# Transitive normality is not quotient-transitive

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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: $H \le K \le G$, $H$ 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:

## 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 $G$.