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.

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

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.


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