No proper nontrivial transitively normal subgroup not implies simple

Statement
We can have a group that is not simple but has no proper nontrivial transitively normal subgroup.

Transitively normal subgroup
A subgroup $$H$$ of a group $$G$$ is termed transitively normal in $$G$$ if whenever $$K$$ is a normal subgroup of $$H$$, $$K$$ is normal in $$G$$.

Simple group
A $$G$$ is termed simple if $$G$$ has no proper nontrivial normal subgroup.

Example of the alternating group
Let $$G$$ be the alternating group on the set $$\{ 1,2,3,4 \}$$. $$G$$ has three normal subgroups: the whole group, the trivial subgroup, and the four-element subgroup:

$$H := \{, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$$.

$$H$$ is not transitively normal, because the two-element subgroup:

$$K := \{, (1,2)(3,4) \}$$

is normal in $$H$$ and not normal in $$G$$. Thus:


 * $$G$$ is not simple since it has a proper nontrivial normal subgroup, namely $$H$$.
 * $$G$$ has no proper nontrivial normal subgroup.

A more generic example
Let $$p$$ be such that $$2^p - 1$$ is a prime (in this case, it is a Mersenne prime). Consider the general affine group $$GA(1,2^p)$$: this is the semidirect product of the additive group of the field of order $$2^p$$ with its multiplicative group. The only proper nontrivial normal subgroup of this group is the additive group.

This subgroup is not transitively normal, because all its proper nontrivial subgroups are normal in it, but none of them are normal in the whole group. That is because the multiplicative group acts transitively on the non-identity elements, and hence cannot preserve any proper nontrivial subgroup.

The alternating group is a special case of the above where $$p = 2$$.