“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’.
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*/S^{1} ≋ ℝ^{+}.
Consider a group homomorphism Φ: (Q_{8}, x) → ℤ_{2} x ℤ_{2} where Q_{8} 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) =[i^{2} = -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 Q_{8}/⟨-1⟩ ≋ ℤ_{2} x ℤ_{2}.
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.
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^{-1}x^{-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.