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)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about left-transitively fixed-depth subnormal subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not left-transitively fixed-depth subnormal subgroup|View examples of subgroups satisfying property normal subgroup and left-transitively fixed-depth subnormal subgroup

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 $k\geq 1$, 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 $k\geq 1$, there exists a group $G$ containing $K$ as a $k$-subnormal subgroup, but in which $H$ is not a $k$-subnormal subgroup.

Related facts

• Left transiter of normal is characteristic
• Cofactorial automorphism-invariant implies left-transitively 2-subnormal
• Normality is not transitive
• There exist subgroups of arbitrarily large subnormal depth
• Ascendant not implies subnormal, descendant not implies subnormal
• Normal not implies right-transitively fixed-depth subnormal

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 $k\geq 1$ 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.