Let (A, *) and (B, ⋄) be two binary algebraic structures. A homomorphism is a structure-preserving map between two algebraic structures of the same type 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 homomorphism that is also a bijection, i.e., one-to-one and onto.
Let (S, *) and (S, ⋄) be two binary algebraic structures. An isomorphism of S with S' is a 1-1 function ϕ mapping 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 it by S ≃ S'.
There is an homomorphism between ℂ* and $\Tau$, ϕ: ℂ* → $\Tau$, ϕ(re^{iθ}) = e^{iθ}.
Let z = re^{iα}, w = se^{iβ} ⇒ ϕ(zw) = e^{i(α+β)} = (e^{iα})(e^{iβ}) = ϕ(z)ϕ(w), but it is not onto because for every point in the unit circle, there are infinite number of complex numbers which map to it.
There is an isomorphism between U = $\Tau$ and ℝ_{2π} with addition modulo n. z_{1} = e^{iθ1} (z_{1} ↔ θ_{1}), z_{2} = e^{iθ2} (z_{1} ↔ θ_{1}) then z_{1} + z_{2} ↔ θ_{1} + θ_{2}.
In other words, the algebraic properties of U and ℝ_{2π} are the same, e.g., z^{4} = 1 has four solutions in U (1, i, -1, -i) and x +_{2π} + x +_{2π} + x +_{2π} x = 0 has four solutions in ℝ_{2π} (0, π, π/2, 3π/2), too.
Let G be a cyclic group with generator a. If the order of G is infinite, then G is isomorphic to (ℤ, +). Otherwise, G is isomorphic to (ℤ_{n}, +_{n}) As a result, U_{10} ≃ ℤ_{4}, U_{7} ≃ ℤ_{6}, and U_{5} ≃ ℤ_{4} because U_{10} and U_{5} are both cyclic of order 4, and U_{7} is cyclic of order 6. Therefore, for every n, there is really only one cyclic group of order n.
U_{7} = ⟨3⟩ = {1, 3, 2, 6, 4, 5}, Φ: ℤ_{6} → U_{7}, Φ(x) = 3^{x} ∀x ∈ ℤ_{6}.
There is no isomorphism from (ℚ, +) to (ℚ^{*}, ·)
Proof by contradiction: Suppose that Φ were such a mapping, there would be an a ∈ ℚ: Φ(a) = -1 ⇒ -1 = Φ(a) = Φ(a/2 + a/2) = Φ(a/2)Φ(a/2) = Φ(a/2)^{2} ⇒ there would be an a ∈ ℚ: -1 = Φ(a/2)^{2} ⊥
U_{12} = {1, 5, 7, 11}. Note that x^{2} = 1 ∀x ∈ U_{12}. Assume there is an isomorphism Φ: U_{10} → U_{12}, Φ(9) = Φ(3·3) = Φ(3)Φ(3) = (Φ(3))^{2} = 1, U_{12}, Φ(1) = Φ(1·1) = Φ(1)Φ(1) = (Φ(1))^{2} = 1. Therefore, Φ(9) = Φ(1), that is, Φ is not one to one, ⊥
Cayley’s Theorem. Every group G is isomorphic to a group of permutations.
Proof.
Theorem. A function is bijective if and only if has an inverse. Therefore, f_{g} is bijective ⇒ f_{g} is a permutation of G.
Definition. The map Φ in the proof of Cayley’s Theorem is the left representation of G.
Suppose that Φ is an isomorphism between two groups, Φ: G → G’. Then,
Proof: Φ(e) = Φ(e·e) = Φ(e)·Φ(e) ⇒ Φ(e) = Φ(e)·Φ(e) ⇒ [By cancellation property, ab = ac ⇒ b = c, and Φ(e) = Φ(e)e’] e’ = Φ(e). Please notice that identities are unique.
Proof: n = 0, 1. Trivial.
n = 2, Φ(a^{2}) = Φ(a·a) = Φ(a)Φ(a) = (Φ(a))^2. Let’s apply induction on n.
Φ(a^{n+1}) = Φ(a^{n}a) = [Φ is isomorphism] Φ(a^{n})Φ(a) = [By Inductive hypothesis] (Φ(a))^{n}Φ(a) = (Φ(a))^{n+1} n
n ≤ 0, -n ≥ 0, e’ = Φ(e) = Φ(a^{n}a^{-n}) = Φ(a^{n})Φ(a^{-n}) = Φ(a^{n})(Φ(a))^{-n} ⇒ [Multiplying both sides by (Φ(a))^{n}] (Φ(a))^{n} = Φ(a^{n})
Isomorphism preserves commutativity, i.e., ∀a, b∈ G, a·b = b·a iff Φ(a)·Φ(b) = Φ(b)·Φ(a)
Isomorphisms between two cyclic groups carries the generator. G = ⟨a⟩ if and only if G’ = ⟨Φ(a)⟩.
Proof: Suppose G = ⟨a⟩ ⇒ [By closure] ⟨Φ(a)⟩ ⊆ G’. Because Φ is onto, ∀b ∈ G’, ∃k ∈ℤ, a^{k} ∈ G: Φ(a^{k}) = b ⇒ b = (Φ(a))^{k} ∈ ⟨Φ(a)⟩ ⇒ ⟨Φ(a)⟩ = G'.
Suppose G’ = ⟨Φ(a)⟩ ⇒ [By closure] ⟨a⟩ ⊆ G. ∀b ∈ G, Φ(b) ∈ ⟨Φ(a)⟩ ⇒ ∃k ∈ℤ, Φ(b) = (Φ(a))^{k} = (Φ(a))^{k}, so Φ(b) = (Φ(a))^{k} ⇒ [Φ is one-to-one] b = a^{k} ∈ G ⇒ ⟨a⟩ = G ∎.
Proof: Suppose that G and G’ are finite, |a| = n and |Φ(a)| = m.
|a| = n ⇒ Φ(a^{n}) = Φ(e) = [Φ carries the identity] e’, and Φ(a^{n}) = Φ(a)^{n} = e’ ⇒ m ≤ n.
|Φ(a)| = m ⇒ (Φ(a))^{m} = Φ(a^{m}) = e’. Therefore, Φ(a^{m}) = Φ(a^{n}) ⇒ [Φ is one-to-one] a^{n} = a^{m} = e ⇒ n ≤ m ⇒ n = m.
Suppose that |a| = n and |Φ(a)| = ∞ ⇒ a^{n} = e ⇒ [Φ carries the identity] Φ(a^{n}) = e’ ⇒ (Φ(a))^{n} = e’ ⊥
Suppose that |a| = ∞ and |Φ(a)| = n ⇒ (Φ(a))^{n} = e’ ⇒ Φ(a^{n}) = e’, but we already know that Φ(e) = e’, and Φ is one-to-one ⇒ a^{n} = e ⊥
Proof: Let A be the set of all solutions to x^{k} = b.
∀a ∈ A: a^{k} = b ⇒ Φ(a^{k}) = Φ(b) ⇒ Φ(a)^{k} = Φ(b). Therefore, the image Φ(A) is a subset of all solutions to x^{b} = Φ(b).
Let q’ ∈ G’ be a solution to x^{k}= Φ(b) in G’: q’^{k} = Φ(b). ∃q ∈ G: Φ(q) = q’, and therefore, (Φ(q))^{k} = Φ(q^{k}) = Φ(b) ⇒ Φ(q^{k}) = Φ(b) ⇒ [Φ is one-to-one] q^{k} = b, so q’∈ Φ(A), so Φ(A) is equal to the set of all solutions to x^{k} = Φ(b), and since Φ is an isomorphism |A|=|Φ(A)|.
ℂ* and ℝ* are not isomorphic because the equation x^{4} = 1 has four solution in ℂ* (1, -1, i,-i), but only two in ℝ*.
Suppose that Φ is an isomorphism between two groups, Φ: G → G’. Then,
Proof: We know that Φ(H) ⊆ G’, is Φ(H) a subgroup of G'?
∀a’, b’ ∈ Φ(H) ⇒ ∃a, b ∈ H: Φ(a) = a’, Φ(b) = b’ ⇒ [H is a subgroup] a·b^{-1} ∈ H ⇒ Φ(a·b^{-1}) = Φ(a)(Φ(b))^{-1} = a’b’^{-1} ∈ Φ(H) ⇒ ∀a’, b’ ∈ Φ(H), a’b’^{-1} ∈ Φ(H) ⇒ Φ(H) ≤ G’ ∎
Definition. An automorphism is an isomorphism from a group to itself.
Examples:
Definition. Let G be a group, and let a be a fixed or given element of G, a ∈ G.An inner automorphism of G (induced by a) is given by the conjugation action of the fixed element a, i.e., Φ_{a} defined by Φ_{a}(x) = a·x·a^{-1} ∀x ∈ G.
Proof. Φ_{a} is an automorphism of G.
Example. Let’s compute the inner automorphism of D_{4} induced by R_{90°}. It is shown in the following Figure 1.a.
Definition. Let G be a group, we use Aut(G) and Inn(G) to express or denote the set of all automorphism and inner automorphism respectively. They are both groups under composition.
Aut(G) = {Φ: G → G’: Φ is an isomorphism}
Inn(G) = {Φ_{g}: G → G’: Φ(x) = g·x·g^{-1}}
Proof (Aut(G) forms a group).
Theorem. Let G be a group. Then, the set Inn(G) of all inner automorphism forms a subgroup of the automorphism group, Inn(G) ≤ Aut(G).
Proof. (Let’s use the subgroup test, H ≤ G iff ∀ x, y ∈ H, xy^{-1} ∈ H)
∀ Φ_{a}, Φ_{b} ∈ Inn(G). We claim (Φ_{b})^{-1} = Φ_{b-1}
∀g ∈ G, (Φ_{b}∘Φ_{b-1})(g) = (Φ_{b})(Φ_{b-1})(g)) = (Φ_{b})(b^{-1}gb) = b(b^{-1}gb)b^{-1} = (bb^{-1})g(bb^{-1}) = g.
∀g ∈ G, (Φ_{b-1}∘Φ_{b})(g) = (Φ_{b-1})(Φ_{b})(g)) = (Φ_{b-1})(bgb^{-1}) = b^{-1}(bgb^{-1})b = (b^{-1}b)g(b^{-1}b) = g. Therefore, ∀g ∈ G, (Φ_{b}∘Φ_{b-1})(g) = (Φ_{b-1}∘Φ_{b})(g) = g.
∀ Φ_{a}, Φ_{b} ∈ Inn(G), ∀g ∈ G, Φ_{a}∘(Φ_{b})^{-1}(g) = Φ_{a}∘Φ_{b-1}(g) = Φ_{a}(Φ_{b-1}(g)) = a(b^{-1}gb)a^{-1} = (ab^{-1})g(ba^{-1}) = (ab^{-1})g(ab^{-1})^{-1} = Φ_{ab-1}(g)
Example. Let’s compute Inn(D^{4}) = {Φ_{R0}, Φ_{R90}, Φ_{R180}, Φ_{R270}, Φ_{H}, Φ_{V}, Φ_{D}, Φ_{D’}} = {Φ_{R0}, Φ_{R90}, Φ_{H}, Φ_{D}}. Observe that ℤ(D_{4}) = {R_{0}, R_{180}}, H = R_{180}V, and D’ = R_{180}D.
Example. Aut(ℤ_{10})
Let be α ∈ Aut(ℤ_{10}). Once we know α(1), we know α(k) for any k, because α(k) = α(1+···+1) = α(1)···α(1) = kα(1)
We know that ∀Φ isomorphism from G to G’, |a| = |Φ(a)| ∀a in G, i.e., isomorphism preserves order (Properties of Isomorphisms acting on elements) ⇒ |α(1)| = |1| = 10 ⇒ There are four candidates for α(1): α(1)=1 (let’s name this isomorphism as α_{1}), α(1)=3 (α_{3}), α(1)=7 (α_{7}), and α(1)=9 (α_{9}).
α_{1} is the identity. is α_{3} an automorphism?
α_{3}(1)=3, α_{3}(k)=3k mod 10.
Therefore α_{3} is an automorphism. The same argument shows that α_{7} and α_{9} are automorphisms, too. Aut(ℤ_{10}) = {α_{1}, α_{3}, α_{7}, α_{9}}
What is it structure? (α_{3}α_{3})(1) = α_{3}(α_{3}(1)) = α_{3}(3) = 3·3 = 9 = α_{9}(1). We can easily demonstrate that α_{3}α_{3} = α_{9}. Further calculations show that α_{3}^{3} = α_{7} and α_{3}^{4} = α_{1}. Thus, Aut(ℤ_{10}) is cyclic, Aut(ℤ_{10}) = ⟨α_{3}⟩. Actually, it is isomorphic to U_{10}.
Theorem. ∀n ∈ ℤ, n > 0, Aut(ℤ_{10})≃U_{n}
Proof.
Let be α ∈ Aut(ℤ_{n}). Once we know α(1), we know α(k) for any k. Any automorphism α is determined by α(1), α(1) ∈ U_{n}.
We claim that Φ: Aut(ℤ_{n}) → U_{n}, Φ(α) = α(1) is an automorphism.
Theorem. G/ℤ(G) ≃ Inn(G).
Proof: (Partial) ℤ(G) = {g ∈ G | gx =xg ∀x ∈ G}
Φ: G → Inn(G), Φ(a) = θ_{g} where θ_{g}(x) = gxg^{-1}