Number of orbits of irreducible representations equals number of orbits of conjugacy classes under any subgroup of automorphism group
From Groupprops
Statement
Suppose is a finite group,
is a splitting field for
, and
is a subgroup of the automorphism group of
. Denote by
the set of conjugacy classes in
and by
the set of (equivalence classes of) irreducible representations of
over
. Then,
acts naturally on both
and
.
The claim is that the number of orbits under the action of on
equals the number of orbits under the action of
on
.
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
Facts used
- 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)
- Orbit-counting theorem (also called Burnside's lemma)
Proof
Given: A finite group , a splitting field
for
, a subgroup
of the automorphism group of
.
denotes the set of conjugacy classes of
,
denotes the set of equivalence classes of irreducible representations of
over
.
To prove: The number of orbits of under the natural action of
equals the number of orbits of
under the action of
.
Proof:
No. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | For any ![]() ![]() ![]() ![]() ![]() |
Fact (1) | Fact-direct | ||
2 | For any ![]() ![]() ![]() ![]() ![]() |
Step (1) | Follows from the previous step, since the fixed points are just the length one cycle. | ||
3 | The number of orbits in ![]() ![]() ![]() ![]() |
Fact (2) (orbit-counting theorem) | Step (2) | [SHOW MORE] |