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The Orbit-Stabilizer Theorem
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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}
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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)
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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)
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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)
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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}
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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.
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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}
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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)
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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}
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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.
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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)).
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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.
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