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Ma/CS 6aClass 20: Subgroups, Orbits, and Stabilizers

By Adam Sheffer

A Group

A group consists of a set 𝐺 and a binary operation ∗, satisfying the following.

◦ Closure. For every 𝑥, 𝑦 ∈ 𝐺𝑥 ∗ 𝑦 ∈ 𝐺.

◦ Associativity. For every 𝑥, 𝑦, 𝑧 ∈ 𝐺𝑥 ∗ 𝑦 ∗ 𝑧 = 𝑥 ∗ 𝑦 ∗ 𝑧 .

◦ Identity. There exists 𝑒 ∈ 𝐺, such that for every 𝑥 ∈ 𝐺

𝑒 ∗ 𝑥 = 𝑥 ∗ 𝑒 = 𝑥.

◦ Inverse. For every 𝑥 ∈ 𝐺 there exists 𝑥−1 ∈ 𝐺such that 𝑥 ∗ 𝑥−1 = 𝑥−1 ∗ 𝑥 = 𝑒.

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Subgroups

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

Find a subgroup of the group of integers under addition.

◦ The subset of even integers.

◦ The subset … ,−2𝑟,−𝑟, 0, 𝑟, 2𝑟, . . for any integer 𝑟 > 1.

Subgroups of a Symmetry Group

Problem. Find a subgroup of the symmetries of the square.

No action Rotation 90∘ Rotation 180∘ Rotation 270∘

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 90∘ Rotation 180∘ Rotation 270∘

Vertical flip Horizontal flip Diagonal flip Diagonal flip 2

Subgroups of a Symmetry Group

No action Rotation 90∘ Rotation 180∘ Rotation 270∘

Vertical flip Horizontal flip Diagonal flip Diagonal flip 2

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Subgroup Conditions

Problem. Let 𝐺 be a group, and let 𝐻 be a non-empty subset of 𝐺 such that

◦ C1. If 𝑥, 𝑦 ∈ 𝐻 then 𝑥𝑦 ∈ 𝐻.

◦ 𝐂𝟐. If 𝑥 ∈ 𝐻 then 𝑥−1 ∈ 𝐻.

Prove that 𝐻 is a subgroup.

◦ Closure. By C1.

◦ Inverse. By C2.

◦ Associativity. By the associativity of 𝐺.

◦ Identity. By C2, 𝑥, 𝑥−1 ∈ 𝐻. By C1, we have 1 = 𝑥𝑥−1 ∈ 𝐻.

Finite Subgroup Conditions

Problem. Let 𝐺 be a finite group, and let 𝐻 be a non-empty subset of 𝐺 such that

◦ C1. If 𝑥, 𝑦 ∈ 𝐻 then 𝑥𝑦 ∈ 𝐻.

◦ 𝐂𝟐. If 𝑥 ∈ 𝐻 then 𝑥−1 ∈ 𝐻.

Prove that 𝐻 is a subgroup.

Proof. Consider 𝑥 ∈ 𝐻.

Since 𝐺 is finite, the series 1, 𝑥, 𝑥2, 𝑥3, … has two identical elements 𝑥𝑖 = 𝑥𝑗 with 𝑖 < 𝑗.

Multiply both sides by 𝑥−𝑖−1 (in 𝐺) to obtain𝑥−1 = 𝑥𝑗−𝑖−1 = 𝑥𝑥𝑥⋯𝑥 ∈ 𝐻.

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Lagrange’s Theorem

Theorem. If 𝐺 is a group of a finite order 𝑛 and 𝐻 is a subgroup of 𝐺 of order 𝑚, then 𝑚|𝑛.

◦ 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.

Reminder: Parity of a Permutation

Theorem. Consider a permutation 𝛼 ∈ 𝑆𝑛. Then

◦ Either every decomposition of 𝛼 consists of an even number of transpositions,

◦ or every decomposition of 𝛼 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 .

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Subgroup of 𝑆𝑛

Consider the group 𝑆𝑛:

◦ Recall. A composition of two even permutations is even.

◦ The subset of even permutations is a subgroup called the alternating group 𝐴𝑛.

◦ Recall. Exactly half of the permutations of 𝑆𝑛are even. That is, the order of 𝐴𝑛 is 𝑛!/2.

Application of Lagrange’s Theorem

Problem. Let 𝐺 be a finite group of order 𝑛 and let 𝑔 ∈ 𝐺 be of order 𝑚. Prove that

𝑚|𝑛 and 𝑔𝑛 = 1.

Proof.

◦ Notice that 1, 𝑔, 𝑔2, … , 𝑔𝑚−1 is a cyclic subgroup of order 𝑚.

◦ By Lagrange’s theorem 𝑚|𝑛.

◦ Write 𝑛 = 𝑚𝑘 for some integer 𝑘. Then𝑔𝑛 = 𝑔𝑚𝑘 = 𝑔𝑚 𝑘 = 1.

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Groups of a Prime Order

Claim. Every group 𝐺 of a prime order 𝑝 is isomorphic to the cyclic group 𝐶𝑝.

Proof.

◦ By the previous slide, every element of 𝐺 ∖ 1 is of order 𝑝.

◦ 𝐺 is cyclic since any element of 𝐺 ∖ 1generates it.

Equivalence Relations

Recall. A binary relation 𝑅 on a set 𝑋 is an equivalence relation if it satisfies the following properties.

◦ Reflexive. For any 𝑥 ∈ 𝑋, we have 𝑥𝑅𝑥.

◦ Symmetric. For any 𝑥, 𝑦 ∈ 𝑋,

𝑥𝑅𝑦 if and only if 𝑦𝑅𝑥.

◦ Transitive. If 𝑥𝑅𝑦 and 𝑦𝑅𝑧 then 𝑥𝑅𝑧.

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Example: Equivalence Relations

Problem. Consider the relation of congruence 𝑚𝑜𝑑 30, defined over the set of integers ℤ. Is it an equivalence relation?

Solution.

◦ Reflexive. For any 𝑥 ∈ ℤ, we have 𝑥 ≡ 𝑥 𝑚𝑜𝑑 30.

◦ Symmetric. For any 𝑥, 𝑦 ∈ ℤ, we have 𝑥 ≡ 𝑦 𝑚𝑜𝑑 30 iff 𝑦 ≡ 𝑥 𝑚𝑜𝑑 30.

◦ Transitive. If 𝑥 ≡ 𝑦 𝑚𝑜𝑑 30 and 𝑦 ≡ 𝑧 𝑚𝑜𝑑 30 then 𝑥 ≡ 𝑧 𝑚𝑜𝑑 30.

Permutations of Objects

We have a set of numbers 𝑋 = 1,2,3, … , 𝑛 and a permutation group 𝐺 of 𝑋.

For example, 𝑋 = 1,2,3,4,5,6

𝐺 = id, 1 2 , 3 4 , 1 2 3 4

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Equivalence Via Permutation Groups Let 𝐺 be a group of permutations of the

set 𝑋. We define a relation on 𝑋:𝑥~𝑦 ⇔ 𝑔 𝑥 = 𝑦 for some 𝑔 ∈ 𝐺.

Claim. ~ is an equivalence relation.

◦ Reflexive. The group 𝐺 contains the identity permutation id. For every 𝑥 ∈ 𝑋 we have id 𝑥 = 𝑥 and thus 𝑥~𝑥.

◦ Symmetric. If 𝑥~𝑦 then 𝑔 𝑥 = 𝑦 for some 𝑔 ∈ 𝐺. This implies that 𝑔−1 ∈ 𝐺 and 𝑥 = 𝑔−1 𝑦 . So 𝑦~𝑥.

Equivalence Via Permutation Groups Let 𝐺 be a group of permutations of the

set 𝑋. We define a relation on 𝑋:𝑥~𝑦 ⇔ 𝑔 𝑥 = 𝑦 for some 𝑔 ∈ 𝐺.

Claim. ~ is an equivalence relation.

◦ Transitive. If 𝑥~𝑦 and 𝑦~𝑧 then 𝑔 𝑥 = 𝑦 and ℎ 𝑦 = 𝑧 for 𝑔, ℎ ∈ 𝐺. Then ℎ𝑔 ∈ 𝐺 and ℎ𝑔 𝑥 = 𝑧, which in turn implies 𝑥~𝑧.

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Orbits

Given a permutation group 𝐺 of a set 𝑋, the equivalence relation ~ partitions 𝑋into equivalence classes or orbits.

◦ For every 𝑥 ∈ 𝑋 the orbit of 𝑥 is 𝐺𝑥 = 𝑦 ∈ 𝑋 | 𝑥~𝑦= 𝑦 ∈ 𝑋 | 𝑔 𝑥 = 𝑦 for some 𝑔 ∈ 𝐺 .

Example: Orbits

Let 𝑋 = 1,2,3,4,5 and let𝐺 = id, 1 2 , 3 4 , 1 2 3 4 .

What are the orbits that 𝐺 induces on 𝑋?

◦ 𝐺1 = 𝐺2 = 1,2 .

◦ 𝐺3 = 𝐺4 = 3,4 .

◦ 𝐺5 = 5 .

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Simple Groups

A trivial group is a group that contains only one element – an identity element.

A simple group is a non-trivial group that does not contain any “well-behaved” subgroups in it.

The finite simple groups are, in a certain sense, the “basic building blocks” of all finite groups.

◦ Somewhat similar to the way prime numbers are the basic building blocks of the integers.

Classification of Finite Simple Groups

“One of the most important mathematical achievements of the 20th century was the collaborative effort, taking up more than 10,000 journal pages” (Wikipedia).

Written by about 100 authors!

Theorem. Every finite simple group is isomorphic to one of the following groups:

◦ A cyclic group.

◦ An alternating group.

◦ A simple Lie group.

◦ 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 “contained” in the monster group are called the happy family.

Stabilizers

Let 𝐺 be a permutation group of the set 𝑋.

Let 𝐺 𝑥 → 𝑦 denote the set of permutations 𝑔 ∈ 𝐺 such that 𝑔 𝑥 = 𝑦.

The stabilizer of 𝑥 is𝐺𝑥 = 𝐺 𝑥 → 𝑥 .

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Example: Stabilizer

Consider the following permutation group of {1,2,3,4}:

𝐺 = {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

◦ 𝐺1 = id, 2 4 .

◦ 𝐺2 = id, 1 3 .

◦ 𝐺3 = id, 2 4 .

◦ 𝐺4 = id, 1 3 .

Stabilizers are Subgroups

Claim. 𝐺𝑥 is a subgroup of 𝐺.

◦ Closure. If 𝑔, ℎ ∈ 𝐺𝑥 then 𝑔 𝑥 = 𝑥 and ℎ 𝑥 = 𝑥. Since 𝑔ℎ 𝑥 = 𝑥 we have 𝑔ℎ ∈ 𝐺𝑥.

◦ Associativity. Implied by the associativity of 𝐺.

◦ Identity. Since id 𝑥 = 𝑥, we have id ∈ 𝐺𝑥.

◦ Inverse. If 𝑔 ∈ 𝐺𝑥 then 𝑔 𝑥 = 𝑥. This implies that 𝑔−1 𝑥 = 𝑥 so 𝑔−1 ∈ 𝐺𝑥.

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Cosets

Let 𝐻 be a subgroup of the group 𝐺. The left coset of 𝐻 with respect to 𝑔 ∈ 𝐺 is

𝑔𝐻 = 𝑎 ∈ 𝐺 | 𝑎 = 𝑔ℎ for some ℎ ∈ 𝐻 .

Example. The coset of the alternating group 𝐴𝑛 with respect to the transposition 1 2 ∈ 𝑆𝑛 is the subset of odd permutations of 𝑆𝑛.

𝐺 𝑥 → 𝑦 are Cosets

Claim. Let 𝐺 be a permutation group and let ℎ ∈ 𝐺 𝑥 → 𝑦 . Then

𝐺 𝑥 → 𝑦 = ℎ𝐺𝑥 .

Proof.

◦ ℎ𝐺𝑥 ⊆ 𝐺 𝑥 → 𝑦 . If 𝑎 ∈ ℎ𝐺𝑥, then 𝑎 = ℎ𝑔 for some 𝑔 ∈ 𝐺𝑥. We have 𝑎 ∈ 𝐺 𝑥 → 𝑦 since

𝑎 𝑥 = ℎ𝑔 𝑥 = ℎ 𝑥 = 𝑦.

◦ 𝐺 𝑥 → 𝑦 ⊆ ℎ𝐺𝑥. If 𝑏 ∈ 𝐺 𝑥 → 𝑦 then ℎ−1𝑏 𝑥 = ℎ−1 𝑦 = 𝑥.

That is, ℎ−1𝑏 ∈ 𝐺𝑥, which implies 𝑏 ∈ ℎ𝐺𝑥.

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Sizes of Cosets and Stabilizers

Claim. Let 𝐺 be a permutation group on 𝑋and let 𝐺𝑥 be the stabilizer of 𝑥 ∈ 𝑋. Then

𝐺𝑥 = ℎ𝐺𝑥 for any ℎ ∈ 𝐺.

◦ Proof. By the Latin square property of 𝐺.

Corollary. The size of 𝐺 𝑥 → 𝑦 :

◦ If 𝑦 is in the orbit 𝐺𝑥 then 𝐺 𝑥 → 𝑦 = 𝐺𝑥 .

◦ If 𝑦 is not in the orbit 𝐺𝑥 then 𝐺 𝑥 → 𝑦 = 0.

The End: A Noah’s 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. “You haven’t cut down any trees.” Noah is puzzled, but does as they wish. Within a month, you can’t walk a step without treading on baby snakes. With difficulty, he tracks down the two parents. “What was all that with the trees?” “Ah,” says one of the snakes, “you didn’t notice which species we are.” Noah still looks blank. “We’re adders, and we can only multiply using logs.”