I was pretty sure that she was planning my murder […] A knife? A too much hands-on approach. Pee on the cooking pot? Too gross. Putting me on a cross and throwing darts at me? Too messianic. A grenade? How will I answer some questions? Where did Dad die, Mum, in the kitchen? -Everywhere, he was everywhere, Apocalypse, 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’.
The mapping γ: G → G/Ker(Φ) is defined as γ(g) = gKer(Φ). It is called the natural mapping and is indeed a homomorphism. The Fundamental theorem on homomorphisms or the first isomorphism theorem shows that Ψγ = Φ.
Proof. Let φ: G/Ker(Φ) → Φ(G), φ(aKer(Φ)) = Φ(a).
Corollary. If Φ is a homomorphism from a finite group G to G', then |Φ(G)| divides both |G| and |G'|.
Proof.
By The First Isomorphism Theorem, G/Ker(Φ) ≋ Φ(G) ⇒ [By Lagrange, |G| = |G/Ker(Φ)]·|Ker(Φ)|] |Φ(G)| divides |G|.
We have previously demonstrated that if H ≤ G, Φ(H) = {Φ(h) | h ∈ H} is a subgroup of G', Φ(H) ≤ G’. In particular, Φ(G) ≤ G’ ⇒ [By Lagrange] |Φ(G)| divides |G’|∎
Let G = U(32) = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31}, U_{16}(32) = {1, 17}, U(16) = {1, 3, 5, 7, 9, 11, 13} and Φ: U(32) → U(16), defined by, Φ(x) = [x]_{mod 16}. Ker(Φ) = {x ∈ U(32)| x ≡ 1 (mod 16)} = U_{16}(32) = {1, 17}. Thus, by the first isomorphism theorem, U(32)/U_{16}(32) ≋ Φ(U(32)) =[Φ is onto] U(16).
Let G be the general linear group, that is, the group of invertibles 2 x 2 matrices with entries from the field of real numbers, and with the group operation being matrix multiplication, G = $GL(2, ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ, ad - bc ≠ 0}} \bigr\}$. The mapping Φ: GL(2, ℝ) → ℝ^{*}, Φ(A) = det(A) is an homomorphism with Ker(Φ) = SL_{2}(ℝ), and im(Φ) = ℝ^{x}. Thus, by the first isomorphism theorem, GL_{2}(ℝ)/SL_{2}(ℝ) ≋ ℝ^{x}.
Let G = $gl_2(ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ}} \bigr\}$, Φ: gl_{2}(ℝ) → ℝ, Φ(A) = tr(A). Φ is an homomorphism, onto, and Ker(Φ) = {$(\begin{smallmatrix}a & b\\ c & -a\end{smallmatrix})$ | a, b, c ∈ ℝ} = sl_{2}(ℝ) = {A ∈ gl_{2}(ℝ) | tr(A) = 0}. Then, gl_{2}(ℝ)/sl_{2}(ℝ) ≋ ℝ.
Let G = $gl_2(ℝ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ}} \bigr\}$, Φ: gl_{2}(ℝ) → sl_{2}(ℝ), $Φ(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})=(\begin{smallmatrix}a-d & b\\ c & d-a\end{smallmatrix})$. Φ is an homomorphism, onto (∀B ∈ sl_{2}(ℝ), B = $(\begin{smallmatrix}a & b\\ c & -a\end{smallmatrix})$, Φ(A) = $Φ(\begin{smallmatrix}a & b\\ c & 0\end{smallmatrix}) = (\begin{smallmatrix}a & b\\ c & -a\end{smallmatrix}) = B$, and ker(Φ) = $\bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: Φ(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})=(\begin{smallmatrix}a-d & b\\ c & d-a\end{smallmatrix})=(\begin{smallmatrix}0 & 0\\ 0 & 0\end{smallmatrix})}}\bigr\}$ ⇒[b = c = 0, a - d = 0 ⇒ a = d] $Ker(Φ) = \bigl\{ {{[\bigl(\begin{smallmatrix}a & 0\\ 0 & a\end{smallmatrix}\bigr)]: a ∈ ℝ}}\bigr\}$ ≋ ℝ ⇒ gl_{2}(ℝ)/ℝ ≋ sl_{2}(ℝ).
Let Φ: (ℤ, +) → (ℤ_{n}, +_{n}), Φ(m) = [m]_{mod n} is a homomorphism, Φ is onto and Ker(Φ) = ⟨n⟩ = nℤ. By the First Isomorphism Theorem, G/Ker(Φ) ≋ Φ(G), that is, ℤ/⟨n⟩ = ℤ/nℤ ≋ ℤ_{n}. Besides, ∀m ∈ ℤ_{n}, the set Φ^{-1}(m) = {…, m - 2n, m - n, m, m + n, m + 2n, …} = m + nℤ is indeed a coset.
Φ: ℤ_{4} → ℤ_{2}, [m]_{4} → [m]_{2} is a homomorphism, Ker(Φ) = ⟨[2]_{4}⟩ and surjective (Φ([2]_{4}) = [0]_{2}, Φ([1]_{4}) = [1]_{2}). By the First Isomorphism Theorem, G/Ker(Φ) ≋ Φ(G), that is, ℤ_{4}/⟨[2]_{4}⟩ ≋ ℤ_{2}.
Φ: ℤ_{6} → ℤ_{15}. There are only two possible homomorphisms: 1. Φ([1]_{6}) = Φ([x]_{6}) =[0]_{15} ⇒ Ker(Φ) = ℤ_{6}, im(Φ) = {[0]_{15}} ⇒ ℤ_{6}/ℤ_{6} ≋ {[0]_{15}} ≤ ℤ_{15}. 2. Φ([1]_{6}) = [5]_{15}, Φ([x]_{6}) =[5x]_{15}. Ker(Φ) = {[0]_{6}, [3]_{6}} = ⟨[3]_{6}⟩ ≋ ℤ_{2}. Im(Φ) = {[0]_{15}, [5]_{15}, [10]_{15}} = ⟨[5]_{15}⟩ ≋ ℤ_{3} ⇒ ℤ_{6}/⟨[3]_{6}⟩ ≋ ℤ_{6}/ℤ_{2} ≋ ⟨[5]_{15}⟩ ≋ ℤ_{3}
The wrapping function takes the real number line and wraps it around the unit circle. W assigns to each θ → e^{iθ} = cosθ + isinθ. W is an homomorphism from (ℝ, +) onto S^{1} = ({z ∈ ℂ: |z| = 1}, ·).
This map is a homomorphism because the multiplication of unit complex numbers corresponds to addition of angles: Φ(θ_{1}+θ_{2}) = e^{i(θ1+θ2)} = e^{iθ1}e^{iθ2} = Φ(θ_{1}) + Φ(θ_{2})
Ker(Φ) = {θ: e^{iθ} = 1} = {2πn: n ∈ ℤ} = 2πℤ. In words, the kernel of this homomorphism is the set of all integer multiples of 2π. By the first isomorphism theorem, we have that S^{1} = {z ∈ ℂ: |z| = 1} ≋ ℝ/2πℤ.
∀p(x) ∈ ℝ[x], p(x) = a_{0} + a_{1}x + a_{2}x^{2} + ··· + a_{n}x^{n}, there exists $a_0x + \frac{a_1}{2}x^2+···+\frac{a_n}{n+1}x^{n+1}$ such that $\frac{d}{dx}(a_0x + \frac{a_1}{2}x^2+···+\frac{a_n}{n+1}x^{n+1})= a_0 + a_1x + ··· a_n x ⇒ im(\frac{d}{dx})=\mathbb{R[x]}$. Besides, $Ker(\frac{d}{dx})$ = ℝ. By the First Isomorphism Theorem, ℝ[x]/ℝ ≋ ℝ[x].
Futhermore, Φ is obviously surjective, and Ker(Φ) = ⟨r⟩ because any arbitrary rotation r^{k} satisfies Φ(r^{k}) = kΦ(r) = 0, and any arbitrary reflection sr^{k}, Φ(sr^{k}) = Φ(s) + kΦ(r) = 1 + k·0 = 1 ⇒ D_{n}/⟨r⟩ ≋ ℤ_{2}. Futhermore, D_{n} = Ker(Φ) ∪ sKer(Φ) = ⟨r⟩ ∪ {s, r, sr, sr^{2}, ···, sr^{n-1}}.
In other words, our homomorphism could have been previously defined by,
$Φ(x) = \begin{cases} 0, &x~is~a~rotation \\ 1, &x~is~a~reflection \end{cases}$
Let G = D_{n} = {e, r, r^{2}, ···, r^{n-1}, r, sr, sr^{2}, ···, sr^{n-1}} = ⟨r, s | r^{n} = s^{2} = e, rs = sr^{n-1}⟩, the dihedral group. Φ: D_{n} → ℤ_{n} be the map given by, r → 1, s → 0. However, this is not an isomorphism because Φ(rs) ≠ Φ(sr^{n-1}). Φ(rs) = Φ(r) + Φ(s) = 1. Φ(sr^{n-1}) = Φ(s) + (n-1)Φ(r) = 0 +(n-1)·1 = n -1, but 1 ≇ n - 1 (mod n) unless n = 2.
Let G = D_{2n} = ⟨r, s | r^{2n} = s^{2} = e, rs = sr^{2n-1}⟩. Φ: D_{2n} → ℤ_{2} be the map given by, r → 1, s → 0. We need to check that the following equations are satisfied:
It is obviously surjective. Ker(Φ) = {e, r^{2}, r^{4}, ···, r^{2n-1}, s, sr^{2}, sr^{4}, ···, sr^{2n-1}} = ⟨r^{2}, s⟩ = ⟨s, r^{2} | s^{2} = (r^{2})^{n} = e, r^{2}s =[rs = sr^{2n-1}, r^{2}s = r(rs) = rsr^{2n-1} = (rs)r^{2n-1} = sr^{2n-1}r^{2n-1}] s(r^{2})^{2n-1}⟩ ≋ D_{n}. Therefore, D_{2n}/Ker(Φ) ≋ D_{2n}/⟨r^{2}, s⟩ ≋ D_{2n}/D_{n} ≋ ℤ_{2}.
is it homomorphism? Φ(r^{6}) = (123456)^{6} = (1) = Φ(e). Φ(s^{2}) = (16)(25)(34)(16)(25)(34) = (1) = Φ(e). Φ(rs) = (123456)(16)(25)(34) = (26)(35). Φ(sr^{5}) = Φ(s)Φ(r)^{5} = (16)(25)(34)(123456)^{5} = (16)(25)(34)(165432) = (26)(35) = Φ(rs).
Besides, im(Φ) = ⟨(123456), (16)(25)(34)⟩ and Ker(Φ) = {e} (it is left as an exercise) ⇒ D_{6} ≋ im(Φ) = ⟨(123456), (16)(25)(34)⟩.
Φ is an homomorphism, onto, and Ker(Φ) = A_{n} ⇒ S_{n}/A_{n} ≋ ℤ_{2}.