# 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., Φa defined by Φa(x) = a·x·a-1 ∀x ∈ G.

Proof. Φa is an automorphism of G.

1. Φa is a group homomorphism, ∀x1, x2 ∈ G, Φa(x1x2) =[By definition] a·(x1x2)·a-1 = a·(x1·a-1·a·x2)·a-1 =[Associativity] (ax1a-1)(ax2a-1) = Φa(x1a(x2)
2. It has an inverse, namely Φa-1. Thus, Φa is bijective, and so is an automorphism∎

We can also prove that is injective Φa(x1) = Φa(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 Φa(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 R90°. 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) = {Φg: 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)

∀ Φa, Φb ∈ Inn(G). We claim (Φb)-1 = Φb-1

∀g ∈ G, (Φb∘Φb-1)(g) = (Φb)((Φb-1)(g)) = (Φb)(b-1gb) = b(b-1gb)b-1 =[Associativity] (bb-1)g(bb-1) = g.

∀g ∈ G, (Φb-1∘Φb)(g) = (Φb-1)((Φb)(g)) = (Φb-1)(bgb-1) = b-1(bgb-1)b =[Associativity] (b-1b)g(b-1b) = g. Therefore, ∀g ∈ G, (Φb∘Φb-1)(g) = (Φb-1∘Φb)(g) = g ⇒ (Φb)-1 = Φb-1

∀ Φa, Φb ∈ Inn(G), ∀g ∈ G, Φa∘(Φb)-1(g) = Φa∘Φb-1(g) = Φab-1(g)) = a(b-1gb)a-1 =[Associativity] (ab-1)g(ba-1) =[The Socks and Shoes Principle] (ab-1)g(ab-1)-1 = Φab-1(g) ⇒ Φa∘(Φb)-1 ∈ Inn(G) ⇒ Inn(G) ≤ Aut(G)

∀Φ ∈ Aut(G), ∀Φa ∈ Inn(G), Φ∘Φa∘Φ-1 ∈ Inn(G)? Claim: Φ∘Φa∘Φ-1 = ΦΦ(a)

∀g ∈ G, (Φ∘Φa∘Φ-1)(g) = (Φ∘Φa)(Φ-1(g)) = Φ(Φa-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 = ΦΦ(a)(g) ⇒ ∀Φ ∈ Aut(G), ∀Φa ∈ Inn(G), Φ∘Φa∘Φ-1 = ΦΦ(a) ∈ Inn(G) ⇒ Inn(G) ◁ Aut(G)

Example. Let’s compute Inn(D4). Observe that the complete list of inner automorphism is {ΦR0, ΦR90, ΦR180, ΦR270, ΦH, ΦV, ΦD, ΦD’} 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(Dn) = $\begin{cases} e, n~is~ odd\\\\ e~ and~ α^{\frac{n}{2}}, n~is~ even \end{cases}$

Therefore, ℤ(D4) = {e, α2} = {R0, R180} ⇒ ΦR180(x) = R180xR180-1 [R180 ∈ ℤ(D4) ↭ R180x = xR180] = x ⇒ ΦR180 = ΦR0.

ΦR270(x) = R270xR270-1 =[R270 = R90R180] R90R180xR180-1R90-1 = R90(R180xR180-1)R90-1 = R90xR90-1 ⇒ ΦR270 = ΦR90.

Inn(D4) = [Figure 1.b, and considering H = R180V, and D’ = R180D] {ΦR0, ΦR90, ΦH, ΦD}.

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]10 = [y]10 ⇒ 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]10 = [b]10
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) = α33(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(ℤn)≋ Un or ℤnx, the group of the integers relatively prime to n under multiplication mod n. In particular, |Aut(ℤn)| = 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(ℤn). α(0) = 0. Once we know α(1), we know α(k) for any k∈ℤn 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) ∈ Un because it needs to be a generator of ℤn, that is, an integer less than n and coprime to n, but these integers are precisely the elements of Un.

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

1. is Φ one-to-one? Let α, β ∈ Aut(ℤn), and suppose Φ(α) = Φ(β) ⇒ α(1) = β(1) ⇒ α(k) = kα(1) = kβ(1) = β(k), ∀k ∈ ℤn ⇒ α = β
2. is Φ onto? Let r ∈ Un, consider α: ℤn → ℤn defined as α(s) = sr (mod n), ∀s ∈ ℤn. It can easily be demonstrated that α ∈ Aut(ℤn), and Φ(α) = α(1) = r.
3. is Φ operation-preserving? ∀α, β ∈ Aut(ℤn), 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

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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