The Orbit-Stabilizer Theorem

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}

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)

Visual Stabilizers

R0(S) R90(S) R180(S) R270(S)

H(S) V(S) D(S) D'(S)

S

p

D4

stab (p) = {R0, H}

Visual Stabilizers

R0(S) R90(S) R180(S) R270(S)

H(S) V(S) D(S) D'(S)

S

p

D4

stab (p) = {R0, D}

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}

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 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}

S

Visual Orbits

D4

orb (p)R0(p)

R90(p) R180(p)

R270(p)

D(p)

H(p) D'(p)

V(p)

More Orbits 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 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 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 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.