# Normal not implies left-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., left-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., left-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 left-transitively fixed-depth subnormal subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property subnormal subgroup but not left-transitively fixed-depth subnormal subgroup|View examples of subgroups satisfying property subnormal subgroup and left-transitively fixed-depth subnormal subgroup

## Statement

It is possible to have a group $K$ and a normal subgroup $H$ such that for every $k \ge 1$, there exists a group $G$ containing $K$ as a $k$-subnormal subgroup, but in which $H$ is not a $k$-subnormal subgroup.

Note that this also gives an example where $K$ is a subnormal subgroup of $H$, and for every $k \ge 1$, there exists a group $G$ containing $K$ as a $k$-subnormal subgroup, but in which $H$ is not a $k$-subnormal subgroup.

## Proof

### Example of the dihedral group

Further information: dihedral group:D8

Let $K$ be the dihedral group of order eight, given by:

$K := \langle a,x \mid a^4 = x^2 = 1, xax = a^{-1} \rangle$.

Let $H$ be the subgroup of $K$ generated by $a^2$ and $x$. $H$ has index two in $K$ and is hence normal. (subgroup of index two is normal).

For any $k \ge 1$ define $G$ as the group:

$G := \langle b,x \mid b^{2^{k+2}} = x^2 = 1, xbx^{-1} = b^{-1} \rangle$.

In other words, $G$ is a dihedral group of order $2^{k+3}$. Consider $K$ as a subgroup of $G$ by identifying $a = b^{2^k}$ and $x = x$. Then:

• The subnormal depth of $K$ in $G$ is $k$.
• The subnormal depth of $H$ in $G$ is $k + 1$.

This completes the proof.