The Normalizer/Centralizer Theorem.

“Beauty is so many things… and you are in all of them,” I said. “Wow! I thought that you were as romantic as a brick to the head or a nuclear bomb’s guts”, Susanne replied, Apocalypse, Anawim, #justothepoint.

“When bias narratives, misleading news without context, and blatant lies spread like wildfire, history needs to be rewritten and truth needs to be cancelled”, Anawim, #justothepoint.

First Isomorphism Theorem. Let Φ be a group homomorphism from G to G’. Then, the function or mapping from G/Ker(Φ) to Φ(G), defined by gKer(Φ) → Φ(g) is an isomorphism, i.e., G/Ker(Φ) ≋ Φ(G). In particular, if Φ is onto, G/Ker(Φ) ≋ G’.

More examples

• Let Φ: ℂ* → ℝ⁺ -the group of positive real numbers under multiplication-, the map defined by a + bi → |a + bi| = $\sqrt{a^2+b^2}$. Φ is an homomorphism, onto, and Ker(Φ) = {z ∈ ℂ*: |z| = 1}. Therefore, ℂ*/Ker(Φ) = ℂ*/S¹ ≋ ℝ⁺.

• Consider a group homomorphism Φ: (Q₈, x) → ℤ₂ x ℤ₂ where Q₈ is the Hamilton Quaternion group, defined by Φ(i) = (1, 0) and Φ(j) = (0, 1).

Φ(k) =[k = ij] Φ(i)Φ(j) = (1, 0) + (1, 0) = (1, 1). Φ(-k) = -Φ(k) = -(1, 1) = (-1, -1) =_{ℤ2 x ℤ2} (1, 1)
Φ(-i) = -Φ(i) = -(1, 0) = (-1, 0) =_{ℤ2 x ℤ2} (1, 0)
Φ(-j) = -Φ(j) = -(0, 1) = (0, -1) =_{ℤ2 x ℤ2} (0, 1)
Φ(-1) =[i² = -1] = Φ(i) + Φ(i) = (1, 0) + (1, 0) =_{ℤ2 x ℤ2} (0, 0). Φ(1) = -Φ(1) = -(0, 0) = (0, 0).

Therefore, Φ is onto, Ker(Φ) = {1, -1} = ⟨-1⟩, and Q₈/⟨-1⟩ ≋ ℤ₂ x ℤ₂.

• Let S¹ be the unit circle in ℂ with multiplication as the group operation, S¹ = {z | |z| = 1}. Let Φ: ℝ → S¹, Φ(r) = e^i2πr is a homomorphism, onto, and Ker(Φ) = ℤ ⇒ ℝ/ℤ ≋ S¹.

Recall

Definitions. Two subgroups H and K of a group G are named conjugate in G if there is an element g ∈ G such that H = gKg^-1. gAg^-1 = {gag^-1 | a ∈ A}. The normalizer of H in the group G is defined as N_G(H) = N(H) = {g ∈ G: gHg^-1 = H} = {g ∈ G: gH = Hg}. Clearly C_G(H) ⊆ N_G(H) and both are subgroups of G.

Let G be a group, and H let be a non-empty subset of G.The centralizer of a group G is the subset of elements of G that commute with every element of H. Formally, C_G(H) = {x ∈ G | xhx^-1 = h ∀h ∈ H} = {x ∈ G | xh = hx ∀ h ∈ H}.

Let (A, *) and (B, ⋄) be two binary algebraic structures. A homomorphism is a structure-preserving map between two algebraic structures of the same type (groups, rings, fields, vector spaces, etc.) or, in other words, a map ϕ: A → B, such that ∀x, y ∈ S : ϕ(x ∗ y) = ϕ(x) ⋄ ϕ(y).

Notice that ϕ may not be one to one (injection), nor onto (surjection). An isomorphism is a bijective homomorphism, i.e., one-to-one and onto. In other words, let (S, *) and (S’, ⋄) be two binary algebraic structures of the same type. An isomorphism of S with S' is a 1-1 function ϕ mapping from S onto S′ such that the homomorphism property holds: ∀x, y ∈ S : ϕ(x ∗ y) = ϕ(x) ⋄ ϕ(y). S and S’ are said to be isomorphic binary structures and we denote or write it by S ≃ S'.

An automorphism is an isomorphism from a group to itself. Let G be a group, and let a be a fixed or given element of G, a ∈ G.An inner automorphism of G (induced or given by a) is defined by the conjugation action of the fixed element a, called the conjugating element i.e., Φ_a: G → G defined by Φ_a(x) = a·x·a^-1 ∀x ∈ G.

The Normalizer/Centralizer Theorem. Let H be a subgroup of G. Consider the mapping from the normalizer of H in G, N(H) to Aut(H) given by γ: g → Φ_g, where Φ_g is the inner automorphism induced or given by g, that is, Φ_g(h) = ghg^-1 ∀h ∈ H. This mapping is indeed a homomorphism with kernel the centralizer of H in G. The centralizer is a normal subgroup of the normalizer and N(H)/C(H) is isomorphic to a subgroup of Aut(H).

Proof.

Let γ be the mapping γ: (N(H),·) → (Aut(H),∘), defined by x → Φ_x where Φ_x(h) = xhx^-1 ∀h ∈ H.

• The mapping is indeed a homomorphism. ∀x, y ∈ N(H), γ(x)γ(y) = Φ_x°Φ_y

Φ_x°Φ_y ⇒ [° represents composition] ∀h ∈ H: Φ_x°Φ_y(h) = Φ_x(Φ_y(h)) =[By definition of inner automorphism] Φ_x(yhy^-1) =[By definition of inner automorphism] x(yhy^-1)x^-1 =[Associative] (xy)h(y^-1x^-1) =[The Socks and Shoes Principle] (xy)h(xy)^-1 = Φ_xy(h)

• It kernel is the centralizer of H in G. x ∈ Ker(γ) ↭ γ(x) = Φ_x = id_H, the identity of Aut(H) ↭ ∀h ∈ H: Φ_x(h) = xhx^-1 = id_H(h) = h ↭ ∀h ∈ H, xhx^-1 = h ↭ ∀h ∈ H, xh = hx ↭ x ∈ C(H) ↭ Ker(γ) = C(H).

  Recall: C_G(H) or simply, C(H) = {x ∈ G | xhx^-1 = h ∀h ∈ H} = {x ∈ G | xh = hx ∀ h ∈ H}.

• Recall: If K is a normal subgroup of G', then Φ^-1(K) = {k ∈ G: Φ(k) ∈ K} is a normal subgroup of G. In particular, the trivial subgroup is obviously normal, {e’} ◁ G’, then Φ^-1({e’}) = Ker(Φ) ◁ G. In words, kernels are normal subgroups of the domains, and therefore C_G(H) ◁ N(H). In words, the centralizer is a normal subgroup of the normalizer.

• By the First Isomorphism Theorem for Groups: N(H)/C(H) is isomorphic to Img(Φ). Futhermore, if Φ homomorphism Φ: G → G’ ⇒ Img(Φ) ≤ G’. Therefore, Img(Φ) ≤ Aut(H), and N(H)/C(H) is isomorphic to a subgroup of Aut(H).

💡 We have already demonstrated that the kernel of a group homomorphism is normal, but is the inverse also true?

Theorem. Every normal subgroup of a group G, N ◁ G, is the kernel of a homomorphism of G.In particular, a normal subgroup N is the kernel of the mapping γ from G to G/N defined by g → gN. Therefore, a subgroup is normal if and only if it is the kernel of a homomorphism.

Proof.

Let’s define a mapping γ: G → G/N, γ(g) = gN.

• is it well-defined? If x and y are in the same coset of N, say gN, x, y ∈ gN ⇒ [A left (or right) coset is uniquely determined or represented by any of its elements, i.e., aH = bH iff a ∈ bH] xN = yN = gN ⇒ γ(x) = γ(y) = γ(g).
• is γ homomorphism? γ(g)γ(h) = gNhN = [By assumption, N is normal in G, N ◁ G, the multiplication of any two left (or right) cosets gN and hN yields the left coset (gh)N] ghN = γ(gh)
• What is it kernel?

1. N ⊆ Ker(γ)? ∀k ∈ N, γ(k) = kN [A coset H absorbs an element if and only if the element belongs to it, aH = H iff a ∈ H] = N (the identity of the factor group) ⇒ k ∈ Ker(γ).
2. Conversely, Ker(γ) ⊆ N? If γ(k) = kN = N ⇒ [A coset H absorbs an element if and only if the element belongs to it, aH = H iff a ∈ H] k ∈ N ⇒[1, 2] Ker(γ) = N ∎

Bibliography

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6. Contemporary Abstract Algebra, Joseph, A. Gallian.
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