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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 ∃h_{1}, h_{2} ∈ H : a = bh_{1}, b = ch_{2} ⇒ a = bh_{1} = (ch_{2})h_{1} = c(h_{2}h_{1}), h_{2}h_{1} ∈ 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: (~_{L} is an equivalent relation & mutatis mutandis ~_{R} is also an equivalent relation)
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.
Let x ∈ gH ⇒ ∃h ∈ H: x = gh.
x = gh ⇒[Multiplying by g^{-1} in both sides of the equation] g^{-1}x = g^{-1}(gh) =[Associative] h ∈ H ⇒[g ~ x ↭ g^{-1}x ∈ H ↭ g^{-1}x = h ↭ ∃h ∈ H: x = gh] 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, …}
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~_{R}b 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}
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, there are only two cosets and ℤ_{8} = ⟨2⟩ ∪ (1 + ⟨2⟩.
U_{12} = {1, 5, 7, 11}, ⟨5⟩ = {1, 5} ≤ U_{12}. Let’s compute the cosets: 1⟨5⟩ = ⟨5⟩ = 5⟨5⟩. 7⟨5⟩ = 11⟨5⟩ =[7·5 = 35 ≡ 1 (mod 12)] {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.
Let G be the symmetric group of degree three, G = S_{3} = {1, (12), (13), (23), (123), (132)}, and let H be the alternating group (the subgroup of all even permutations) = 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)}.
(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)}.
(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.
R_{0}K = K = R_{180}K,
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 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:
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^{-1}h ∈ H ⇒[a, a^{-1}h ∈ H ∈ H, H ≤ G] a(a^{-1}h) = h ∈ aH∎
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.
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 ∎
Proof:
aH = bH ↭ b^{-1}(aH) = b^{-1}(bH) ↭ (b^{-1}a)H = (b^{-1}b)H = H ↭[aH = H iff a ∈ H] b^{-1}a ∈ H∎
However, it is not true that aH = bH ↭ ab^{-1} ∈ H. Counterexample: G = S_{3}, H = ⟨(12)⟩ = {1, (12)}, a = (123), b = (13), a^{-1}b = (132)(13) = (12) ∈ H ⇒ (123)H = (13)H = {(123), (13)}. However, ab^{-1} = (123)(13) = (23) ∉ 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 cardinality (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}ah = a^{-1}ah’ ⇒ 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.
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}
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 ∎