# Normalizer of 2-subnormal subgroup may have 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

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.

## Facts used

1. There exist subgroups of arbitrarily large subnormal depth
2. Subnormality satisfies image condition: The image of a $k$-subnormal subgroup under a surjective homomorphism is $k$-subnormal in the image.
3. 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 $V$.
• 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 $G$.
• 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 $K$.