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Ma/CS 6a Class 20: Subgroups, Orbits, and Stabilizers
By Adam Sheffer
A Group
A group consists of a set πΊ and a binary operation β, satisfying the following.
β¦ Closure. For every π₯, π¦ β πΊ, we have π₯ β π¦ β πΊ.
β¦ Associativity. For every π₯, π¦, π§ β πΊ, we have π₯ β π¦ β π§ = π₯ β π¦ β π§ .
β¦ Identity. The exists π β πΊ, such that for every π₯ β πΊ, we have
π β π₯ = π₯ β π = π₯.
β¦ Inverse. For every π₯ β πΊ there exists π₯β1 β πΊ such that π₯ β π₯β1 = π₯β1 β π₯ = π.
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Reminder: Subgroups
A subgroup of a group πΊ is a group with the same operation as πΊ, and whose set of members is a subset of πΊ.
No action Rotation 90β Rotation 180β Rotation 270β
Vertical flip Horizontal flip Diagonal flip Diagonal flip 2
Lagrangeβs Theorem
Theorem. If πΊ is a group of a finite order π and π» is a subgroup of πΊ of order π, then π|π.
β¦ We will not prove the theorem.
Example. The symmetry group of the square is of order 8.
β¦ The subgroup of rotations is of order 4.
β¦ The subgroup of the identity and rotation by 180β is of order 2.
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Reminder: Parity of a Permutation
Theorem. Consider a permutation πΌ β ππ. Then
β¦ Either every decomposition of πΌ consists of an even number of transpositions,
β¦ or every decomposition of πΌ consists of an odd number of transpositions.
1 2 3 4 5 6 :
β¦ 1 3 1 2 4 6 4 5 .
β¦ 1 4 1 6 1 5 3 4 2 4 1 4 .
Subgroup of Even Permutations
Consider the group ππ:
β¦ Recall. A product of two even permutations is even.
β¦ The subset of even permutations is a subgroup. It is called the alternating group π΄π.
β¦ Recall. Exactly half of the permutations of ππ are even. That is, the order of π΄π is half the order of ππ.
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Atlas of Finite Groups
(only in class)
Application of Lagrangeβs Theorem
Problem. Let πΊ be a finite group of order π and let π β πΊ be of order π. Prove that
π|π and ππ = 1.
Proof.
β¦ Notice that 1, π, π2, β¦ , ππβ1 is a cyclic subgroup of order π.
β¦ By Lagrangeβs theorem π|π.
β¦ Write π = ππ for some integer π. Then ππ = πππ = ππ π = 1.
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Groups of a Prime Order
Claim. Every group πΊ of a prime order π is isomorphic to the cyclic group πΆπ.
Proof.
β¦ By Lagrangeβs theorem, πΊ has no subgroups.
β¦ Thus, by the previous slide, every element of πΊ β 1 is of order π.
β¦ πΊ is cyclic since any element of πΊ β 1 generates it.
Symmetries of a Tiling Given a repetitive tiling of the plane, its
symmetries are the transformations of the plane that
β¦ Map the tiling to itself (ignoring colors).
β¦ Preserve distances.
These are combinations of translations, rotations, and reflections.
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Example: Square Tiling
What symmetries does the square tiling has?
β¦ Translations in every direction.
β¦ Rotations around a vertex by 0β, 90β, 180β, 270β.
β¦ Rotations around the center of a square by 0β, 90β, 180β, 270β.
β¦ Reflections across vertical, horizontal and diagonal lines.
β¦ Rotations around the center of an edge by 180β.
Wallpaper Groups
Given a tiling, its set of symmetries is a group called a wallpaper group (not accurate! More technical conditions).
β¦ Closure. Composing two symmetries results in a transformation that preserves distances and takes the lattice to itself.
β¦ Associativity. Holds.
β¦ Identity. The βno operationβ element.
β¦ Inverse. Since symmetries are bijections 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 βbehaviorβ.
Equivalence Relations
Recall. A relation π on a set π is an equivalence relation if it satisfies the following properties.
β¦ Reflexive. For any π₯ β π, we have π₯π π₯.
β¦ Symmetric. For any π₯, π¦ β π, we have π₯π π¦ if and only if π¦π π₯.
β¦ Transitive. If π₯π π¦ and π¦π π§ then π₯π π§.
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Example: Equivalence Relations
Problem. Consider the relation of congruence πππ 31, defined over the set of integers β€. Is it an equivalence relation?
Solution.
β¦ Reflexive. For any π₯ β β€, we have π₯ β‘ π₯ πππ 31.
β¦ Symmetric. For any π₯, π¦ β β€, we have π₯ β‘ π¦ πππ 31 iff π¦ β‘ π₯ πππ 31.
β¦ Transitive. If π₯ β‘ π¦ πππ 31 and π¦ β‘ π§ πππ 31 then π₯ β‘ π§ πππ 31.
Equivalence Via Permutation Groups Let πΊ be a group of permutations of the
set π. We define a relation on π: π₯~π¦ β π π₯ = π¦ for some π β πΊ.
Claim. ~ is an equivalence relation.
β¦ Reflexive. The group πΊ contains the identity permutation id. For every π₯ β π we have id π₯ = π₯ and thus π₯~π₯.
β¦ Symmetric. If π₯~π¦ then π π₯ = π¦ for some π β πΊ. This implies that πβ1 β πΊ and π₯ = πβ1 π¦ . So π¦~π₯.
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Equivalence Via Permutation Groups Let πΊ be a group of permutations of the
set π. We define a relation on π: π₯~π¦ β π π₯ = π¦ for some π β πΊ.
Claim. ~ is an equivalence relation.
β¦ Transitive. If π₯~π¦ and π¦~π§ then π π₯ = π¦ and β π¦ = π§ for π, β β πΊ. Then βπ β πΊ and βπ π₯ = π§, which in turn implies π₯~π§.
Orbits
Given a permutation group πΊ of a set π, the equivalence relation ~ partitions π into equivalence classes or orbits.
β¦ For every π₯ β π the orbit of π₯ is πΊπ₯ = π¦ β π | π₯~π¦
= π¦ β π | π π₯ = π¦ for some π β πΊ .
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Example: Orbits
Let π = 1,2,3,4,5 and let πΊ = id, 1 2 , 3 4 , 1 2 3 4 .
What are the equivalence classes that πΊ induces on π?
β¦ πΊ1 = πΊ2 = 1,2 .
β¦ πΊ3 = πΊ4 = 3,4 .
β¦ πΊ5 = 5 .
Stabilizers
Let πΊ be a permutation group of the set π.
Let πΊ π₯ β π¦ denote the set of permutations π β πΊ such that π π₯ = π¦.
The stabilizer of π₯ is πΊπ₯ = πΊ π₯ β π₯ .
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Example: Stabilizer
Consider the following permutation group of {1,2,3,4}:
πΊ = {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
β¦ πΊ1 = id, 2 4 .
β¦ πΊ2 = id, 1 3 .
β¦ πΊ3 = id, 2 4 .
β¦ πΊ4 = id, 1 3 .
Stabilizers are Subgroups
Claim. πΊπ₯ is a subgroup of πΊ.
β¦ Closure. If π, β β πΊπ₯ then π π₯ = π₯ and β π₯ = π₯. Since πβ π₯ = π₯ we have πβ β πΊπ₯.
β¦ Associativity. Implied by the associativity of πΊ.
β¦ Identity. Since id π₯ = π₯, we have ππ β πΊπ₯.
β¦ Inverse. If π β πΊπ₯ then π π₯ = π₯. This implies that πβ1 π₯ = π₯ so πβ1 β πΊπ₯.
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Cosets
Let π» be a subgroup of the group πΊ. The left coset of π» with respect to π β πΊ is
ππ» = π β πΊ | π = πβ for some β β π» .
Example. The coset of the alternating group π΄π with respect to a transposition π₯ π¦ β ππ is the subset of odd
permutations of ππ.
πΊ π₯ β π¦ are Cosets
Claim. Let πΊ be a permutation group and let β β πΊ π₯ β π¦ . Then
πΊ π₯ β π¦ = βπΊπ₯ .
Proof.
β¦ βπΊπ₯ β πΊ π₯ β π¦ . If π β βπΊπ₯, then π = βπ for some π β πΊπ₯. We have π β πΊ π₯ β π¦ since
π π₯ = βπ π₯ = β π₯ = π¦.
β¦ πΊ π₯ β π¦ β βπΊπ₯. If π β πΊ π₯ β π¦ then ββ1π π₯ = ββ1 π¦ = π₯.
That is, ββ1π β πΊπ₯, which implies π β βπΊπ₯.
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Sizes of Cosets and Stabilizers
Claim. Let πΊ be a permutation group on π and let πΊπ₯ be the stabilizer of π₯ β π. Then
πΊπ₯ = βπΊπ₯ for any β β πΊ.
β¦ Proof. By the Latin square property of πΊ.
Corollary. The size of πΊ π₯ β π¦ :
β¦ If π¦ is in the orbit πΊπ₯ then πΊ π₯ β π¦ = πΊπ₯ .
β¦ If π¦ is not in the orbit πΊπ₯ then πΊ π₯ β π¦ = 0.
Sizes of Orbits and Stabilizers
Theorem. Let πΊ be a group of permutations of the set π. For every π₯ β π we have
πΊπ₯ β πΊπ₯ = πΊ .
The orbit of π₯ The stabilizer of π₯
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Example: Orbits and Stabilizers
Consider the following permutation group of {1,2,3,4}:
πΊ = {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 }.
β¦ We have πΊ = 8.
β¦ We have the orbit πΊ1 = 1,2,3,4 . So πΊ1 = 4.
β¦ We have the stabilizer πΊ1 = id, 2 4 . So πΊ1 = 2.
β¦ Combining the above yields πΊ = 8 = πΊ1 β πΊ1 .
A Useful Table
Let πΊ = π1, π2, β¦ , ππ be a group of permutations of π = π₯1, π₯2, β¦ , π₯π .
β¦ For an element π₯ β π, we build the following table, where implies that ππ π₯ = π₯π.
ππ ππ ππ ππ ππ ππ ππ β¦ ππ
π1
π2
π3
β¦
ππ
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Table Properties 1
How many βs are in the table?
β¦ Since ππ π₯ has a unique value, each row contains exactly one .
β¦ The total number of βs in the table is πΊ .
ππ ππ ππ ππ ππ ππ ππ β¦ ππ
π1
π2
π3
β¦
ππ
Table Properties 2
How many βs are in the column of π₯π?
β¦ If π₯π is not in the orbit πΊπ₯, then 0.
β¦ If π₯π is in the orbit πΊπ₯, then πΊ π₯ β π¦ = πΊπ₯ .
ππ ππ ππ ππ ππ ππ ππ β¦ ππ
π1
π2
π3
β¦
ππ
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Proving the Theorem
Theorem. Let πΊ be a group of permutations of the set π. For every π₯ β π we have
πΊπ₯ β πΊπ₯ = πΊ .
Proof.
β¦ Counting by rows, the number of βs in the table is πΊ .
β¦ Counting by columns, there are πΊπ₯ non-empty columns, each containing πΊπ₯ βs.
β¦ That is, πΊ = πΊπ₯ β πΊπ₯ .
Double Counting
Our proof technique was to count the same value (the number of βs in the table) in two different ways.
This technique is called double counting and is very useful in combinatorics.
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