# Transitive normality is not centralizer-closed

## 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.

## 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 $G$.