Transitive normality is not centralizer-closed

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

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 facts

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

Proof

Further information: symmetric group:S3

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 $G$ .