# Definition.

Let G be a group, H ≤ G a subgroup. We define an equivalent relation on G, ∀a, b ∈ G, a ≡ b mod H if ∃h ∈ H : a = bh. The equivalent classes are called cosets.

Proof: Reflexivity. a ≡ a mod H if ∃h ∈ H : a = ah, but h = e ∈ H because H is a subgroup.

Symmetry. ∀a, b ∈ G, a ≡ b mod H if ∃h ∈ H : a = bh ⇒ b = ah-1, h-1∈ H because H is a subgroup.

Transitivity. ∀a, b, c ∈ G, a ≡ b mod H, b ≡ c mod H ∃h1, h2 ∈ H : a = bh1, b = ch2 ⇒ a = c(h1h2), h1h2 ∈ H because H is a subgroup, so a ≡ c mod H.

Let G be a group, H < G a subgroup. We define two equivalent relations on G, ∀a, b ∈ G, a ~L b if and only if a-1b ∈ H, a ~R b if and only if ab-1 ∈ H

Proof:

1. Reflexive, a-1a = e ∈ H ⇒ a~a.
2. Symmetric. Suppose a ~ b ⇒ a-1b ∈ H ⇒ [H ≤ G] (a-1b)-1 ∈ H ⇒ [Sock and shoes theorem] (a-1b)-1 = b-1a ∈ H ↭ b ~ a.
3. Transitivity. Suppose a ~ b and b ~ c ⇒ a-1b, b-1c ∈ H ⇒ [H ≤ G] a-1bb-1c ∈ H, a-1c ∈ H ⇒ a ~ c.

Alternative definition. Let G be a group, H < G a subgroup. 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.

We have demonstrated that this is a equivalent relation, we claim gH = [g], and g is the representative of the coset gH.

Proof.

• gH ⊆ [g]

Let x ∈ gH ⇒ ∃h ∈ H: x = gh.

g-1x = g-1(gh) = h ∈ H ⇒ g ~ x ⇒ x ∈ [g]∎

• [g] ⊆ gH

Let x ∈ [g] ⇒ g ~ x ⇒ g-1x ∈ H ⇒ ∃h ∈ H: g-1x = h ⇒ x = gh ∈ gH ∎

We use |aH|, |Ha| to express or denote the number of elements in aH or Ha respectively.

# Examples

• Let H = 3ℤ < ℤ. 3ℤ = {…, -9, -6, -3, 0, 3, 6, 9, …}

a = 0, 0H = 0 + H (additive notation) = 3ℤ.

1 + 3ℤ = {…, -8, -5, -2, 1, 4, 7, 10, …}

2 + 3ℤ = {…, -7, -4, -1, 2, 5, 8, 11, …}

This three left cosets constitute the partition of ℤ into left cosets of 3ℤ. Since ℤ is Abelian, left cosets and right cosets are the same. Besides, a~Rb if and only if ab-1 ∈ nℤ, a-b ∈ nℤ (addition notation) is the same as the relation of congruence modulo n, that is, a ≡ b mod n (a-b divisible by n).

Therefore, the partition of ℤ into cosets of nℤ is the partition of ℤ into residue classes modulo n and we refer it as cosets modulo nℤ.

Example. Let H = 4ℤ ≤ ℤ. Then, ℤ = 4ℤ ∪ (1 + 4ℤ) ∪ (2 + 4ℤ) ∪ (3 + 4ℤ) = {4n: n ∈ ℤ} + {1 + 4n: n ∈ ℤ} + {2 + 4n: n ∈ ℤ} ∪ {3 + 4n: n ∈ ℤ}

• 8, ⟨2⟩ = {0, 2, 4, 6} ≤ ℤ8. Then, 0 + ⟨2⟩ = ⟨2⟩. 1 + ⟨2⟩ = {1, 3, 5, 7}. 2 + ⟨2⟩ = ⟨2⟩. 3 + ⟨2⟩ = = {3, 5, 7, 1} = 1 + ⟨2⟩. Thus, ℤ8 = ⟨2⟩ ∪ (1 + ⟨2⟩)

• U12 = {1, 5, 7, 11}, ⟨5⟩ = {1, 5}. Let’s compute the cosets: 1⟨5⟩ = ⟨5⟩ = 5⟨5⟩. 7⟨5⟩ = 11⟨5⟩ = {7, 11}. Thus, U12 = ⟨5⟩ ∪ 7⟨5⟩.

• H = ⟨3⟩ = {0, 3} ≤ ℤ6. The partition of ℤ6 into cosets of {0, 3} is 0 + H = H + 0 = {0, 3}, 1 + H = H + 1 = {1, 4}, and 2 + H = H + 2 = {2, 5}.

6 is Abelian, the left and right cosets are the same thing.

• G = S3, H = A3 = ⟨(123)⟩ ≤ S3. H = {1, (123), (132)}. Let’s compute the left cosets of H in G.

(id)H = (123)H = (132)H = H = {1, (123), (132)},
(12)H = (23)H = (13)H = {(12), (23), (13)}.

Futhermore, (id)H = (123)H = (132)H = H(id) = H(123) = H(132)
(12)H = (23)H = (13)H = H(12) = H(23) = H(13) = {(12), (23), (13)}.

• G = S3 = {1, (12), (13), (23), (123), (132)}, H = ⟨(12)⟩ = {1, (12)}. Let’s compute the left cosets of H in G.

(1)H = (12)H = H = {1, (12)},
(123)H = (13)H = {(123), (13)},
(132)H = (23)H = {(132), (23)}.

G = H ∪ (13)H ∪ (23)H

However, H(123) = H(23) = {(123), (23)}, and H(132) = H(13) = {(132), (13)}, and in particular, (123)H ≠ H(123).

• G = S3 = {id, (12), (13), (23), (123), (132)}, H = {id, (12)}. Let’s compute the left cosets of H in G.

(id)H = H,
(123)H = {(123), (123)(12)} = {(123), (13)},
(132)H = {(132), (132)(12)} = {(132), (23)}

G = H ∪ (123)H ∪ (132)H.

• G = D4, K = {R0, R180}.

R0K = K = R180,
R90K = {R90R0, R90R180} = {R90, R270} = R270K,
VK = {V, H} = HK,
DK = {D, D’} = D’K.

G = K ∪ R90K ∪ VK ∪ DK

• G = D4, K = {R0, R180, V, H}.

R0K = R180K = VK = HK = K
R90K = R270K = DK = D’K = {R90, R270, R90V, R90H} = {R90, R270, D, D'}

G = K ∪ R90K

# Properties of Cosets

Let H be a subgroup of G, and let a, b ∈ G. Then, the following statements hold true:

• The left (or right) coset of H containing a does contain a, a ∈ aH. Proof: a = ae ∈ aH [e ∈ H because H is a subgroup]
• A coset H absorbs an element if and only if the element belongs to it, aH = H iff a ∈ H.

Proof. Suppose aH = H ⇒ a = ae ∈ aH = H.

Suppose a ∈ H, aH ⊆ H because H is a subgroup, so there is closure.

Let’s prove that H ⊆ aH, let h ∈ H, h ∈ aH? Since a, h ∈ H ⇒ a-1h ∈ H ⇒ a(a-1h) = h ∈ aH

• A left (or right) coset is uniquely determined or represented by any one of its elements, i.e., aH = bH iff a ∈ bH.

Proof. If aH = bH ⇒ a = ae ∈ aH = bH.

If a ∈ bH ⇒ ∃h ∈ H: a = bh ⇒ aH = (bh)H = b(hH) = [aH = H iff a ∈ H] bH.

• Two left (or right) cosets are either identical or disjoint, i.e., aH = bH or aH ∩ bH = ∅.

Let’s suppose aH ∩ bH ≠ ∅ ⇒ ∃ c ∈ aH ∩ bH ⇒ c ∈ aH and c ∈ bH ⇒ [aH = bH iff a ∈ bH] cH = aH, cH = bH ⇒ aH = bH ∎

• aH = bH iff a-1b ∈ H.

Proof: aH = bH iff H = a-1bH ⇒ [aH = H iff a ∈ H] aH = bH iff a-1b ∈ H ∎

• Let G be a group, H ≤ G, g ∈ G. Then, |H| = |gH|

Proof.

We define a mapping Φ: H → gH, Φ(h) = gh which is well-defined.

1. Let’s prove that his mapping is injective, let Φ(h) = Φ(h’) ↭ gh = gh’ ⇒ [H ≤ G, the cancellation laws applies] h = h'.
2. Let’s prove that his mapping is onto. Let x ∈ gH ⇒ ∃h ∈ H: x = gh, and therefore Φ(h) = gh = x.

Φ is bijective ⇒ |H| = |gH| ∎ The same reasoning can be used to demonstrate that Φ: H → Hg defined by Φ(h) = hg is bijective, and therefore |H| = |gH| = |Hg|, that is not to say that gH = Hg, but they have the same number of elements.

• Any two left (or right) cosets have the same size or number of elements, i.e., |aH| = |bH|

Proof: (for educational purposes) Let G be a group, H < G a subgroup. Let’s define f: aH → bH by f(ah) = bh.

is it well defined? Suppose ah = ah’ ⇒ a-1aah = a-1ah’ ⇒ h = h'.

Likewise, I can define, g: bH → aH by g(bh) = ah. Since f and g are clearly inverses, f or g are bijections, and aH and bH have the same number of elements.

Another way of proving is saying that the map is (ba-1): aH → bH by f(ah) = bh. Its inverse is multiplication by ab-1.

• aH = Ha iff H = aHa-1.

Proof: aH = Ha iff (aH)a-1 = H if and only if (aH)a-1 = (Ha)a-1 = H(aa-1) = H.

• H itself is the only coset of H that is a subgroup of G, i.e. aH ≤ G iff a ∈ H.

Proof: aH ≤ G ⇒ e ∈ H, aH (every group contains the identity) ⇒ e ∈ aH ∩ eH ⇒ [Two left (or right) cosets are either identical or disjoint, i.e., aH = bH or aH ∩ bH = ∅.] aH = eH = H ⇒ [aH = H iff a ∈ H] a ∈ H.

Conversely, let a ∈ H ⇒ [aH = H iff a ∈ H] aH = H ≤ G ∎

Lemma. Let G be a group, H ≤ G, 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 = g1h-1 ⇒ [x ∈ g1H = g2H] ∃h’ ∈ H: x-1 = g2h’ ⇒ x = 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 = h-1g1-1 ∈ Hg1-1 = Hg2-1 ⇒ ∃h’ ∈ H: x-1 = h’g2-1 ⇒ x = g2h’-1 ∈ g2H

3 ⇒ 4, let’s suppose g1H ⊆ g2H ⇒ g1 = g1e ∈ g1H ⊆ g2H

4 ⇒ 5, let’s suppose g1 ∈ g2H ⇒ ∃h ∈ H: g1 = g2h ⇒ g1-1g2 = h-1g2-1g2 = h-1 ∈ H.

5 ⇒ 1, let’s suppose g1-1g2 ∈ H, ∀x ∈ g1H, ∃h, h’ ∈ H: x = g1h, g1-1g2 = h’ ⇒ h’-1 = 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 ⇒ 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, 77 mod 30} = {7, 17}
We don't need to calculate 11H or 17H, we already know that 1H = 11H and 7H = 17H. 13H = {13, 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]

Definition. Let G be a group, H < G a subgroup. The set of cosets of H in G is denoted as G/H and called quotient set of G by H. G/H = {aH | a ∈ G}. The index of H in G 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. 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, by a similar argument Hg’-1 ⊆ Hg-1, Hg-1 = Hg’-1.

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