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

Statement

The following are equal for a finite group $G$:

1. The number of characters of $G$ taking values in $\mathbb{R}$ arising from irreducible representations of $G$ over $\mathbb{R}$.
2. The number of characters of $G$ taking values in $\mathbb{R}$ arising from representations of $G$ over $\mathbb{C}$ such that no proper nonzero subrepresentation takes values entirely in $\mathbb{R}$.
3. The number of equivalence classes of $G$ under real conjugacy. Each such class arises as the union of a conjugacy class and the conjugacy class of inverse elements.
4. The number of homomorphisms from $\mathbb{Z}$ to $G$, up to equivalence of automorphisms of $\mathbb{Z}$ and inner automorphisms of $G$.

Caveats and corollaries

The number of irreducible representations over reals is not the same as the number of irreducible representations over the complex numbers that can be realized over the reals. The latter number is either smaller or equal, and it is equal when the group is an ambivalent group, which means that every element is conjugate to its inverse.

Also, although the counts in (1) and (2) are equal, it is possible for a real character to arise from an irreducible representation over the complex numbers that is not realized over the reals. However, some multiple of that representation can be realized over the reals. This explains the equality of counts in (1) and (2). The smallest multiple used is termed the Schur index.

Related facts

• Number of irreducible representations equals number of conjugacy classes
• Number of irreducible representations over complex numbers with real character values equals number of conjugacy classes of real elements
• Number of orbits of irreducible representations need not equal number of orbits of conjugacy classes under automorphism group

Particular cases

Small finite groups

Note that

Total number of irreducible representations over complex numbers = (Number of irreducible representations over the complex numbers with real character value) + 2(Number of conjugate pairs of complex numbers with non-real character value)

and

Number of irreducible representations over reals = (Number of irreducible representations over the complex numbers with real character value) + (Number of conjugate pairs of complex numbers with non-real character value)

Group	Order	Second part of GAP ID	Total number of irreducible representations = number of conjugacy classes	Number of irreducible representations over the complex numbers with real character value = number of conjugacy classes of real elements	Number of conjugate pairs of complex numbers with non-real character value = Number of pairs of conjugacy classes of non-real elements and their inverses	Total number of irreducible representations over the real numbers = Number of equivalence classes of elements up to real conjugacy
cyclic group:Z2	2	1	2	2	0	2
cyclic group:Z3	3	1	3	1	1	2
cyclic group:Z4	4	1	4	2	1	3
Klein four-group	4	2	4	4	0	4
cyclic group:Z5	5	1	5	1	2	3
symmetric group:S3	6	1	3	3	0	3
cyclic group:Z7	7	1	7	1	3	4
cyclic group:Z8	8	1	8	2	3	5
direct product of Z4 and Z2	8	2	8	4	2	6
dihedral group:D8	8	3	5	5	0	5
quaternion group	8	4	5	5	0	5
elementary abelian group:E8	8	5	8	8	0	8

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 $K=\mathbb{R}$ (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 cycles for the permutation of $C(G)$ induced by $g\mapsto g^{-1}$ is the number of equivalence classes in $G$ under real conjugacy.				By definition
3	The number of cycles for the permutation of $I(G)$ induced by complex conjugation is the number of irreducible representations over the reals.				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 cycles. Steps (2) and (3) now complete the proof.