Automorphisms

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Recall

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'.

In particular, two groups (G, *) and (G', ⋄) are isomorphic if there exist a bijective homomorphism, i.e., a one-to-one and onto map Φ: G → G’ such that the group operation is preserved, that is, ∀x, y ∈ G : ϕ(x ∗ y) = ϕ(x) ⋄ ϕ(y). Isomorphic groups are equivalent, that is, they share the same algebraic and group theoretic properties.

Definition. An automorphism is an isomorphism from a group to itself.

Examples:

1. There are two automorphism of ℤ: the identity, and the mapping n → -n.
2. The conjugation map, i.e., Φ: ℂ → ℂ, Φ(a + bi) = a - bi is an automorphism of (ℂ, +).
3. Let ℝ² = {(a, b) | a, b ∈ ℝ}, Φ: ℝ² → ℝ², Φ(a, b) = Φ(b, a), the reflection across the vertical line y = x, is an automorphism of (ℝ², +).
4. f: ℝ → ℝ, ∀x ∈ ℝ, f(x) = αx, f is a group automorphism ↭ α ≠ 0.
5. There is an automorphism Φ: ℤ₅ → ℤ₅, for each choice of Φ(1) ∈ {1, 2, 3, 4}. Notice that Φ(0) = 0 (Φ always carries the identity).

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 or given by a) is defined by the conjugation action of the fixed element a, called the conjugating element i.e., Φ_a defined by Φ_a(x) = a·x·a^-1 ∀x ∈ G.

Proof. Φ_a is an automorphism of G.

1. Φ_a is a group homomorphism, ∀x₁, x₂ ∈ G, Φ_a(x₁x₂) =[By definition] a·(x₁x₂)·a^-1 = a·(x₁·a^-1·a·x₂)·a^-1 =[Associativity] (ax₁a^-1)(ax₂a^-1) = Φ_a(x₁)Φ_a(x₂)
2. It has an inverse, namely Φ_a^-1. Thus, Φ_a is bijective, and so is an automorphism∎

We can also prove that is injective Φ_a(x₁) = Φ_a(x₂) ↭ a·x₁·a¹ = a·x₂·a^-1 ↭ [Cancellation laws] x₁ = x₂. Besides, ∀x∈ G, since G is closed a^-1·x·a ∈ G, then Φ_a(a^-1·x·a) = [By definition] a·(a^-1·x·a)·a^-1 = [Associativity] (a·a^-1)·x·(a·a^-1) = x ⇒ surjective∎

Example. Let’s compute the inner automorphism of D₄ 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 automorphisms and inner automorphisms 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).

1. Let be id: G → G, id(a) = a, ∀a ∈ G. Then, id∘Φ = Φ∘id = Φ, and therefore id ∈ Aut(G)
2. Associativity follows from the associativity of the compositions of functions.
3. ∀Φ ∈ Aut(G), Φ is an isomorphism ⇒ Φ has an inverse for composition of functions ∘, say Φ^-1. is Φ^-1 a group homomorphism? ∀a, b ∈ G, ∃a’, b’: Φ(a’) = a, Φ(b’) = b ⇒ Φ^-1(ab) = Φ^-1(Φ(a’)Φ(b’)) =[Φ is a group homomorphism] Φ^-1(Φ(a’b’)) = a’b’ = Φ^-1(a)Φ^-1(b). Φ is bijective ⇒ Φ^-1 is bijective ⇒ Φ^-1 ∈ Aut(G).
4. ∀Φ, Φ’ ∈ Aut(G), (Φ ∘ Φ’)(ab) = Φ(Φ’(ab)) = [In particular, Φ’ is a group homomorphism] Φ(Φ’(a)Φ’(b)) = [Φ is group homomorphism] Φ(Φ’(a))Φ(Φ’(b)) = (Φ ∘ Φ’)(a)(Φ ∘ Φ’)(b) ⇒ Φ ∘ Φ’ ∈ Aut(G)∎

Theorem. Let G be a group. Then, the set Inn(G) of all inner automorphisms forms a normal 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^-1gb) = b(b^-1gb)b^-1 =[Associativity] (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 =[Associativity] (b^-1b)g(b^-1b) = g. Therefore, ∀g ∈ G, (Φ_b∘Φ_b^-1)(g) = (Φ_b^-1∘Φ_b)(g) = g ⇒ (Φ_b)^-1 = Φ_b^-1

∀ 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^-1gb)a^-1 =[Associativity] (ab^-1)g(ba^-1) =[The Socks and Shoes Principle] (ab^-1)g(ab^-1)^-1 = Φ_ab^-1(g) ⇒ Φ_a∘(Φ_b)^-1 ∈ Inn(G) ⇒ Inn(G) ≤ Aut(G)

∀Φ ∈ Aut(G), ∀Φ_a ∈ Inn(G), Φ∘Φ_a∘Φ^-1 ∈ Inn(G)? Claim: Φ∘Φ_a∘Φ^-1 = Φ_Φ(a)

∀g ∈ G, (Φ∘Φ_a∘Φ^-1)(g) = (Φ∘Φ_a)(Φ^-1(g)) = Φ(Φ_a(Φ^-1(g))) = Φ(a·Φ^-1(g)·a^-1) =[In particular, Φ is a group homomorphism] Φ(a)·Φ(Φ^-1(g))·Φ(a^-1) =[∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] Φ(a)·g·Φ(a)^-1 = Φ_Φ(a)(g) ⇒ ∀Φ ∈ Aut(G), ∀Φ_a ∈ Inn(G), Φ∘Φ_a∘Φ^-1 = Φ_Φ(a) ∈ Inn(G) ⇒ Inn(G) ◁ Aut(G) ∎

Example. Let’s compute Inn(D₄). Observe that the complete list of inner automorphism is {Φ_R0, Φ_R90, Φ_R180, Φ_R270, Φ_H, Φ_V, Φ_D, Φ_D’} where R₀, R₉₀, R₁₈₀ and R₂₇₀ are rotations of a square, H and V denote the horizontal reflections, and D and D’ denote the diagonal reflections. This list may have repeated automorphisms.

Recall, ℤ(D_n) = $ \begin{cases} e, n~is~ odd\\\\ e~ and~ α^{\frac{n}{2}}, n~is~ even \end{cases}$

Therefore, ℤ(D₄) = {e, α²} = {R₀, R₁₈₀} ⇒ Φ_R180(x) = R₁₈₀xR₁₈₀^-1 [R₁₈₀ ∈ ℤ(D₄) ↭ R₁₈₀x = xR₁₈₀] = x ⇒ Φ_R180 = Φ_R0.

Φ_R270(x) = R₂₇₀xR₂₇₀^-1 =[R₂₇₀ = R₉₀R₁₈₀] R₉₀R₁₈₀xR₁₈₀^-1R₉₀^-1 = R₉₀(R₁₈₀xR₁₈₀^-1)R₉₀^-1 = R₉₀xR₉₀^-1 ⇒ Φ_R270 = Φ_R90.

Inn(D₄) = [Figure 1.b, and considering H = R₁₈₀V, and D’ = R₁₈₀D] {Φ_R0, Φ_R90, Φ_H, Φ_D}.

Example. Let’s compute Aut(ℤ₁₀)

Let be α ∈ Aut(ℤ₁₀). Once we know α(1), we know α(k) for any k, because α(k) = α(1+··_k··+1) = α(1)+··_k··+α(1) = kα(1)

We have already demonstrated that ∀Φ isomorphism from G to G’, |a| = |Φ(a)| ∀a in G, i.e., isomorphisms preserve order and also map generators to generators -Properties of Isomorphisms acting on elements- ⇒ |α(1)| = |1| = 10 ⇒ There are four candidates for α(1) -4 generators of ℤ₁₀, numbers less than 10 and coprime with 10-, namely α(1)=1 (let’s name this isomorphism as α₁), α(1)=3 (α₃), α(1)=7 (α₇), and α(1)=9 (α₉).

A similar argument will reveal that the possible automorphisms of ℤ₈ are α(1) = 1 (id), α(1) = 3 (α₃), α(1) = 5 (α₅), and α(1) = 7 (α₇).

α₁ is the identity. is α₃ an automorphism?

α₃(1)=3, α₃(k) = kα₃(1) = 3k mod 10.

1. Is it well defined? [x]_ℤ10 = [y]_ℤ10 ⇒ x mod 10 = y mod 10 ⇒ 3x mod 10 = 3y mod 10.
2. Is α₃ onto? Yes, it is, because α(1) = 3 and 3 is a generator of ℤ₁₀.
3. Is α₃ one to one? 3a mod 10 = 3b mod 10 ⇒ (3a-3b) mod 10 = 0 mod 10 ⇒ a-b mod 10 = 0 mod 10 ⇒ [a]_ℤ10 = [b]_ℤ10
4. Is it a homomorphism? α₃(a + b) = 3(a + b) = 3a + 3b = α₃(a)α₃(b)

Therefore α₃ is an automorphism. The same argument shows that α₇ and α₉ are automorphisms, too. Aut(ℤ₁₀) = {α₁, α₃, α₇, α₉}

What is it structure? (α₃α₃)(1) = α₃(α₃(1)) = α₃(3) = 3·3 = 9 = α₉(1) ⇒ α₃α₃ = α₉. Further calculations show that α₃³ = α₇ and α₃⁴ = α₁. Thus, Aut(ℤ₁₀) is cyclic, Aut(ℤ₁₀) = ⟨α₃⟩. Actually, it is isomorphic to U₁₀ = {1, 3, 7, 9}.

Theorem. ∀n ∈ ℤ, n > 0, Aut(ℤ_n)≋ U_n or ℤ_n^x, the group of the integers relatively prime to n under multiplication mod n. In particular, |Aut(ℤ_n)| = Φ(n) where Φ(n) is Euler’s totient function, which gives the number of positive integers less than or equal to n that are relatively prime to n.

Proof.

Let α ∈ Aut(ℤ_n). α(0) = 0. Once we know α(1), we know α(k) for any k∈ℤ_n because α(k) = α(1+1+··_k··+1) = α(1)+α(1)+··_k··+α(1) = kα(1). Any automorphism α is determined by α(1). Besides, α(1) ∈ U_n because it needs to be a generator of ℤ_n, that is, an integer less than n and coprime to n, but these integers are precisely the elements of U_n.

We claim that Φ: Aut(ℤ_n) → U_n, defined by Φ(α) = α(1) is an isomorphism.

1. is Φ one-to-one? Let α, β ∈ Aut(ℤ_n), and suppose Φ(α) = Φ(β) ⇒ α(1) = β(1) ⇒ α(k) = kα(1) = kβ(1) = β(k), ∀k ∈ ℤ_n ⇒ α = β
2. is Φ onto? Let r ∈ U_n, consider α: ℤ_n → ℤ_n defined as α(s) = sr (mod n), ∀s ∈ ℤ_n. It can easily be demonstrated that α ∈ Aut(ℤ_n), and Φ(α) = α(1) = r.
3. is Φ operation-preserving? ∀α, β ∈ Aut(ℤ_n), Φ(αβ) = (αβ)(1) = α(β(1)) = α(1+1+··_β(1)··+1) =[In particular, α is a group homomorphism] α(1)+α(1)+··_β(1)··+α(1) = α(1)β(1) = Φ(α)Φ(β) ∎

Bibliography

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