Let a group $G$ act on a finite set $\Omega$. Suppose that the corresponding permutation character has a regular component. Does it follow that $\Omega$ has a regular $G$-orbit? (The converse is obviously true.)
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3 Answers
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No, this is false in general, and the smallest counterexample is the Klein 4-group $G=C_2\times C_2$. It can act transitively on a set of order 2 in three different ways, with 3 different $C_2$'s in the kernel. Call these sets $\Omega_1, \Omega_2, \Omega_3$, and their permutation characters $1+\chi_1, 1+\chi_2, 1+\chi_3$. Then the character of $\Omega=\Omega_1\coprod\Omega_2\coprod\Omega_3$ is $3+\chi_1+\chi_2+\chi_3$, which is 1+1+regular character of $G$, but $\Omega$ does not have a regular orbit.
answered May 12, 2012 at 6:24
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Not every non-cyclic finite group has such a relation. E.g., the quaternion group of order 8 doesn't.
(I meant this as a comment on Alex's answer, but I don't have enough reputation to comment.)
answered May 12, 2012 at 9:06
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$\begingroup$ I believe that all non-cyclic groups $G$ have the property that some two non-isomorphic permutation actions have the same character. But there's no guarantee that (exactly) one of the two permutation actions contains a regular orbit. $\endgroup$ Commented May 12, 2012 at 14:49
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No, it does not. As a simple example, let $G\cong S_3$, $\Omega = G/C_2 \sqcup G/C_2 \sqcup G/C_3$. You can easily check that the corresponding permutation character is isomorphic to the regular character plus two copies of the trivial. In other words, $\mathbb{C}[G/1] \oplus \mathbb{C}[G/G]^{\oplus 2} \cong \mathbb{C}[G/C_3] \oplus \mathbb{C}[G/C_2]^{\oplus 2}$.
One systematic source of such counterexamples are Brauer relations in which the regular set $G/1$ enters with non-zero coefficient, of which both Tim's and my answer are particular examples (for more on Brauer relations see e.g. this MO question, as well as google).
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$\begingroup$ @Alex: Thank you! I did not realize that there is a nontrivial and interesting theory behind all this. I still have one question though. You say that my question is equvalent to the existence of a nontrivial Brauer relation for $G$. Of course, if such a relation exists then the answer to the original question is negative. However, the converse is not so obvious to me. Suppose that the answer is negative, i.e. my permutation character $\chi=\sum_i\mathbb{C}[G/H_i]$, where all $H_i≠1$, has also the form $\chi$=$\rho$+$\theta$, where $\rho$ is the regular character and $\theta$ is ... (tbc) $\endgroup$– AnvitaCommented May 14, 2012 at 3:00
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$\begingroup$ (contd)...some other character. In order for this to be a Brauer relation, $\theta$ must be a permutation character. But how do we know that it really is one? $\endgroup$– AnvitaCommented May 14, 2012 at 3:01
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$\begingroup$ @Anvita You are right. Corrected. $\endgroup$– Alex B.Commented May 16, 2012 at 18:05