Transitive normality is not centralizer-closed

Statement with symbols
Suppose $$H$$ is a transitively normal subgroup of a group $$G$$. Then, it is not necessary that $$C_G(H)$$ (the centralizer of $$H$$ in $$G$$) is transitively normal.

Related centralizer-closed properties

 * Normality is centralizer-closed: Since transitively normal implies normal, this shows that the centralizer of a transitively normal subgroup, even though not necessarily transitively normal, is still normal.
 * Central factor is centralizer-closed: Since central factor implies transitively normal, this shows that the centralizer of a central factor is transitively normal.

Related metaproperty dissatisfactions for transitively normal subgroups

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

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

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

Then, $$H$$ is a normal subgroup of prime order, hence is transitively normal. On the other hand, the centralizer of $$H$$ in $$G$$ is $$K = A \times C$$, which is not transitively normal. To see this, let $$\sigma$$ be an isomorphism from $$A$$ to $$C$$. Consider:

$$L := \{ (a, \sigma(a)) \mid a \in A \}$$.

$$L$$ is a normal subgroup of $$K$$ (since $$K$$ is abelian). On the other hand, conjugating an element of $$L$$ by an element of $$1 \times B \setminus C $$ sends $$(a,\sigma(a))$$ to $$(a,\sigma(a)^2)$$, which is not in $$L$$. Hence, $$L$$ is not normal in $$G$$. Thus, $$K$$ is not transitively normal in $$G$$.