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The Orbit-Stabilizer Theorem

The Orbit-Stabilizer Theorem. Stabilizers. Let G be a group of permutations on a set S. For each i in S, let stab G (i) be the set of permutations π in G that fix i. That is, stab G (i) = {π in G | π(i) = i}. stab (p). D 4. p. S. Visual Stabilizers.

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The Orbit-Stabilizer Theorem

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  1. The Orbit-Stabilizer Theorem

  2. 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. stab (p) D4 p S 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

  4. stab (p) D4 R0(S) R90(S) R180(S) R270(S) p S H(S) V(S) D(S) D'(S) Visual Stabilizers = {R0, H}

  5. stab (p) D4 R0(S) R90(S) R180(S) R270(S) p S H(S) V(S) D'(S) D(S) Visual Stabilizers = {R0, D}

  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. orb (p) D4 S Visual Orbits D(p) V(p) R0(p) R270(p) R90(p) R180(p) H(p) D'(p)

  10. 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}

  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.

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