Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

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.

share|improve this answer
add comment

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.)

share|improve this answer
Good point, thank you! –  Alex B. May 12 '12 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 '12 at 14:49
add comment

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).

share|improve this answer
@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 '12 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 '12 at 3:01
@Anvita You are right. Corrected. –  Alex B. May 16 '12 at 18:05
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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