1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Oct 22 '17 at 11:22
1
$\begingroup$

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 $\mathrm{Sym}(B)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.