Number of irreducible representations over complex numbers with real character values equals number of conjugacy classes of real elements

Statement

Suppose $G$ is a finite group. Then, the following numbers are equal:

1. The number of irreducible representations of $G$ over the complex numbers whose characters are real-valued. Note that this includes both real representations (representations realized over $\R$), and quaternionic representations, which are not realized over $\mathbb{R}$ but whose double is realized over $\R$ (so they have Schur index 2).
2. The number of conjugacy classes in $G$ of real elements, i.e., elements that are conjugate to their inverses.

Related facts

• Number of irreducible representations equals number of conjugacy classes
• Number of irreducible representations over reals equals number of equivalence classes under real conjugacy

Facts used

1. Application of Brauer's permutation lemma to Galois automorphism on conjugacy classes and irreducible representations (follows in turn from Brauer's permutation lemma): Suppose $G$ is a finite group and $r$ is an integer relatively prime to the order of $G$. Suppose $K$ is a field and $L$ is a splitting field of $G$ of the form $K(\zeta)$ where $\zeta$ is a primitive $d^{th}$ root of unity, with $d$ also relatively prime to $r$ (in fact, we can arrange $d$ to divide the order of $G$ because sufficiently large implies splitting). Suppose there is a Galois automorphism of $L/K$ that sends $\zeta$ to $\zeta^r$. Consider the following two permutations:
   • The permutation on the set of conjugacy classes of $G$, denoted $C(G)$, induced by the mapping $g \mapsto g^r$.
   • The permutation on the set of irreducible representations of $G$ over $L$, denoted $I(G)$, induced by the Galois automorphism of $L$ that sends $\zeta$ to $\zeta^r$.

Then, these two permutations have the same cycle type. In particular, they have the same number of cycles, and the same number of fixed points, as each other.

Proof

Given: A finite group $G$

To prove: The number of irreducible representations of $G$ over the real numbers equals the number of equivalence classes of elements of $G$ under real conjugacy.

Proof: Let $C(G)$ be the set of conjugacy classes of $G$ and $I(G)$ be the set of irreducible representations of $G$ over $\mathbb{C}$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The cycle type of the permutation of $C(G)$ induced by $g \mapsto g^{-1}$ is the same as the cycle type of the permutation of $I(G)$ induced by post-composing with complex conjugation.	Fact (1)			[SHOW MORE] Set $K = \R$ (the base field), $r = -1$, and $d = 4$ so $\zeta = i$ and $L = \mathbb{C}$. Then we get the result from Fact (1).
2	The number of fixed points for the permutation of $C(G)$ induced by $g \mapsto g^{-1}$ is the number of conjugacy classes of real elements in $G$.				By definition
3	The number of fixed points for the permutation of $I(G)$ induced by complex conjugation is the number of irreducible representations over the complex numbers with real character values.				PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
4	The result follows			Steps (1), (2), (3)	By Step (1), the permutation of $C(G)$ and of $I(G)$ have the same cycle type, hence the same number of fixed points. Steps (2) and (3) now complete the proof.