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 ∃h_{1}, h_{2} ∈ H : a = bh_{1}, b = ch_{2} ⇒ a = c(h_{1}h_{2}), h_{1}h_{2} ∈ 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^{-1}b ∈ H, a ~_{R} b if and only if ab^{-1} ∈ H
Proof:
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.
Let x ∈ gH ⇒ ∃h ∈ H: x = gh.
g^{-1}x = g^{-1}(gh) = h ∈ H ⇒ g ~ x ⇒ x ∈ [g]∎
Let x ∈ [g] ⇒ g ~ x ⇒ g^{-1}x ∈ H ⇒ ∃h ∈ H: g^{-1}x = h ⇒ x = gh ∈ gH ∎
We use |aH|, |Ha| to express or denote the number of elements in aH or Ha respectively.
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~_{R}b 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}
ℤ_{8}, ⟨2⟩ = {0, 2, 4, 6} ≤ ℤ_{8}. Then, 0 + ⟨2⟩ = ⟨2⟩. 1 + ⟨2⟩ = {1, 3, 5, 7}. 2 + ⟨2⟩ = ⟨2⟩. 3 + ⟨2⟩ = = {3, 5, 7, 1} = 1 + ⟨2⟩. Thus, ℤ_{8} = ⟨2⟩ ∪ (1 + ⟨2⟩)
U_{12} = {1, 5, 7, 11}, ⟨5⟩ = {1, 5}. Let’s compute the cosets: 1⟨5⟩ = ⟨5⟩ = 5⟨5⟩. 7⟨5⟩ = 11⟨5⟩ = {7, 11}. Thus, U_{12} = ⟨5⟩ ∪ 7⟨5⟩.
H = ⟨3⟩ = {0, 3} ≤ ℤ_{6}. The partition of ℤ_{6} 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}.
ℤ_{6} is Abelian, the left and right cosets are the same thing.
G = S_{3}, H = A_{3} = ⟨(123)⟩ ≤ S_{3}. 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)}.
(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).
(id)H = H,
(123)H = {(123), (123)(12)} = {(123), (13)},
(132)H = {(132), (132)(12)} = {(132), (23)}
G = H ∪ (123)H ∪ (132)H.
R_{0}K = K = R_{180},
R_{90}K = {R_{90}R_{0}, R_{90}R_{180}} = {R_{90}, R_{270}} = R_{270}K,
VK = {V, H} = HK,
DK = {D, D’} = D’K.
G = K ∪ R_{90}K ∪ VK ∪ DK
R_{0}K = R_{180}K = VK = HK = K
R_{90}K = R_{270}K = DK = D’K = {R_{90}, R_{270}, R_{90}V, R_{90}H} = {R_{90}, R_{270}, D, D'}
G = K ∪ R_{90}K
Let H be a subgroup of G, and let a, b ∈ G. Then, the following statements hold true:
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^{-1}h ∈ H ⇒ a(a^{-1}h) = h ∈ aH
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.
Let’s suppose aH ∩ bH ≠ ∅ ⇒ ∃ c ∈ aH ∩ bH ⇒ c ∈ aH and c ∈ bH ⇒ [aH = bH iff a ∈ bH] cH = aH, cH = bH ⇒ aH = bH ∎
Proof: aH = bH iff H = a^{-1}bH ⇒ [aH = H iff a ∈ H] aH = bH iff a^{-1}b ∈ H ∎
Proof.
We define a mapping Φ: H → gH, Φ(h) = gh which is well-defined.
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.
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^{-1}aah = a^{-1}ah’ ⇒ 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}.
Proof: aH = Ha iff (aH)a^{-1} = H if and only if (aH)a^{-1} = (Ha)a^{-1} = H(aa^{-1}) = 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, g_{1}, g_{2} ∈ G. The following statements are equivalent:
Proof: 1 ⇒ 2, let’s suppose g_{1}H = g_{2}H
x ∈ Hg_{1}^{-1} ⇒ ∃h ∈ H, x = hg_{1}^{-1} ⇒ x^{-1} = g_{1}h^{-1} ⇒ [x ∈ g_{1}H = g_{2}H] ∃h’ ∈ H: x^{-1} = g_{2}h’ ⇒ x = 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} = h^{-1}g_{1}^{-1} ∈ Hg_{1}^{-1} = Hg_{2}^{-1} ⇒ ∃h’ ∈ H: x^{-1} = h’g_{2}^{-1} ⇒ x = g_{2}h’^{-1} ∈ 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
4 ⇒ 5, let’s suppose g_{1} ∈ g_{2}H ⇒ ∃h ∈ H: g_{1} = g_{2}h ⇒ g_{1}^{-1}g_{2} = h^{-1}g_{2}^{-1}g_{2} = h^{-1} ∈ 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_{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_{1}H ⊆ g_{2}H ⇒ g_{1}H = g_{2}H
Let d = det(A), d ≠ 0. If B ∈ SL(2, ℝ) ⇒ det(B) = 1 ⇒ det(AB) = det(A)det(B) = d·1 = d.
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 ~_{R}b if and only if a^{-1}b ∈ 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}
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, by a similar argument Hg^{’-1} ⊆ Hg^{-1}, Hg^{-1} = Hg^{’-1}.