“I will fart in your general direction,” I muttered in a barely audible and intelligible voice, and I proceeded doing so against what the school etiquette demanded. [···]
The teacher looked at the sky, a cry of anguish for God’s mercy and help that was relentlessly begged, but never answered. “Obviously,” she thought, “it’s God’s revenge for being a committed and militant atheist, proud gay, and communist.” What an unforgiven decision to be a teacher - just three good reasons, June, July, and August!, Anawim, Apocalypse, #justtothepoint.
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. Analogously, Ha = {ha | h ∈ H} is the right coset of H in G.
Definition. Let G be a group, H ≤ G a subgroup. The set of cosets of a subgroup H in a group G, written or denoted as G/H, is called the quotient set of G by H. G/H = {aH | a ∈ G}. Do the cosets form a group?
If we want the set of cosets to form a group, then given two cosets, say xH and yH, their product need to be a coset, (xH)(yH) = xyH. Let’s see when this happens. Let’s take two representatives, x·n_{1}∈ xH, y·n_{2} ∈ yH.
(x·n_{1})·(y·n_{2}) ∈ xyH ⇒ (x·n_{1})·(y·n_{2}) = x· y·n_{3}
x^{-1}·(x·n_{1})·(y·n_{2}) = x^{-1}·x· y·n_{3}
n_{1}·y·n_{2} = y·n_{3}
y^{-1}·n_{1}·y·n_{2} = y^{-1}·y·n_{3}
y^{-1}·n_{1}·y·n_{2} = n_{3}
y^{-1}·n_{1}·y = n_{3}·n_{2}^{-1} ∈ H for any y ∈ G.
Therefore, H ≤ G and we want the set of cosets to form a group, then we need that y^{-1}·n_{1}·y = n_{3}·n_{2}^{-1} ∈ H for any y ∈ G ↭[n_{3}, n_{2}∈ H, H ≤ G ⇒ n_{3}·n_{2}^{-1} ∈ H] ∀ y ∈ G, n_{1} ∈ H, y^{-1}·n_{1}·y ∈ H When this condition is true, we say that H is a normal subgroup of G.
Definition. A subgroup H of a group G, H ≤ G, is called a normal subgroup of G if aH = Ha, ∀a ∈ G. That is, a normal subgroup of a group G is one in which the right and left cosets are precisely the same. The usual notation for this relation is H ◁ G. It means ∀a ∈ G, h ∈ H, ∃h’, h’’∈ H: ah = h’a and ha = ah’’.
Normal Subgroup Test or alternative definition. A subgroup H of a group G is normal in G if and only if is invariant under conjugation by members of the group G. H ◁ G iff gHg^{-1} ⊆ H, ∀g ∈ G, i.e., ∀g ∈ G, h ∈ H, ghg^{-1} ∈ H.
Actually, for any arbitrary subgroup H of G, the set ghg^{-1} will be another subgroup of G, so when H is normal in G, then ghg^{-1} will be a subgroup of H. In fact, it will be equal to H.
Proof:
⇒) H ◁ G ⇒ ∀a ∈ G, ∀h ∈ H, ∃h’ ∈ H: ah = h’a ⇒ a^{-1}h’a = h ∈ H ⇒ ∀a ∈ G, aHa^{-1} ⊆ H
⇐) ∀g ∈ G, gHg^{-1} ⊆ H ⇒ [g = a] aHa^{-1} ⊆ H ⇒ ∀a ∈ G, aH ⊆ Ha
∀g ∈ G, gHg^{-1} ⊆ H ⇒ [g = a^{-1}] a^{-1}Ha ⊆ H ⇒ ∀a ∈ G, Ha ⊆ aH ⇒[We have previously demonstrated that aH ⊆ Ha] aH = Ha ⇒ H ◁ G ∎
Theorem. Let G be a group, N ≤ G, the following statements are equivalent.
Proof. We only need to prove (2) ⇒ (3)
Suppose, gHg^{-1} ⊆ H, H ⊆ gHg^{-1}?
∀h ∈ H, g ∈ G, ghg^{-1} ∈ H ⇒ ∃h’ ∈ H: ghg^{-1} = h’ ⇒ h = g^{-1}h’g = g^{-1}h’(g^{-1})^{-1} ⇒ ∀h ∈ H, h = g^{-1}h’(g^{-1})^{-1} ∈ g^{-1}H(g^{-1})^{-1} ⇒ ∀g ∈ G, H ⊆ g^{-1}H(g^{-1})^{-1}, and we know (2) g^{-1}H(g^{-1})^{-1} ⊆ H ⇒ ∀g ∈ G, g^{-1}H(g^{-1})^{-1} = H ↭ ∀g ∈ G, (g^{-1})^{-1}Hg^{-1} = H ↭ ∀g∈ G, gHg^{-1} = H∎
Examples:
Recall. The index of a subgroup H (H ≤ G) is the number of its distinct left cosets, or equivalently, the number of its distinct right cosets. It is denoted as |G : H|. When G is a finite group, the following equation is satisfied |G : H| = |G|/|H|.
If |G : H| = 2 ⇒ There are only two distinct left cosets, say H and aH (a ∉ H), and two distinct right cosets, say H and Hb (b ∉ H). Since H and aH makes a partition of G ⇒ they are obviously disjoint. The same applies to H and Hb ⇒ aH = Hb, and this coset contains all the elements of G that are not in H, a ∈ G \ H ⇒ aH = Ha = Hb = bH ⇒ H ◁ G.
The subgroup of rotations ⟨r⟩ = {e, r, r^{2}} is a cyclic subgroup of order three. It is also normal, ⟨r⟩ ◁ D_{3}.
Proof:
∀a ∈ G, aH = Ha? We only really need to check this equation for all a ∈ D_{3} that are not in the subgroup, but even better as D_{3} = ⟨r, s⟩, we only need to check it for the generator that is not part of the subgroup ⟨r⟩, that is, s.
s⟨r⟩ = {s, sr, sr^{2}}
⟨r⟩s = {s, rs, r^{2}s}, but we know that rs = sr^{2}. Futhermore, recall that r^{k}s = sr^{n-k}, 1 ≤ k ≤ n-1 ⇒ [n = 3, k = 2] r^{2}s = sr ⇒ s⟨r⟩ = ⟨r⟩s ⇒ ⟨r⟩ ◁ D_{3}. Futhermore, D_{3}/⟨r⟩ = {⟨r⟩, s⟨r⟩} ≋ ℤ_{2}. We can define this isomorphism ⟨r⟩ → 0, and s⟨r⟩ → 1.
⟨r⟩ | s⟨r⟩ | |
---|---|---|
⟨r⟩ | ⟨r⟩ | s⟨r⟩ |
s⟨r⟩ | s⟨r⟩ | ⟨r⟩ -ss⟨r⟩ = s^{2}⟨r⟩ = e⟨r⟩ = ⟨r⟩- |
r⟨s⟩ = {r, rs} = {r, sr^{2}} ≠ ⟨s⟩r = {r, sr} because rs ≠ r, rs ≠ sr.
The alternating group of even permutations A_{3} = {(), (123) -(13)(12)-, (132) -(12)(13)-} is a normal subgroup of S_{3}, A_{3} ◁ S_{3}, e.g., (12)A_{3} = {(12), (12)(123), (12)(132)} = {(12), (23), (13)} = A_{3}(12), even though (12)(123) = (23) ≠ (123)(12) = (13). The cosets are A_{3} = {(), (123), (132)} and (12)A_{3} = {(12), (23), (13)}. The factor group S_{3}/A_{3} has the following multiplication table: (and is isomorphic to ℤ_{2})
A_{3} | (12)A_{3} | |
---|---|---|
A_{3} | A_{3} | (12)A_{3}, e.g., (123)(12) = (13) ∈ (12)A_{3} |
(12)A_{3} | (12)A_{3} | A_{3}, e.g., (12)(12)=() |
H < S_{3} = {(1), (12)} is not normal because (123)H = {(123)(1), (123)(12)} = {(123), (13)} ≠ H(123) = {(1)(123), (12)(123)} = {(123), (23)}
H = ()H = (12)(34)H = (13)(24)H = (14)(23)H = H(1) = H(12)(34) = H(13)(24) = H(14)(23)
(123)H = {(123), (134), (243), (142)}. (123)H = (134)H = (243)H = (142)H = H(123) = H(134) = H(243) = H(142)
(123)(12)(34) = (134), (123)(13)(24) = (243), (123)(14)(23) = (142)
(132)H = {(132), (234), (124), (143)}. (132)H = (234)H = (124)H = (143)H = H(132) = H(234) = H(124) = H(143). And therefore, H ◁ A4 ∎ In general, A_{n} ◁ S_{n}
For any subgroup H of G, the set gHg^{-1} is always another subgroup of G. However, when H is normal, then gHg^{-1} is a subgroup of H. In fact, it is H itself.
∀x ∈ GL(2, ℝ), h ∈ SL(2, ℝ), xhx^{-1} ∈ SL(2, ℝ)? det(xhx^{-1}) =[h ∈ SL(2, ℝ) ⇒ det(h) = 1] det(x)det(h)det(x)^{-1} = det(x)det(x)^{-1} = 1 ⇒ xhx^{-1} ∈ SL(2, ℝ)∎
Theorem. If H is a normal subgroup of a group G, H ◁ G, then the set of left (or right) cosets of H in G form a group G/H = {aH | a ∈ G} ≤ G under the operation (aH)(bH) = abH. This group is called the factor or quotient group of G and H of order [G : H]. It comes with a natural group homomorphism Φ: G → G/H such that Φ is onto and Ker(Φ)= H.
Proof:
The converse of this theorem is also true, that is, if the operation (aH)(bH) = abH defines a group on the set of left (or right) cosets of H in G, then H is a normal subgroup.
0 + 3ℤ = 3ℤ = {0, ±3, ±6, ±9, …}
1 + 3ℤ = {1, 4, 7, …; -2, -5, -8, …}
2 + 3ℤ = {2, 5, 8, …; -1, -4, -7, …}
The group ℤ/3ℤ is given by the Cayley table below.
+ | 0 + 3ℤ | 1 + 3ℤ | 2 + 3ℤ |
---|---|---|---|
0 + 3ℤ | 0 + 3ℤ | 1 + 3ℤ | 2 + 3ℤ |
1 + 3ℤ | 1 + 3ℤ | 2 + 3ℤ | 0 + 3ℤ |
2 + 3ℤ | 2 + 3ℤ | 0 + 3ℤ | 1 + 3ℤ |
0 + 4ℤ = 4ℤ = {0, ±4, ±8, ±12, …}
1 + 4ℤ = {1, 5, 9, 13, …; -3, -7, -11, …}
2 + 4ℤ = {2, 6, 10, 14, …; -2, -6, -10, …}
3 + 4ℤ = {3, 7, 11, 15, …; -1, -5, -9, …}
(1 + 4Z)^{2} = (1 + 4Z) + (1 + 4Z) = 2 + 4Z
(1 + 4Z)^{3} = (1 + 4Z) + (1 + 4Z) + (1 + 4ℤ) = 3 + 4Z
(1 + 4Z)^{4} = (1 + 4Z) + (1 + 4Z) + (1 + 4ℤ) + (1 + 4ℤ) = 4ℤ ⇒ ℤ/4ℤ = {0 + 4ℤ, 1 + 4ℤ, 2 + 4ℤ, 3 + 4ℤ} = ⟨1 + 4ℤ⟩. The factor group is a cyclic group of order 4. The natural homomorphism is Φ: ℤ → ℤ/4ℤ, Φ(m) = m + 4ℤ = r + 4ℤ where m = 4p +r, r is the remainder, 0 ≤ r < 4. Φ is onto and Ker(Φ) = 4ℤ.
There are no other cosets because if k ∈ ℤ, ∃q, r: 0 ≤ r < 4, k = 4q + r ⇒ k ∈ r + 4ℤ for some r, 0 ≤ r < 4.
This group is given by the Cayley table 1.a below.
Notice that ℤ/4ℤ ≋ ℤ_{4}. In general, ∀n > 0, nℤ ◁ ℤ. The cosets of ℤ/nℤ are: nℤ = {0, ±n, ±2n, ±3n,…}, 1 + ℤ, 2 + nℤ,…, (n-1) + nℤ. The sum of the coset (k + nℤ) + (l + nℤ) = (k + l) + nℤ. Notice that we have written it additively. Futhermore, ℤ/nℤ ≋ ℤ_{n}.
e | a | |
---|---|---|
e | e | a |
a | a | e |
Φ is onto and Ker(Φ) = {m ∈ ℤ: Φ(m) = e} = {even integers} = 2ℤ.
$D_4/K$ = {$K, R_{90}K, HK, DK$} Its multiplication table is given in Table 2.3. Observe that the multiplication table for D_{4} can be simplified into small boxes that are cosets of $K$. We can represent an entire box by a single element of the box!