Number of irreducible representations over rationals equals number of equivalence classes under rational conjugacy

Statement

Suppose $G$ is a finite group whose splitting field is a cyclic extension of the rationals. Then, the following numbers are equal:

1. The number of irreducible linear representations of $G$ over the field of rational numbers.
2. The number of linear representations of $G$ over $\mathbb{C}$ with rational character values and for which no proper nonzero subrepresentation has rational character values.
3. The number of equivalence classes in $G$ under rational conjugacy.
4. The number of conjugacy classes of cyclic subgroups in $G$.

Note that the result is not true for arbitrary finite groups.

Caveats

The number of irreducible representations over rationals is not the same as the number of irreducible representations over the complex numbers that can be realized over the rationals. The latter number is either smaller or equal, and it is equal when the group is a rational group, which means that any two elements generating the same cyclic subgroup are conjugate.

Also, although the counts in (1) and (2) are equal, it is possible for a rational character to arise from an irreducible representation over the complex numbers that is not realized over the rationals. 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 reals equals number of equivalence classes under real conjugacy
• 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

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 $\mathbb{C}$. We basically set $K=\mathbb{Q}$, $L=\mathbb{Q}(\zeta )$ where $\zeta$ is a primitive $d^{th}$ root of unity so that $L$ is a cyclic extension of $K$, and now ...