Let $G$ be a group.
A self-inverse conjugacy class $C$ of $G$ is a conjugacy class where if $g \in C$, $g^{-1} \in C$.
A real character $\chi$ of a group is one such that $\chi(g) \in \mathbb{R} \quad \forall g \in G$.
By considering the actions induced on the rows and on the columns of the character table by complex conjugation, I want to show that the number of irreducible real characters is equal to the number of self-inverse conjugacy classes.
Can I get a hint on how to do this? I am struggling to begin.
Warning: This question is asking nearly the exact same thing, but I do not believe this should be marked as a duplicate for these reasons:
- The author asks the question only for finite groups. I ask it for any group.
- The only answer is not accepted, highly upvoted, and also relies on the Brauer permutation lemma. I would like to see a hint to a solution using the direction I have given (i.e. considering the action on the character table). The comments on the proof also do not indicate a full answer.