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Ma/CS 6aClass 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 πΊπ₯ β π¦ β πΊ.
β¦ Associativity. For every π₯, π¦, π§ β πΊπ₯ β π¦ β π§ = π₯ β π¦ β π§ .
β¦ Identity. There exists π β πΊ, such that for every π₯ β πΊ
π β π₯ = π₯ β π = π₯.
β¦ Inverse. For every π₯ β πΊ there exists π₯β1 β πΊsuch that π₯ β π₯β1 = π₯β1 β π₯ = π.
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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.
Find a subgroup of the group of integers under addition.
β¦ The subset of even integers.
β¦ The subset β¦ ,β2π,βπ, 0, π, 2π, . . for any integer π > 1.
Subgroups of a Symmetry Group
Problem. Find a subgroup of the symmetries of the square.
No action Rotation 90β Rotation 180β Rotation 270β
Vertical flip Horizontal flip Diagonal flip Diagonal flip 2
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Subgroups of a Symmetry Group
Problem. Find a subgroup of the subgroup.
No action Rotation 90β Rotation 180β Rotation 270β
Vertical flip Horizontal flip Diagonal flip Diagonal flip 2
Subgroups of a Symmetry Group
No action Rotation 90β Rotation 180β Rotation 270β
Vertical flip Horizontal flip Diagonal flip Diagonal flip 2
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Subgroup Conditions
Problem. Let πΊ be a group, and let π» be a non-empty subset of πΊ such that
β¦ C1. If π₯, π¦ β π» then π₯π¦ β π».
β¦ ππ. If π₯ β π» then π₯β1 β π».
Prove that π» is a subgroup.
β¦ Closure. By C1.
β¦ Inverse. By C2.
β¦ Associativity. By the associativity of πΊ.
β¦ Identity. By C2, π₯, π₯β1 β π». By C1, we have 1 = π₯π₯β1 β π».
Finite Subgroup Conditions
Problem. Let πΊ be a finite group, and let π» be a non-empty subset of πΊ such that
β¦ C1. If π₯, π¦ β π» then π₯π¦ β π».
β¦ ππ. If π₯ β π» then π₯β1 β π».
Prove that π» is a subgroup.
Proof. Consider π₯ β π».
Since πΊ is finite, the series 1, π₯, π₯2, π₯3, β¦ has two identical elements π₯π = π₯π with π < π.
Multiply both sides by π₯βπβ1 (in πΊ) to obtainπ₯β1 = π₯πβπβ1 = π₯π₯π₯β―π₯ β π».
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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.
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 .
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Subgroup of ππ
Consider the group ππ:
β¦ Recall. A composition of two even permutations is even.
β¦ The subset of even permutations is a subgroup called the alternating group π΄π.
β¦ Recall. Exactly half of the permutations of ππare even. That is, the order of π΄π is π!/2.
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. Thenππ = 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 the previous slide, every element of πΊ β 1 is of order π.
β¦ πΊ is cyclic since any element of πΊ β 1generates it.
Equivalence Relations
Recall. A binary 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,
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 30, 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 30.
β¦ Symmetric. For any π₯, π¦ β β€, we have π₯ β‘ π¦ πππ 30 iff π¦ β‘ π₯ πππ 30.
β¦ Transitive. If π₯ β‘ π¦ πππ 30 and π¦ β‘ π§ πππ 30 then π₯ β‘ π§ πππ 30.
Permutations of Objects
We have a set of numbers π = 1,2,3, β¦ , π and a permutation group πΊ of π.
For example, π = 1,2,3,4,5,6
πΊ = id, 1 2 , 3 4 , 1 2 3 4
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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¦ 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₯.
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§.
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Orbits
Given a permutation group πΊ of a set π, the equivalence relation ~ partitions πinto 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 .
Example: Orbits
Let π = 1,2,3,4,5 and letπΊ = id, 1 2 , 3 4 , 1 2 3 4 .
What are the orbits that πΊ induces on π?
β¦ πΊ1 = πΊ2 = 1,2 .
β¦ πΊ3 = πΊ4 = 3,4 .
β¦ πΊ5 = 5 .
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Simple Groups
A trivial group is a group that contains only one element β an identity element.
A simple group is a non-trivial group that does not contain any βwell-behavedβ subgroups in it.
The finite simple groups are, in a certain sense, the βbasic building blocksβ of all finite groups.
β¦ Somewhat similar to the way prime numbers are the basic building blocks of the integers.
Classification of Finite Simple Groups
βOne of the most important mathematical achievements of the 20th century was the collaborative effort, taking up more than 10,000 journal pagesβ (Wikipedia).
Written by about 100 authors!
Theorem. Every finite simple group is isomorphic to one of the following groups:
β¦ A cyclic group.
β¦ An alternating group.
β¦ A simple Lie group.
β¦ One of the 26 sporadic groups.
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The Monster Group
One of the 26 sporadic groups is the monster group.
It has an order of 808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000.
The 6 sporadic groups that are not βcontainedβ in the monster group are called the happy family.
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₯ isπΊπ₯ = 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 id 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 the transposition 1 2 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 πand 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.
The End: A Noahβs Ark JokeThe Flood has receded and the ark is safely aground atop Mount Ararat. Noah tells all the animals to go forth and multiply. Soon the land is teeming with every kind of living creature in abundance, except for snakes. Noah wonders why. One morning two miserable snakes knock on the door of the ark with a complaint. βYou havenβt cut down any trees.β Noah is puzzled, but does as they wish. Within a month, you canβt walk a step without treading on baby snakes. With difficulty, he tracks down the two parents. βWhat was all that with the trees?β βAh,β says one of the snakes, βyou didnβt notice which species we are.β Noah still looks blank. βWeβre adders, and we can only multiply using logs.β