# Normal not implies right-transitively fixed-depth subnormal

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup) need not satisfy the second subgroup property (i.e., right-transitively fixed-depth subnormal subgroup)
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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subnormal subgroup) need not satisfy the second subgroup property (i.e., right-transitively fixed-depth subnormal subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about subnormal subgroup|Get more facts about right-transitively fixed-depth subnormal subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property subnormal subgroup but not right-transitively fixed-depth subnormal subgroup|View examples of subgroups satisfying property subnormal subgroup and right-transitively fixed-depth subnormal subgroup

## Statement

There exists a group $G$ and a normal subgroup $K$ of $G$ such that $K$ is not right-transitively fixed-depth subnormal in $G$. In other words, for any $k \ge 1$, there exists a $k$-subnormal subgroup $H$ of $K$ such that $H$ is not $k$-subnormal in $G$.

## Proof

### Example of the infinite dihedral group

Let $G$ be the infinite dihedral group, i.e., the group given by: $G = \langle a,x \mid x^2 = 1, xax^{-1} = a^{-1} \rangle$.

Let $K = \langle a^2, x \rangle$. $K$ is a subgroup of index two in $G$, and is normal. For any $k \ge 1$, consider the subgroup: $H = \langle a^{2^{k+1}}, x \rangle$.

Then:

• The subnormal depth of $H$ in $K$ is $k$. In particular, $H$ is $k$-subnormal in $K$.
• The subnormal depth of $H$ in $G$ is $k + 1$. In particular, $H$ is not $k$-subnormal in $G$.