# A relation between self-paired orbitals of a group action and its associated permutation representation

Let $$G$$ be a finite group. Suppose $$G$$ acts on a finite set $$X$$. Consider the permutation representaion or character associated with the action, call it $$\pi$$. Since permutation character is the character of a real representation , any irreducible of type -1, must appear in it with even multiplicity. In particular we see that all the orbitals are self-paired if and only if all the irreducibles in $$\pi$$ are of type 1 and multiplicity 1.

I mention here, that type -1, type 1, refers to the Frobenius-Schur index. Also , by a orbital, we mean that orbits of the action of $$G$$ on $$X\times X$$, which is naturally defined using $$G$$ acting on $$X$$. Also self-paired orbital $$\Delta$$ is one in which $$(a,b)\in \Delta \implies (b,a)\in \Delta$$.

This is a paragraph from the book "Permutation Groups", by Peter Cameron( Pg-46). I don't understand the claims in the paragraph. First of all why should a character of type -1, must occur even number of times.

Next, I also don't understand why all the orbitals are self-paired if and only if all the irreducibles in $$\pi$$ are of type 1 and multiplicity 1. I have proved one part of this.

If $$\pi=\sum_{\chi \in Irr(G)} m_{\chi}\chi$$, we have that number of self-paired orbitals $$s= \frac{1}{|G|}\sum_{g\in G} \pi(g^{2}) = \sum_{\chi \in Irr(G)} \epsilon_{\chi}m_{\chi}$$, where $$\epsilon_{\chi}$$ is the Frobenius-Schur index of $$\chi$$. Let $$r$$ be the total number of orbitals of the action. Now, Let us assume that all the irreducible characters in $$\pi$$ are of multplicity 1 and type 1. Let us assume there are $$k$$ irreducible constituents of $$\pi$$. Then by the above formula, $$s=k$$. Also $$\langle \pi, \pi \rangle= \sum_{\chi \in Irr(G)}m_{\chi}^{2}=k$$. So $$\langle \pi, \pi \rangle=s$$. But, $$\langle \pi, \pi \rangle=r$$, is a standard result and henceforth, $$r=s$$, and so all are self-paired orbitals. But I couldn't determine the other direction.

Thanks in advance for any kind of help!!!