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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!!!

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