Cosets II

Truth is not always black or white, either/or, and there are always two sides to every story. Everyone thinks they’re the good guy. There’s no happy endings, no saints or pure villains, and no atheists in foxholes. Debt is slavery, destiny is a nasty bitch, and size does matter - those who disagree have small penises. The truth is never what you’ve been told, it’s far away from bias narratives. It’s grey, it sets us free, it needs context, history, and perspective, and it lies somewhere in the middle, hidden in the details and complexity of reality. Personal integrity, being minimalistic, avoiding debt and addictions like eating shit, living within your means, and more importantly, the memory of a gold-fish is the only path to happiness, The Ten Immutable, Absolute, and Transcendent Laws of Physics, Psychology, Philosophy, and Reality, Apocalypse, Anawim, #justtothepoint.

Recall

Let G be an arbitrary group, and let H be a subgroup of G, H ≤ G. A left coset of H in G is a subset of the form aH = {ah | h ∈ H} for some a ∈ G. The collection of left cosets is denoted by G/H. Analogously, Ha = {ha | h ∈ H} is the right coset of H in G. The collection of right cosets is denoted by H\G.  

Lemma. Let G be a group, let H be a subgroup of G, H ≤ G, and let g₁, g₂ ∈ G. The following statements are equivalent:

1. g₁H = g₂H
2. Hg₁^-1 = Hg₂^-1
3. g₁H ⊆ g₂H
4. g₁ ∈ g₂H
5. g₁^-1g₂ ∈ H.

Proof: 1 ⇒ 2, let’s suppose g₁H = g₂H

∀x ∈ Hg₁^-1 ⇒ ∃h ∈ H, x = hg₁^-1 ⇒ x^-1 =[The Socks and Shoes Principle] (g₁^-1)^-1h^-1 = g₁h^-1 ⇒ [x^-1 = g₁h^-1, x^-1 ∈ g₁H = g₂H] ∃h’ ∈ H: x^-1 = g₂h’ ⇒ x =[The Socks and Shoes Principle] h’^-1g₂^-1 ∈ Hg₂^-1 ⇒ Hg₁^-1 ⊆ Hg₂^-1

Mutatis mutandis, the same reasoning applies to Hg₂^-1 ⊆ Hg₁^-1, and therefore Hg₁^-1 = Hg₂^-1

2 ⇒ 3) Let’s suppose Hg₁^-1 = Hg₂^-1. ∀x ∈ g₁H, ∃h ∈ H, x = g₁h ⇒ x^-1 =[The Socks and Shoes Principle] h^-1g₁^-1 ∈ Hg₁^-1 = Hg₂^-1 ⇒ ∃h’ ∈ H: x^-1 = h’g₂^-1 ⇒ x =[The Socks and Shoes Principle] g₂h’^-1 ∈ g₂H. Therefore, ∀x ∈ g₁H, x ∈ g₂H ⇒ g₁H ⊆ g₂H∎

3 ⇒ 4) Let’s suppose g₁H ⊆ g₂H ⇒ g₁ = g₁e ∈ g₁H ⊆ g₂H, and therefore g₁ ∈ g₂H∎

4 ⇒ 5) Let’s suppose g₁ ∈ g₂H ⇒ ∃h ∈ H: g₁ = g₂h ⇒ g₁^-1g₂ = (g₂h)^-1g₂ =[The Socks and Shoes Principle] (h^-1g₂^-1)g₂ =[Associativity] h^-1(g₂^-1g₂) = h^-1 ∈ H ⇒ g₁^-1g₂ ∈ H∎

5 ⇒ 1) Let’s suppose g₁^-1g₂ ∈ H, ∀x ∈ g₁H, ∃h, h’ ∈ H: x = g₁h, g₁^-1g₂ = h’ ⇒ h’^-1 = (g₁^-1g₂)^-1 =[The Socks and Shoes Principle] g₂^-1g₁ ⇒[h’^-1 = g₂^-1g₁ and multiplying both sides by g₂, g₂h’^-1 = g₂(g₂^-1g₁)] g₁ = g₂h’^-1 ⇒ x = g₁h = g₂h’^-1h ∈ g₂H ⇒ g₁H ⊆ g₂H

∀x ∈ g₂H, ∃h, h’ ∈ H: x = g₂h, g₁^-1g₂ = h’ ⇒ g₂ = g₁h’ ⇒ x = g₂h = g₁h’h ∈ g₁H ⇒ g₂H ⊆ g₁H ⇒[We have previously demonstrated that g₁H ⊆ g₂H] g₁H = g₂H∎

Examples

• Let G = $GL(2, ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ, ad - bc ≠ 0}} \bigr\}$, and H = $SL(2, ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ, ad - bc = 1}} \bigr\}$. Then, ∀A ∈ G, the coset AH is the set of all 2 x 2 matrices with the same determinant as A.

Let d = det(A), d ≠ 0. If B ∈ SL(2, ℝ) ⇒ det(B) = 1 ⇒ det(AB) = det(A)det(B) = d·1 = d.

• Let G = U₃₀ = {1, 7, 11, 13, 17, 19, 23, 29} and let H = {1, 11}.

1H = {1, 11}. 7H = {7, 7*11 mod 30} = {7, 17}
We don't need to calculate 11H or 17H, we already know that 1H = 11H and 7H = 17H. Next, we choose a representative not already appearing in the previously calculated cosets, say 13. 13H = {13, 23 (13*11 mod 30 = 143 mod 30 = 23)}.

19H = {19, 29}, so we have found the partitioning of G into equivalence classes under the equivalence relation defined by a ~_Rb if and only if a^-1b ∈ H or aH = bH, G/H = {1H, 7H, 13H, 19H} where 1H = 11H = {1, 11}, 7H = 17H = {7, 17}, 13H = 23H = {13, 23}, 19H = 29H = {19, 29}.

Definition. Let G be a group, and let H be a subgroup of G, H ≤ G. The set of cosets of H in G is denoted as G/H. It is called the quotient set of G by or mod H. G/H = {aH | a ∈ G}. The index of H in G is the number of distinct left (right) cosets of H in G. It is denoted as [G:H] = |G/H| where we say [G:H] = ∞ if G/H is infinite.

Example: k ∈ ℤ, k >0, kℤ is a subgroup of ℤ (kℤ ≤ ℤ). The quotient set is ℤ/kℤ. ℤ/kℤ = {[0], [1], … [k-1]} and [ℤ : kℤ] = k.

Theorem. Let H be a subgroup of G. Then, [G/H] = [H\G].

Proof.

Let Φ: G/H → H\G defined by Φ(gH) = Hg^-1

• Φ is well-defined. Suppose gH = g’H, we claim Hg^-1 = Hg^’-1

Let x ∈ Hg^-1 ⇒ ∃h ∈ H: x = hg^-1 ⇒ x^-1 = [Shoes and sock theorem] gh^-1 ∈ gH = g’H ⇒ ∃k ∈ H: x^-1 = g’k ⇒ x = [Shoes and socks theorem] k^-1g’^-1 ∈ Hg’^-1 ⇒ Hg^-1 ⊆ Hg’^-1. Mutatis mutandis, that is, by a completely similar argument, Hg^’-1 ⊆ Hg^-1, hence Hg^-1 = Hg^’-1.

• Let ψ: H\G → G/H, ψ(Hg) = g^-1H. By similar reasoning, this mapping is well-defined. It is indeed the inverse of Φ (Notice: ψ(Φ(gH)) = ψ(Hg^-1) = (g^-1)^-1H = gH and Φ(ψ(Hg)) = Φ(g^-1H) = H(g^-1)^-1 = Hg) ⇒ Φ is invertible and bijective ⇒ [G/H] = [H\G]∎

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.