Transitive normality is not finite-join-closed

Verbal statement
The join of two transitively normal subgroups of a group need not be transitively normal.

Statement with symbols
Suppose $$G$$ is a group and $$H, K$$ are two transitively normal subgroups of $$G$$. Then, the join $$\langle H, K \rangle$$, which in this case equals the product $$HK$$, need not be a transitively normal subgroup.

Related metaproperty dissatisfactions for transitively normal subgroups

 * Transitive normality is not finite-intersection-closed
 * Transitive normality is not centralizer-closed

Related properties that are join-closed

 * Normality is strongly join-closed: In particular, this implies that any join of transitively normal subgroups is a normal subgroup, even though it need not be transitively normal.

Proof
Let $$A$$ be the cyclic group of order three. Let $$B$$ be the symmetric group of degree three, and $$C$$ be the subgroup of order three in $$B$$. Define:

$$G = A \times B, \qquad H = A \times 1, \qquad K = 1 \times C$$.

Then, we have:


 * $$H$$ and $$K$$ are both normal subgroups of prime order. In particular, the only normal subgroups they have are the whole group and the trivial subgroup, both of which are normal in $$G$$. Thus, $$H$$ and $$K$$ are both transitively normal.
 * $$HK$$ is not transitively normal: $$HK = A \times C$$. $$HK$$ is elementary abelian of order nine. Pick an isomorphism $$\sigma:A \to C$$ and consider the subgroup $$L$$ of $$HK$$ given by $$L = \{ (a, \sigma(a) \mid a \in A \}$$. Then, $$L$$ is normal in $$HK$$, since $$HK$$ is abelian. However, $$L$$ is not normal in $$G$$, because conjugation by an element in $$1 \times (B \setminus C)$$ sends $$(a,\sigma(a))$$ to $$(a,\sigma(a)^2)$$, which is not in $$L$$. Thus, $$HK$$ is not transitively normal.

Note that in this example, $$H$$ is the center of $$G$$ and is also a direct factor of $$G$$. This shows that taking the join of a transitively normal subgroup with the center or with a direct factor does not guarantee a transitively normal subgroup.