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

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## 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.

## 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]
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]
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.