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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.)

flag	asked May 12 2012 at 5:38 Anvita 297●1●6

Tim Dokchitser · Accepted Answer · 2012-05-12 06:24:22Z UTC

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.

Jeremy Rickard · Answer 2 · 2012-05-12 09:20:03Z UTC

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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.)

edited May 12 2012 at 9:20

answered May 12 2012 at 9:06

Jeremy Rickard
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	Good point, thank you! – Alex Bartel May 12 2012 at 10:28
	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. – Andreas Blass May 12 2012 at 14:49

Alex Bartel · Answer 3 · 2012-05-16 18:05:18Z UTC

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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).

edited May 16 2012 at 18:05

answered May 12 2012 at 6:25

Alex Bartel
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	@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) – Anvita May 14 2012 at 3:00
	(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? – Anvita May 14 2012 at 3:01
	@Anvita You are right. Corrected. – Alex Bartel May 16 2012 at 18:05

Permutation characters and regular orbits

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