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 $C(G)$.

Related facts

Similar facts

• Number of irreducible representations equals number of conjugacy classes (extreme case where the subgroup has no outer automorphisms)
• Number of orbits of irreducible representations equals number of orbits under automorphism group (extreme case where the subgroup is the whole automorphism group).
• Cyclic quotient of automorphism group by class-preserving automorphism group implies same orbit sizes of conjugacy classes and irreducible representations under automorphism group

Opposite facts

• Orbit sizes for irreducible representations may differ from orbit sizes for conjugacy classes under action of automorphism group

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] For every $a \in A$, the number of fixed points for both actions is the same. By Fact (2), the number of orbits has a formula that depends only on the number of fixed points of each element, so this formula yields the same number for both the group actions.