# Number of orbits of irreducible representations equals number of orbits of conjugacy classes under any subgroup of automorphism group

## Statement

Suppose $G$ is a finite group, $K$ is a splitting field for $G$, and $A$ is a subgroup of the automorphism group of $G$. Denote by $C(G)$ the set of conjugacy classes in $G$ and by $R(G)$ the set of (equivalence classes of) irreducible representations of $G$ over $K$. Then, $A$ acts naturally on both $C(G)$ and $R(G)$.

The claim is that the number of orbits under the action of $A$ on $R(G)$ equals the number of orbits under the action of $A$ on $C(G)$.

## Facts used

1. Application of Brauer's permutation lemma to group automorphism on conjugacy classes and irreducible representations, which in turn uses Brauer's permutation lemma (actually, we don't need the full strength of the statement about cycle types for our purpose, we simply need the fixed point version, which can be deduced even more directly)
2. Orbit-counting theorem (also called Burnside's lemma)

## Proof

Given: A finite group $G$, a splitting field $K$ for $G$, a subgroup $A$ of the automorphism group of $G$. $C(G)$ denotes the set of conjugacy classes of $G$, $R(G)$ denotes the set of equivalence classes of irreducible representations of $G$ over $K$.

To prove: The number of orbits of $C(G)$ under the natural action of $A$ equals the number of orbits of $R(G)$ under the action of $A$.

Proof:

No. Assertion/construction Facts used Given data used Previous steps used Explanation
1 For any $a \in A$, the cycle type for the permutation induced by $a$ on $C(G)$ is the same as the cycle type of the permutation induced by $a$ on $R(G)$. Fact (1) Fact-direct
2 For any $a \in A$, the number of fixed points for the permutation induced by $a$ on $C(G)$ equals the number of fixed points of the permutation induced by $a$ on $R(G)$. Step (1) Follows from the previous step, since the fixed points are just the length one cycle.
3 The number of orbits in $C(G)$ under the action of $A$ equals the number of orbits in $R(G)$ under the action of $A$. Fact (2) (orbit-counting theorem) Step (2) [SHOW MORE]