Vrftjkry

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 $Xtimes X$, 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 $s= frac{1}{|G|}sum_{gin G} pi(g^{2}) = sum_{chi in Irr(G)} epsilon_{chi}m_{chi}$, 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!!!