We can have a 2-subnormal subgroup that is not normal, and whose normalizer is abnormal -- very much the opposite of being subnormal.

Facts used

 * 1) uses::There exist subgroups of arbitrarily large subnormal depth
 * 2) uses::Subnormality satisfies image condition: The image of a $$k$$-subnormal subgroup under a surjective homomorphism is $$k$$-subnormal in the image.
 * 3) uses::Subnormality satisfies inverse image condition: The inverse image of a $$k$$-subnormal subgroup for any homomorphism is $$k$$-subnormal.

Proof
The construction is as follows:


 * 1) (uses: Fact (1)): Find a finite group $$G$$ and a subgroup $$H$$ of $$G$$ such that $$H$$ is $$(k + 1)$$-subnormal in $$G$$ but is not $$k$$-subnormal in $$G$$.
 * 2) Let $$V$$ be the additive group of the group ring of $$G$$ over any finite field, and consider the semidirect product $$K := V \rtimes G$$ where $$G$$ acts by left multiplication. Then, $$V$$ is an Abelian normal subgroup of $$G$$, so any subgroup of $$V$$ is 2-subnormal in $$G$$.
 * 3) Let $$W$$ be the subspace of $$V$$ spanned by the basis elements corresponding to $$H$$. $$W$$ is 2-subnormal in $$G$$, and its normalizer in $$K = V \rtimes G$$ is $$V \rtimes H$$. Thus, we have a 2-subnormal subgroup whose normalizer in the whole group is the subgroup $$V \rtimes H$$.
 * 4) Consider the projection map from $$K = V \rtimes G$$ to $$G$$ with kernel $$V$$.
 * 5) * By fact (2), we observe that if $$V \rtimes H$$ is $$k$$-subnormal in $$K$$, then $$H$$ is $$k$$-subnormal in $$G$$. Thus, $$V \rtimes H$$ is not $$k$$-subnormal in $$G$$.
 * 6) * By fact (3), we find that since $$H$$ is $$(k+1)$$-subnormal in $$G$$, its full inverse image $$V \rtimes H$$ is $$(k+1)$$-subnormal in $$K$$.