The Orbit-Stabilizer Theorem

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The Orbit-Stabilizer Theorem
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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 p in G that fix i. That is,
• stabG(i) p in G p(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

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Visual Stabilizers
R0, H
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Visual Stabilizers
R0, D
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More Stabilizers

• Let G be the group of permutations
• ? (1) p2 (124)
• p3 (142) p4 (35)
• p5 (124)(35) p6 (142)(35)
• stabG(1) ?, p4
• stabG(3) ?, p2, p3

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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) p(s) p in G

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Visual Orbits
D(p)
V(p)
R0(p)
R270(p)
R90(p)
R180(p)
H(p)
D'(p)
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More Orbits

• Let G be the group of permutations
• ? (1) p2 (124)
• p3 (142) p4 (35)
• p5 (124)(35) p6 (142)(35)
• stabG(1) ?, p4 orbG(1) 1,2,4
• stabG(3) ?, p2, p3 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) GstabG(i).
• But G GstabG(i) stabG(i) by Lagrange,
  and the result follows.

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The correspondence

• Let H stabG(i), and choose any permutation p in
  H.
• The correspondence between orbG(i) and the
  cosets of stabG(i) is given by ? where
• ?(p(i)) pH.
• In case there is another permutation ? with
  ?(i) p(i), we need to show ? is really a
  function, that is, that ?(p(i)) ?(?(i)).

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Show ? is one-to-one and onto

• But if p(i) ?(i), then ? -1p(i) ?(i) i
• so ? -1p is in stabG(i) H.
• Hence pH ?H
• So ?(p(i)) ?(?(i)) as required.
• To show ? is one-to-one,
• reverse the steps.
• Clearly ? is onto.
• This completes the proof.