11/13/2014

Ma/CS 6a Class 20: , Orbits, and Stabilizers

By Adam Sheffer

A Group

 A group consists of a set 퐺 and a binary operation ∗, satisfying the following. ◦ Closure. For every 푥, 푦 ∈ 퐺, we have 푥 ∗ 푦 ∈ 퐺. ◦ Associativity. For every 푥, 푦, 푧 ∈ 퐺, we have 푥 ∗ 푦 ∗ 푧 = 푥 ∗ 푦 ∗ 푧 . ◦ Identity. The exists 푒 ∈ 퐺, such that for every 푥 ∈ 퐺, we have 푒 ∗ 푥 = 푥 ∗ 푒 = 푥. ◦ Inverse. For every 푥 ∈ 퐺 there exists 푥−1 ∈ 퐺 such that 푥 ∗ 푥−1 = 푥−1 ∗ 푥 = 푒.

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Reminder: Subgroups

 A of a group 퐺 is a group with the same operation as 퐺, and whose set of members is a subset of 퐺.

No action Rotation 90∘ Rotation 180∘ Rotation 270∘

Vertical flip Horizontal flip Diagonal flip Diagonal flip 2

Lagrange’s Theorem

 Theorem. If 퐺 is a group of a finite 푛 and 퐻 is a subgroup of 퐺 of order 푚, then 푚|푛. ◦ We will not prove the theorem.

 Example. The symmetry group of the square is of order 8. ◦ The subgroup of rotations is of order 4. ◦ The subgroup of the identity and rotation by 180∘ is of order 2.

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Reminder: Parity of a

 Theorem. Consider a permutation 훼 ∈ 푆푛. Then ◦ Either every decomposition of 훼 consists of an even number of transpositions, ◦ or every decomposition of 훼 consists of an odd number of transpositions.

 1 2 3 4 5 6 : ◦ 1 3 1 2 4 6 4 5 . ◦ 1 4 1 6 1 5 3 4 2 4 1 4 .

Subgroup of Even

 Consider the group 푆푛: ◦ Recall. A product of two even permutations is even. ◦ The subset of even permutations is a subgroup. It is called the 퐴푛. ◦ Recall. Exactly half of the permutations of 푆푛 are even. That is, the order of 퐴푛 is half the order of 푆푛.

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Atlas of Finite Groups

 (only in class)

Application of Lagrange’s Theorem

 Problem. Let 퐺 be a finite group of order 푛 and let 𝑔 ∈ 퐺 be of order 푚. Prove that 푚|푛 and 𝑔푛 = 1.

 Proof. ◦ Notice that 1, 𝑔, 𝑔2, … , 𝑔푚−1 is a cyclic subgroup of order 푚. ◦ By Lagrange’s theorem 푚|푛. ◦ Write 푛 = 푚푘 for some integer 푘. Then 𝑔푛 = 𝑔푚푘 = 𝑔푚 푘 = 1.

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Groups of a Prime Order

 Claim. Every group 퐺 of a prime order 푝 is isomorphic to the cyclic group 퐶푝.

 Proof. ◦ By Lagrange’s theorem, 퐺 has no subgroups. ◦ Thus, by the previous slide, every element of 퐺 ∖ 1 is of order 푝. ◦ 퐺 is cyclic since any element of 퐺 ∖ 1 generates it.

Symmetries of a Tiling  Given a repetitive tiling of the plane, its symmetries are the transformations of the plane that ◦ Map the tiling to itself (ignoring colors). ◦ Preserve distances.  These are combinations of translations, rotations, and reflections.

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Example: Square Tiling

 What symmetries does the square tiling has? ◦ Translations in every direction. ◦ Rotations around a vertex by 0∘, 90∘, 180∘, 270∘. ◦ Rotations around the center of a square by 0∘, 90∘, 180∘, 270∘. ◦ Reflections across vertical, horizontal and diagonal lines. ◦ Rotations around the center of an edge by 180∘.

Wallpaper Groups

 Given a tiling, its set of symmetries is a group called a wallpaper group (not accurate! More technical conditions). ◦ Closure. Composing two symmetries results in a transformation that preserves distances and takes the lattice to itself. ◦ Associativity. Holds. ◦ Identity. The “no operation” element. ◦ Inverse. Since symmetries are from the plane to itself, inverses are well defined.

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Wallpaper Groups

 There are exactly 17 different wallpaper groups.  That is, the set of all repetitive tilings of the plane can be divided into 17 classes. Two tilings of the same class have the same “behavior”.

Equivalence Relations

 Recall. A relation 푅 on a set 푋 is an equivalence relation if it satisfies the following properties. ◦ Reflexive. For any 푥 ∈ 푋, we have 푥푅푥. ◦ Symmetric. For any 푥, 푦 ∈ 푋, we have 푥푅푦 if and only if 푦푅푥. ◦ Transitive. If 푥푅푦 and 푦푅푧 then 푥푅푧.

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Example: Equivalence Relations

 Problem. Consider the relation of congruence 푚표푑 31, defined over the set of integers ℤ. Is it an equivalence relation?  Solution. ◦ Reflexive. For any 푥 ∈ ℤ, we have 푥 ≡ 푥 푚표푑 31. ◦ Symmetric. For any 푥, 푦 ∈ ℤ, we have 푥 ≡ 푦 푚표푑 31 iff 푦 ≡ 푥 푚표푑 31. ◦ Transitive. If 푥 ≡ 푦 푚표푑 31 and 푦 ≡ 푧 푚표푑 31 then 푥 ≡ 푧 푚표푑 31.

Equivalence Via Permutation Groups  Let 퐺 be a group of permutations of the set 푋. We define a relation on 푋: 푥~푦 ⇔ 𝑔 푥 = 푦 for some 𝑔 ∈ 퐺.  Claim. ~ is an equivalence relation. ◦ Reflexive. The group 퐺 contains the identity permutation id. For every 푥 ∈ 푋 we have id 푥 = 푥 and thus 푥~푥. ◦ Symmetric. If 푥~푦 then 𝑔 푥 = 푦 for some 𝑔 ∈ 퐺. This implies that 𝑔−1 ∈ 퐺 and 푥 = 𝑔−1 푦 . So 푦~푥.

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Equivalence Via Permutation Groups  Let 퐺 be a group of permutations of the set 푋. We define a relation on 푋: 푥~푦 ⇔ 𝑔 푥 = 푦 for some 𝑔 ∈ 퐺.  Claim. ~ is an equivalence relation. ◦ Transitive. If 푥~푦 and 푦~푧 then 𝑔 푥 = 푦 and ℎ 푦 = 푧 for 𝑔, ℎ ∈ 퐺. Then ℎ𝑔 ∈ 퐺 and ℎ𝑔 푥 = 푧, which in turn implies 푥~푧.

Orbits

 Given a permutation group 퐺 of a set 푋, the equivalence relation ~ partitions 푋 into equivalence classes or orbits. ◦ For every 푥 ∈ 푋 the orbit of 푥 is 퐺푥 = 푦 ∈ 푋 | 푥~푦 = 푦 ∈ 푋 | 𝑔 푥 = 푦 for some 𝑔 ∈ 퐺 .

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Example: Orbits

 Let 푋 = 1,2,3,4,5 and let 퐺 = id, 1 2 , 3 4 , 1 2 3 4 .  What are the equivalence classes that 퐺 induces on 푋? ◦ 퐺1 = 퐺2 = 1,2 . ◦ 퐺3 = 퐺4 = 3,4 . ◦ 퐺5 = 5 .

Stabilizers

 Let 퐺 be a permutation group of the set 푋.  Let 퐺 푥 → 푦 denote the set of permutations 𝑔 ∈ 퐺 such that 𝑔 푥 = 푦.  The stabilizer of 푥 is 퐺푥 = 퐺 푥 → 푥 .

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Example: Stabilizer

 Consider the following permutation group of {1,2,3,4}: 퐺 = {id, 1 2 3 4 , 1 3 2 4 , 1 4 3 2 , 2 4 , 1 3 , 1 2 3 4 , 1 4 2 3 }.

 The stabilizers are

◦ 퐺1 = id, 2 4 . ◦ 퐺2 = id, 1 3 . ◦ 퐺3 = id, 2 4 . ◦ 퐺4 = id, 1 3 .

Stabilizers are Subgroups

 Claim. 퐺푥 is a subgroup of 퐺. ◦ Closure. If 𝑔, ℎ ∈ 퐺푥 then 𝑔 푥 = 푥 and ℎ 푥 = 푥. Since 𝑔ℎ 푥 = 푥 we have 𝑔ℎ ∈ 퐺푥. ◦ Associativity. Implied by the associativity of 퐺.

◦ Identity. Since id 푥 = 푥, we have 𝑖푑 ∈ 퐺푥. ◦ Inverse. If 𝑔 ∈ 퐺푥 then 𝑔 푥 = 푥. This implies −1 −1 that 𝑔 푥 = 푥 so 𝑔 ∈ 퐺푥.

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Cosets

 Let 퐻 be a subgroup of the group 퐺. The left coset of 퐻 with respect to 𝑔 ∈ 퐺 is 𝑔퐻 = 푎 ∈ 퐺 | 푎 = 𝑔ℎ for some ℎ ∈ 퐻 .

 Example. The coset of the alternating group 퐴푛 with respect to a transposition 푥 푦 ∈ 푆푛 is the subset of odd permutations of 푆푛.

퐺 푥 → 푦 are Cosets

 Claim. Let 퐺 be a permutation group and let ℎ ∈ 퐺 푥 → 푦 . Then 퐺 푥 → 푦 = ℎ퐺푥.

 Proof.

◦ ℎ퐺푥 ⊆ 퐺 푥 → 푦 . If 푎 ∈ ℎ퐺푥, then 푎 = ℎ𝑔 for some 𝑔 ∈ 퐺푥. We have 푎 ∈ 퐺 푥 → 푦 since 푎 푥 = ℎ𝑔 푥 = ℎ 푥 = 푦.

◦ 퐺 푥 → 푦 ⊆ ℎ퐺푥. If 푏 ∈ 퐺 푥 → 푦 then ℎ−1푏 푥 = ℎ−1 푦 = 푥. −1 That is, ℎ 푏 ∈ 퐺푥, which implies 푏 ∈ ℎ퐺푥.

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Sizes of Cosets and Stabilizers

 Claim. Let 퐺 be a permutation group on 푋 and let 퐺푥 be the stabilizer of 푥 ∈ 푋. Then 퐺푥 = ℎ퐺푥 for any ℎ ∈ 퐺. ◦ Proof. By the Latin square property of 퐺.

 Corollary. The size of 퐺 푥 → 푦 : ◦ If 푦 is in the orbit 퐺푥 then 퐺 푥 → 푦 = 퐺푥 . ◦ If 푦 is not in the orbit 퐺푥 then 퐺 푥 → 푦 = 0.

Sizes of Orbits and Stabilizers

 Theorem. Let 퐺 be a group of permutations of the set 푋. For every 푥 ∈ 푋 we have 퐺푥 ⋅ 퐺푥 = 퐺 .

The orbit of 푥 The stabilizer of 푥

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Example: Orbits and Stabilizers

 Consider the following permutation group of {1,2,3,4}: 퐺 = {id, 1 2 3 4 , 1 3 2 4 , 1 4 3 2 , 2 4 , 1 3 , 1 2 3 4 , 1 4 2 3 }.

◦ We have 퐺 = 8. ◦ We have the orbit 퐺1 = 1,2,3,4 . So 퐺1 = 4.

◦ We have the stabilizer 퐺1 = id, 2 4 . So 퐺1 = 2. ◦ Combining the above yields 퐺 = 8 = 퐺1 ⋅ 퐺1 .

A Useful Table

 Let 퐺 = 𝑔1, 𝑔2, … , 𝑔푛 be a group of permutations of 푋 = 푥1, 푥2, … , 푥푚 . ◦ For an element 푥 ∈ 푋, we build the following table, where  implies that 𝑔푖 푥 = 푥푗.

풙ퟏ 풙ퟐ 풙ퟑ 풙ퟒ 풙ퟓ 풙ퟔ 풙ퟕ … 풙풎

𝑔1 

𝑔2 

𝑔3  …

𝑔푛 

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Table Properties 1

 How many ’s are in the table?

◦ Since 𝑔푖 푥 has a unique value, each row contains exactly one . ◦ The total number of ’s in the table is 퐺 .

풙ퟏ 풙ퟐ 풙ퟑ 풙ퟒ 풙ퟓ 풙ퟔ 풙ퟕ … 풙풎

𝑔1 

𝑔2 

𝑔3  …

𝑔푛 

Table Properties 2

 How many ’s are in the column of 푥푖? ◦ If 푥푖 is not in the orbit 퐺푥, then 0. ◦ If 푥푖 is in the orbit 퐺푥, then 퐺 푥 → 푦 = 퐺푥 .

풙ퟏ 풙ퟐ 풙ퟑ 풙ퟒ 풙ퟓ 풙ퟔ 풙ퟕ … 풙풎

𝑔1 

𝑔2 

𝑔3  …

𝑔푛 

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Proving the Theorem

 Theorem. Let 퐺 be a group of permutations of the set 푋. For every 푥 ∈ 푋 we have 퐺푥 ⋅ 퐺푥 = 퐺 .

 Proof. ◦ Counting by rows, the number of ’s in the table is 퐺 . ◦ Counting by columns, there are 퐺푥 non- empty columns, each containing 퐺푥 ’s. ◦ That is, 퐺 = 퐺푥 ⋅ 퐺푥 .

Double Counting

 Our proof technique was to count the same value (the number of ’s in the table) in two different ways.  This technique is called double counting and is very useful in combinatorics.

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The End: Alhambra

 Alhmbra is a palace and fortress complex located in Granada, Spain. ◦ The Islamic art on the walls is claimed to contain all 17 wallpaper groups. ◦ Mathematicians like to visit the palace and look for as many types as they can find.

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