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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 g_{1}, g_{2} ∈ G. The following statements are equivalent:

- g
_{1}H = g_{2}H - Hg
_{1}^{-1}= Hg_{2}^{-1} - g
_{1}H ⊆ g_{2}H - g
_{1}∈ g_{2}H - g
_{1}^{-1}g_{2}∈ H.

Proof:
1 ⇒ 2, let’s suppose g_{1}H = g_{2}H

∀x ∈ Hg_{1}^{-1} ⇒ ∃h ∈ H, x = hg_{1}^{-1} ⇒ x^{-1} =[The Socks and Shoes Principle] (g_{1}^{-1})^{-1}h^{-1} = g_{1}h^{-1} ⇒ [x^{-1} = g_{1}h^{-1}, x^{-1} ∈ g_{1}H = g_{2}H] ∃h’ ∈ H: x^{-1} = g_{2}h’ ⇒ x =[The Socks and Shoes Principle] h’^{-1}g_{2}^{-1} ∈ Hg_{2}^{-1} ⇒ Hg_{1}^{-1} ⊆ Hg_{2}^{-1}

Mutatis mutandis, the same reasoning applies to Hg_{2}^{-1} ⊆ Hg_{1}^{-1}, and therefore Hg_{1}^{-1} = Hg_{2}^{-1}

2 ⇒ 3) Let’s suppose Hg_{1}^{-1} = Hg_{2}^{-1}. ∀x ∈ g_{1}H, ∃h ∈ H, x = g_{1}h ⇒ x^{-1} =[The Socks and Shoes Principle] h^{-1}g_{1}^{-1} ∈ Hg_{1}^{-1} = Hg_{2}^{-1} ⇒ ∃h’ ∈ H: x^{-1} = h’g_{2}^{-1} ⇒ x =[The Socks and Shoes Principle] g_{2}h’^{-1} ∈ g_{2}H. Therefore, ∀x ∈ g_{1}H, x ∈ g_{2}H ⇒ g_{1}H ⊆ g_{2}H∎

3 ⇒ 4) Let’s suppose g_{1}H ⊆ g_{2}H ⇒ g_{1} = g_{1}e ∈ g_{1}H ⊆ g_{2}H, and therefore g_{1} ∈ g_{2}H∎

4 ⇒ 5) Let’s suppose g_{1} ∈ g_{2}H ⇒ ∃h ∈ H: g_{1} = g_{2}h ⇒ g_{1}^{-1}g_{2} = (g_{2}h)^{-1}g_{2} =[The Socks and Shoes Principle] (h^{-1}g_{2}^{-1})g_{2} =[Associativity] h^{-1}(g_{2}^{-1}g_{2}) = h^{-1} ∈ H ⇒ g_{1}^{-1}g_{2} ∈ H∎

5 ⇒ 1) Let’s suppose g_{1}^{-1}g_{2} ∈ H, ∀x ∈ g_{1}H, ∃h, h’ ∈ H: x = g_{1}h, g_{1}^{-1}g_{2} = h’ ⇒ h’^{-1} = (g_{1}^{-1}g_{2})^{-1} =[The Socks and Shoes Principle] g_{2}^{-1}g_{1} ⇒[h’^{-1} = g_{2}^{-1}g_{1} and multiplying both sides by g_{2}, g_{2}h’^{-1} = g_{2}(g_{2}^{-1}g_{1})] g_{1} = g_{2}h’^{-1} ⇒ x = g_{1}h = g_{2}h’^{-1}h ∈ g_{2}H ⇒ g_{1}H ⊆ g_{2}H

∀x ∈ g_{2}H, ∃h, h’ ∈ H: x = g_{2}h, g_{1}^{-1}g_{2} = h’ ⇒ g_{2} = g_{1}h’ ⇒ x = g_{2}h = g_{1}h’h ∈ g_{1}H ⇒ g_{2}H ⊆ g_{1}H ⇒[We have previously demonstrated that g_{1}H ⊆ g_{2}H] g_{1}H = g_{2}H∎

- 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 = U
_{30}= {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 ~ _{R}b if and only if a^{-1}b ∈ 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^{-1}g’^{-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
^{-1}H. By similar reasoning, this mapping is well-defined. It is indeed the inverse of Φ (Notice: ψ(Φ(gH)) = ψ(Hg^{-1}) = (g^{-1})^{-1}H = gH and Φ(ψ(Hg)) = Φ(g^{-1}H) = H(g^{-1})^{-1}= Hg) ⇒ Φ is invertible and bijective ⇒ [G/H] = [H\G]∎

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.

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