6
$\begingroup$

The question is in the title, but let me clarify the terminology. I consider a permutation group $\Sigma\subseteq\mathrm{Sym}(\Omega)$ on a finite set $\Omega$.

  • $\Sigma$ is regular if it acts transitively and freely on $\Omega$, i.e., for any two $i,j\in \Omega$ there is a unique $\sigma\in\Sigma$ with $\sigma(i)=j$.
  • $\Sigma$ is multiplicity-free if its permutation character (the character of the linear representation of $\Sigma$ by permutation matrices) is the sum of distinct irreducible characters.

Examples are the permutation groups generated by a single cyclic permutation.

$\endgroup$
1
  • 3
    $\begingroup$ The regular action is multiplicity free iff the group is abelian $\endgroup$ – Benjamin Steinberg Mar 3 at 1:31
9
$\begingroup$

These are the abelian regular permutation groups. The permutation character in this case is the character of the regular representation and in the regular representation a character appears with multiplicity equal to the dimension of the irreducible representation. So it can be multiplicity free iff all irreducibles are one dimensional which is equivalent to abelian.

$\endgroup$

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.