The Orbit-Stabilizer Theorem
Stabilizers
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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}
Visual Stabilizers
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
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.
p
S

Find stab (p)
D4
Visual Stabilizers
stab (p) = {R0, H}
D4
p
R0(S)
R90(S)
R180(S)
R270(S)
H(S)
V(S)
D(S)
D'(S)
S
Visual Stabilizers
stab (p) = {R0, D}
D4
p
R0(S)
R90(S)
R180(S)
R270(S)
H(S)
V(S)
D(S)
D'(S)
S
More Stabilizers
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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}
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.
Orbits
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
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}
Visual Orbits
orb (p)
D4
D(p) V(p)
R0(p)
R270(p)
R90(p)
H(p)
R180(p)
D'(p)
S
More Orbits
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
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Let G be the group of permutations:
 = (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}
Orbit-Stabilizer Theorem
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
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.
The correspondence
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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)).
Show  is one-to-one and onto
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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.