Cosets

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Definition.

Let G be an arbitrary group, and let H be a subgroup of G, H ≤ G . 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, and obviously a = ae.

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 = bh₁ = (ch₂)h₁ = c(h₂h₁), h₂h₁ ∈ 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: (~_L is an equivalent relation & mutatis mutandis ~_R is also an equivalent relation)

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 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.

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.

x = gh ⇒[Multiplying by g^-1 in both sides of the equation] g^-1x = g^-1(gh) =[Associative] h ∈ H ⇒[g ~ x ↭ g^-1x ∈ H ↭ g^-1x = h ↭ ∃h ∈ H: x = gh] 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, …}

These 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 ↭[Additive notation] a-b ∈ nℤ 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}

• ℤ₈, ⟨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, there are only two cosets and ℤ₈ = ⟨2⟩ ∪ (1 + ⟨2⟩.

• U₁₂ = {1, 5, 7, 11}, ⟨5⟩ = {1, 5} ≤ U₁₂. Let’s compute the cosets: 1⟨5⟩ = ⟨5⟩ = 5⟨5⟩. 7⟨5⟩ = 11⟨5⟩ =[7·5 = 35 ≡ 1 (mod 12)] {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}.

  ℤ₆ is Abelian ⇒ the left and right cosets are the same.

• Let G be the symmetric group of degree three, G = S₃ = {1, (12), (13), (23), (123), (132)}, and let H be the alternating group (the subgroup of all even permutations) = 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)}.

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

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), i.e., right and left cosets are not always identical.

• Let G be the dihedral group of order 8, G = D₄, K = {R₀, R₁₈₀}. D₄ = {id (R₀), rotations (R₉₀, R₁₈₀, R₂₇₀), reflections or mirror images (V, H, D, D’)}.

R₀K = K = R₁₈₀K,
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 G be an arbitrary group, let H be a subgroup of G, H ≤ G, and let a, b be two arbitrary elements of G, a, b ∈ G. Then, the following statements hold true:

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

Proof.

⇒) Suppose aH = H ⇒[H ≤ G ⇒ e ∈ H] a = ae ∈ aH = H ⇒ a ∈ H.

⇐) Suppose a ∈ H ⇒ aH ⊆ H because H is a subgroup, so there is closure under the group operation.

Let’s prove that H ⊆ aH, let h ∈ H, h ∈ aH? Since a, h ∈ H ⇒ a^-1h ∈ H ⇒[a, a^-1h ∈ H ∈ H, H ≤ G] a(a^-1h) = h ∈ aH∎

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

Proof.

⇒) If aH = bH ⇒ a = ae ∈ aH = bH ⇒ a ∈ bH.

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

• Any two left (or right) cosets are either identical or disjoint, i.e., aH = bH or aH ∩ bH = ∅. The left (right) cosets form a partition of G, because every element of G belong to one and only one left (right) coset. Futhermore, the identity is only in one coset, and that coset is H itself and this is the only coset that is a subgroup.

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

• aH = bH iff a^-1b ∈ H.

Proof:

aH = bH ↭ b^-1(aH) = b^-1(bH) ↭ (b^-1a)H = (b^-1b)H = H ↭[aH = H iff a ∈ H] b^-1a ∈ H∎

However, it is not true that aH = bH ↭ ab^-1 ∈ H. Counterexample: G = S₃, H = ⟨(12)⟩ = {1, (12)}, a = (123), b = (13), a^-1b = (132)(13) = (12) ∈ H ⇒ (123)H = (13)H = {(123), (13)}. However, ab^-1 = (123)(13) = (23) ∉ H.

• Let G be a group, H ≤ G, and g an arbitrary element of 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 this mapping is injective, let Φ(h) = Φ(h’) ↭ gh = gh’ ⇒ [H ≤ G, the cancellation laws applies] h = h'.
2. Let’s prove that this 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 cardinality (number of elements).

• Any two left (or right) cosets have the same cardinality or number of elements, i.e., ∀a, b ∈ G, |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^-1ah = a^-1ah’ ⇒ h = h’ ⇒ bh = bh'.

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

• aH = Ha iff H = aHa^-1.

Proof:

aH = Ha ↭ (aH)a^-1 = (Ha)a^-1 = H(aa^-1) = H ↭ aHa^-1 = H∎

Notice that these equalities are not that of elements, but that of sets, e.g., aH = {ah: h ∈ H}, (aH)a^-1 = {(ah)a^-1 : h ∈ H} =_{Associativity} {aha^-1 : h ∈ H} and a^-1(aH) = {a^-1(ah): h ∈ H} =_{Associativity} H = (Ha)a^-1 = {(ha)a^-1 : h ∈ 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, e ∈ aH (every group contains the identity) ⇒ [A coset H absorbs an element if and only if the element belongs to it, aH = H iff a ∈ H.] 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 ∎

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