1. The Orbit-Stabilizer Theorem

2. stabG(i) = {π in G | π(i) = i}
Stabilizers
Let G be a group of permutations on a set S.
For each i in S, let stabG(i) be the set of permutations π in G that fix i. That is,
stabG(i) = {π in G | π(i) = i}

3. Visual Stabilizers
Think of D4 as a group of permutations acting on a square region, S.
Let p be the point in S indicated by the red dot.
Find S p D4
stab (p)

4. Visual Stabilizers stab (p) = {R0, H} D4 R270(S) R0(S) R90(S) R180(S)
Think of D4 as a group of permutations on a square region, S
H(S)
V(S)
D(S)
D'(S)

5. Visual Stabilizers stab (p) = {R0, D} D4 R0(S) R90(S) R180(S) R270(S)
Think of D4 as a group of permutations on a square region, S
H(S)
V(S)
D(S)
D'(S)

6. More Stabilizers Let G be the group of permutations:
 = (1) π2 = (124)
π3 = (142) π4 = (35)
π5 = (124)(35) π6 = (142)(35)
stabG(1) = {, π4}
stabG(3) = {, π2, π3}

7. stabG(a) is a subgroup of G.
Proof: Let us use the two-step test.
(a) = a, so stabG(a) is not empty.
Choose any  in stabG(a). Then
(a) = (a) since  in stabG(a).
= a since  in stabG(a).
So stabG(a) is closed.
Since (a) = a,  -1(a) = a, so stabG(a) is closed under inverses.
By the two-step test, stabG(a) ≤ G.

8. Orbits Let G be a group of permutations on a set S.
For each s in S, the orbit of s under G, denoted orbG(s) = {π(s) | π in G}

9. Visual Orbits orb (p) D(p) V(p) R0(p) S R270(p) R90(p) R180(p) H(p)
Think of D4 as a group of permutations on a square region, S
H(p)
D'(p)

10. More Orbits Let G be the group of permutations:  = (1) π2 = (124)
 = (1) π2 = (124)
π3 = (142) π4 = (35)
π5 = (124)(35) π6 = (142)(35)
stabG(1) = {, π4} orbG(1) = {1,2,4}
stabG(3) = {, π2, π3} orbG(3) = {3,5}

11. Orbit-Stabilizer Theorem
Let G be a finite group of permutations on a set S. Then for any i in S,
|G| =|orbG(i)| |stabG(i)|
Proof: We will show there is a one-to-one correspondence between orbG(i) and the cosets of stabG(i).
Then |orbG(i)| = |G:stabG(i)|.
But |G| = |G:stabG(i)| |stabG(i)| by Lagrange, and the result follows.

12. The correspondence
Let H = stabG(i), and choose any permutation π in H.
The correspondence between orbG(i) and the cosets of stabG(i) is given by  where
(π(i)) = πH.
In case there is another permutation  with (i) = π(i), we need to show  is really a function, that is, that (π(i)) = ((i)).

13. Show  is one-to-one and onto
But if π(i) = (i), then  -1π(i) = (i) = i
so  -1π is in stabG(i) = H.
Hence πH = H
So (π(i)) = ((i)) as required.
To show  is one-to-one,
reverse the steps.
Clearly  is onto.
This completes the proof.