11/13/2014 Ma/CS 6a Class 20: Subgroups, 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. 1 11/13/2014 Reminder: Subgroups A subgroup 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 order 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. 2 11/13/2014 Reminder: Parity of a Permutation 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 Permutations 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 alternating group 퐴푛. ◦ Recall. Exactly half of the permutations of 푆푛 are even. That is, the order of 퐴푛 is half the order of 푆푛. 3 11/13/2014 Atlas of Finite Groups (only in class) Application of Lagrange’s Theorem Problem. Let 퐺 be a finite group of order 푛 and let ∈ 퐺 be of order 푚. Prove that 푛 푚|푛 and = 1. Proof. ◦ Notice that 1, , 2, … , 푚−1 is a cyclic subgroup of order 푚. ◦ By Lagrange’s theorem 푚|푛. ◦ Write 푛 = 푚푘 for some integer 푘. Then 푛 = 푚푘 = 푚 푘 = 1. 4 11/13/2014 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. 5 11/13/2014 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 bijections from the plane to itself, inverses are well defined. 6 11/13/2014 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. 7 11/13/2014 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 ⇔ 푥 = 푦 for some ∈ 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 푥 = 푦 for some ∈ 퐺. This implies that −1 ∈ 퐺 and 푥 = −1 푦 . So 푦~푥. 8 11/13/2014 Equivalence Via Permutation Groups Let 퐺 be a group of permutations of the set 푋. We define a relation on 푋: 푥~푦 ⇔ 푥 = 푦 for some ∈ 퐺. Claim. ~ is an equivalence relation. ◦ Transitive. If 푥~푦 and 푦~푧 then 푥 = 푦 and ℎ 푦 = 푧 for , ℎ ∈ 퐺. Then ℎ ∈ 퐺 and ℎ 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 | 푥 = 푦 for some ∈ 퐺 . 9 11/13/2014 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 ∈ 퐺 such that 푥 = 푦. The stabilizer of 푥 is 퐺푥 = 퐺 푥 → 푥 . 10 11/13/2014 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 , ℎ ∈ 퐺푥 then 푥 = 푥 and ℎ 푥 = 푥. Since ℎ 푥 = 푥 we have ℎ ∈ 퐺푥. ◦ Associativity. Implied by the associativity of 퐺. ◦ Identity. Since id 푥 = 푥, we have 푑 ∈ 퐺푥. ◦ Inverse. If ∈ 퐺푥 then 푥 = 푥. This implies −1 −1 that 푥 = 푥 so ∈ 퐺푥. 11 11/13/2014 Cosets Let 퐻 be a subgroup of the group 퐺. The left coset of 퐻 with respect to ∈ 퐺 is 퐻 = 푎 ∈ 퐺 | 푎 = ℎ 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 = ℎ for some ∈ 퐺푥. We have 푎 ∈ 퐺 푥 → 푦 since 푎 푥 = ℎ 푥 = ℎ 푥 = 푦. ◦ 퐺 푥 → 푦 ⊆ ℎ퐺푥. If 푏 ∈ 퐺 푥 → 푦 then −1 −1 ℎ 푏 푥 = ℎ 푦 = 푥. −1 That is, ℎ 푏 ∈ 퐺푥, which implies 푏 ∈ ℎ퐺푥. 12 11/13/2014 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 13 11/13/2014 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 = 1, 2, … , 푛 be a group of permutations of 푋 = 푥1, 푥2, … , 푥푚 . ◦ For an element 푥 ∈ 푋, we build the following table, where implies that 푖 푥 = 푥푗. 풙ퟏ 풙ퟐ 풙ퟑ 풙ퟒ 풙ퟓ 풙ퟔ 풙ퟕ … 풙풎 1 2 3 … 푛 14 11/13/2014 Table Properties 1 How many ’s are in the table? ◦ Since 푖 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 1 2 3 … 푛 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 .