Transitive normality is not quotient-transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., transitively normal subgroup) not satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup property).
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Statement

Property-theoretic statement

The property of being a transitively normal subgroup is not a quotient-transitive subgroup property.

Statement with symbols

It is possible to have the following situation: $H\leq K\leq G$, $H$ is a transitively normal subgroup of a group $G$, and $K/H$ is a transitively normal subgroup of $G/H$, and $K$ is not a transitively normal subgroup of $G$.

Definitions used

Transitively normal subgroup

Further information: Transitively normal subgroup

We say that $H$ is a transitively normal subgroup of $G$ if whenever $K$ is a normal subgroup of $H$, $K$ is also a normal subgroup of $G$.

Related facts

Failure of quotient-transitivity for related properties

The same example used here applies to all these related properties:

• Central factor is not quotient-transitive
• SCAB is not quotient-transitive
• Conjugacy-closed normality is not quotient-transitive

Proof

Example of the dihedral group

Further information: dihedral group:D8

Let $G$ be the dihedral group, given by:

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

Define subgroups:

$H=\langle a^{2}\rangle ,\qquad K=\langle a^{2},x\rangle$.

$H$ is a subgroup of order two, hence it is transitively normal in $G$. $K/H$ is a subgroup of order two in $G/H$, hence it is transitively normal in $G$.

However, $K$ is not transitively normal in $G$, because the subgroup $\langle x\rangle$ of $K$ is normal in $K$ but not in $G$.