Centralizer of a regular permutation group

I was reading the presentation "Regular and semiregular permutation groups and their centralizers and normalizers" by Tim Kohl, which is available online. For this post I will use the same definitions: a semi-regular group is one that acts without fixed points, a regular group is a semi-regular transitive group.

Consider $B = \operatorname{Sym}_X$, $N \leq B$ a regular subgroup and $C = C_B(N)$. Then it says that $C$ is also regular (slide 11).

I can see why $C$ is semi-regular, but why should it be transitive?

• The reason for this, which is in the presentation, is that the image of the left regular permutation representation of a group is centralized by the image of the right regular representation. – Derek Holt Oct 22 '17 at 11:22

If $B$ acts on $X$ regularly then given any $x\in X$ there is a bijection $B\to X$ given by $b\mapsto bx$, which can be used to define an isomorphism $\mathrm{Sym}(X)\to\mathrm{Sym}(B)$. The image of $B$ under this map will be $\lambda(B)$, where $\lambda:B\to \mathrm{Sym}(B)$ is the left regular action. Suppose $\sigma\in C_{\mathrm{Sym}(B)}(\lambda(B))$. Then observe $\sigma(b)=\sigma(b\cdot 1)=(\sigma\circ\lambda_b)(1)=(\lambda_b\circ\sigma)(1)=b\sigma(1)$, so if we write $c=\sigma(1)$, we have $\sigma(b)=bc$, i.e. $\sigma=\rho_c$ where $\rho:B\to\mathrm{Sym}(B)$ is the right regular action. Thus we conclude
$$C_{\mathrm{Sym}(B)}(\lambda(B))=\rho(B)$$
Of course, $\rho(B)$ is also a regular subgroup of $\mathrm{Sym}(B)$.