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Isomorphisms. Cayley's Theorem.

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'.

Examples

  1. First, we define the function ϕ: ℤ → 2ℤ, ϕ(n) = 2n.
  2. Let’s see that ϕ is a one-to-one function. ϕ(n)=ϕ(m) ⇒ 2n = 2m ⇒ n = m.
  3. ϕ is onto S’. If n ∈ 2ℤ, then n = 2m, ϕ(m)=2(n/2)=n.
  4. ϕ(m+n) = 2(m+n) = 2m + 2n = ϕ(m) + ϕ(n).
  1. First, we define the function ϕ: ℝ → ℝ+. ϕ(x) = ex.
  2. Let’s see that ϕ is a one-to-one function. ϕ(n)=ϕ(m) ⇒ en = em ⇒ ln(en) = ln(em) ⇒ n = m.
  3. ϕ is onto S’. If r ∈ ℝ+ ⇒ r’= ln(r) ∈ ℝ, and obviously ϕ(r’) = ϕ(ln(r)) = r.
  4. ϕ(r+s) = er+s = eres = ϕ(r)ϕ(s).
  1. First, we define the function ϕ: ℝ+ → ℝ. ϕ(r) = log(r).
  2. Let’s see that ϕ is a one-to-one function. ϕ(n)=ϕ(m) ⇒ log(n) = log(m) ⇒ elog(n) = elog(m) ⇒ n = m.
  3. ϕ is onto S’. If r ∈ ℝ ⇒ r’= er ∈ ℝ+, and obviously ϕ(r’) = ln(er) = r.
  4. ϕ(r+s) = log(r+s) = log(r) + log(s) = ϕ(r) + ϕ(s).

There is an homomorphism between ℂ* and $\Tau$, ϕ: ℂ* → $\Tau$, ϕ(re) = e.

Let z = re, w = se ⇒ ϕ(zw) = ei(α+β) = (e)(e) = 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 with addition modulo n. z1 = e1 (z1 ↔ θ1), z2 = e2 (z1 ↔ θ1) then z1 + z2 ↔ θ1 + θ2.

In other words, the algebraic properties of U and ℝ are the same, e.g., z4 = 1 has four solutions in U (1, i, -1, -i) and x + + x + + x + x = 0 has four solutions in ℝ (0, π, π/2, 3π/2), too.

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

U12 = {1, 5, 7, 11}. Note that x2 = 1 ∀x ∈ U12. Assume there is an isomorphism Φ: U10 → U12, Φ(9) = Φ(3·3) = Φ(3)Φ(3) = (Φ(3))2 = 1, U12, 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, ⊥

  1. ΦM(A) = MAM-1 ∈ SL(2, ℝ)? det(MAM-1) = (det M)(det A)(det M)-1 = 1 · 1 · 1 = 1.
  2. ΦM(A) = ΦM(B) ⇒ MAM-1 = MBM-1 ⇒ [∴ SL(2, ℝ) is a group and cancellation laws do apply] A = B.
  3. B ∈ SL(2, ℝ), we must find a matrix A: ΦM(A) = MAM-1 = B ⇒ A = M-1BM.
  4. ΦM(AB) = M(AB)M-1 = MA(M-1M)BM-1 = (MAM-1)(MBM-1) = ΦM(A)ΦM(B).

Cayley’s Theorem

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, fg is bijective ⇒ fg is a permutation of G.

  1. Associative. The operation of function composition is associative.
  2. ∀g, h, x ∈ G: fg·fh(x) = fg(fh(x)) = fg(h·x) = g·(h·x) = (g·h)·x = fgh(x), gh ∈ G (G is a group, so fgh ∈ K).
  3. fe is the identity.
  4. ∀g ∈ G, (fg)-1 = fg-1.
  1. First, we define the function Φ: G → K, Φ(g) = fg.
  2. Φ(g) = Φ(h) ⇒ fg = fh ⇒ ∀x ∈ G, fg(x) = fh(x) ⇒ ∀x ∈ G, g·x = h·x. The identity element, e ∈ G ⇒ g·e = h·e ⇒ g = h.
  3. Φ is trivially onto.
  4. Φ(g·h) = fg·h = fg·fh = Φ(g)·Φ(h)

Definition. The map Φ in the proof of Cayley’s Theorem is the left representation of G.

Examples of the regular group representation.

Properties of Isomorphisms Acting on Elements

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(a2) = Φ(a·a) = Φ(a)Φ(a) = (Φ(a))^2. Let’s apply induction on n.

Φ(an+1) = Φ(ana) = [Φ is isomorphism] Φ(an)Φ(a) = [By Inductive hypothesis] (Φ(a))nΦ(a) = (Φ(a))n+1 n

n ≤ 0, -n ≥ 0, e’ = Φ(e) = Φ(ana-n) = Φ(an)Φ(a-n) = Φ(an)(Φ(a))-n ⇒ [Multiplying both sides by (Φ(a))n] (Φ(a))n = Φ(an)

Proof: Suppose G = ⟨a⟩ ⇒ [By closure] ⟨Φ(a)⟩ ⊆ G’. Because Φ is onto, ∀b ∈ G’, ∃k ∈ℤ, ak ∈ G: Φ(ak) = 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 = ak ∈ G ⇒ ⟨a⟩ = G ∎.

Proof: Suppose that G and G’ are finite, |a| = n and |Φ(a)| = m.

|a| = n ⇒ Φ(an) = Φ(e) = [Φ carries the identity] e’, and Φ(an) = Φ(a)n = e’ ⇒ m ≤ n.

|Φ(a)| = m ⇒ (Φ(a))m = Φ(am) = e’. Therefore, Φ(am) = Φ(an) ⇒ [Φ is one-to-one] an = am = e ⇒ n ≤ m ⇒ n = m.

Suppose that |a| = n and |Φ(a)| = ∞ ⇒ an = e ⇒ [Φ carries the identity] Φ(an) = e’ ⇒ (Φ(a))n = e’ ⊥

Suppose that |a| = ∞ and |Φ(a)| = n ⇒ (Φ(a))n = e’ ⇒ Φ(an) = e’, but we already know that Φ(e) = e’, and Φ is one-to-one ⇒ an = e ⊥

Proof: Let A be the set of all solutions to xk = b.

∀a ∈ A: ak = b ⇒ Φ(ak) = Φ(b) ⇒ Φ(a)k = Φ(b). Therefore, the image Φ(A) is a subset of all solutions to xb = Φ(b).

Let q’ ∈ G’ be a solution to xk= Φ(b) in G’: q’k = Φ(b). ∃q ∈ G: Φ(q) = q’, and therefore, (Φ(q))k = Φ(qk) = Φ(b) ⇒ Φ(qk) = Φ(b) ⇒ [Φ is one-to-one] qk = b, so q’∈ Φ(A), so Φ(A) is equal to the set of all solutions to xk = Φ(b), and since Φ is an isomorphism |A|=|Φ(A)|.

ℂ* and ℝ* are not isomorphic because the equation x4 = 1 has four solution in ℂ* (1, -1, i,-i), but only two in ℝ*.

Properties of Isomorphisms Acting on Groups

Suppose that Φ is an isomorphism between two groups, Φ: G → G’. Then,

  1. Φ is bijective ⇒ [A function is bijective if and only if has an inverse] Φ-1 exists and is also bijective.
  2. Suppose a’, b’ ∈ G’ ⇒ ∃a, b ∈ G: Φ(a) = a’, Φ(b) = b’⇒ Φ-1(a’b’) = Φ-1(Φ(a)Φ(b)) = [Φ is a isomorphism] Φ-1(Φ(ab)) = ab = Φ-1(a’)Φ-1(b’).
  1. Suppose G is Abelian, ∀a’, b’ ∈ G’ ⇒ ∃a, b ∈ G: Φ(a) = a’, Φ(b) = b’, a’b’ = Φ(a)Φ(b) = [Φ is a isomorphism] Φ(ab) = [G is Abelian] Φ(ba) = Φ(a)Φ(b) = a’b'
  2. Suppose G’ is Abelian, ∀a, b ∈ G, Φ(ab) = [Φ is a isomorphism] Φ(a)Φ(b) = [G’ is Abelian] Φ(b)Φ(a) = [Φ is a isomorphism] = Φ(ba), therefore Φ(ab) = Φ(ba) ⇒ [Φ is one-to-one] ab = ba
  1. Suppose G is cyclic. G = ⟨g⟩. Let a’∈ G ⇒ ∃a ∈ G, n ∈ ℤ, Φ(a) = a’ and a = gn ⇒ Φ(gn) = (Φ(g))n = a’ ⇒ G’ = ⟨Φ(g)⟩.
  2. Suppose G’ is cyclic. G’ = ⟨g’⟩ ⇒ ∃ g ∈ G: g’ = Φ(g). Let a ∈ G ⇒ Φ(a) = g’n (n ∈ ℤ) = (Φ(g))n = Φ(gn) ⇒ Φ(a) = Φ(gn) ⇒ [Φ is one-to-one] a = gn ⇒ G = ⟨g⟩

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’ ∎

Automorphism

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

Examples:

  1. Φ: ℂ → ℂ, Φ(a + bi) = a - bi is an automorphism of (ℂ, +).
  2. Let ℝ2 = {(a, b) | a, b ∈ ℝ}, Φ: ℝ → ℝ, Φ(a, b) = Φ(b, a), the reflection across the vertical line y = x, is an automorphism of (ℝ2, +).

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

Proof. Φa is an automorphism of G.

  1. Φa is an endomorphism of G, ∀x1, x2 ∈ G, Φa(x1x2) = a·(x1x2)·a-1 = (ax1a-1)(ax2a-1) = Φa(x1a(x2)
  2. It has an inverse, namely Φa-1. Thus, Φa is bijective, and so is an automorphism.

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 automorphism and inner automorphism 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 ∘, 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-1(Φ(a’b’)) = a’b’ = Φ-1(a)Φ-1(b).
  4. ∀Φ, Φ’ ∈ Aut(G), (Φ ∘ Φ’)(ab) = Φ(Φ’(ab)) = [Φ’ is an isomorphism] Φ(Φ’(a)Φ’(b)) = [Φ is an isomorphism] Φ(Φ’(a))Φ(Φ’(b)) = (Φ ∘ Φ’)(a)(Φ ∘ Φ’)(b)

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)

∀ 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 = (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 = (b-1b)g(b-1b) = g. Therefore, ∀g ∈ G, (Φb∘Φb-1)(g) = (Φb-1∘Φb)(g) = g.

∀ Φa, Φb ∈ Inn(G), ∀g ∈ G, Φa∘(Φb)-1(g) = Φa∘Φb-1(g) = Φab-1(g)) = a(b-1gb)a-1 = (ab-1)g(ba-1) = (ab-1)g(ab-1)-1 = Φab-1(g)

Example. Let’s compute Inn(D4) = {ΦR0, ΦR90, ΦR180, ΦR270, ΦH, ΦV, ΦD, ΦD’} = {ΦR0, ΦR90, ΦH, ΦD}. Observe that ℤ(D4) = {R0, R180}, H = R180V, and D’ = R180D.  

Example. Aut(ℤ10)

Let be α ∈ Aut(ℤ10). 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 α1), α(1)=3 (α3), α(1)=7 (α7), and α(1)=9 (α9).

α1 is the identity. is α3 an automorphism?

α3(1)=3, α3(k)=3k mod 10.

  1. is it well defined? x = y ⇒ 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 = b (ℤ10)
  4. 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). We can easily demonstrate that 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.

Theorem. ∀n ∈ ℤ, n > 0, Aut(ℤ10)≃Un

Proof.

Let be α ∈ Aut(ℤn). Once we know α(1), we know α(k) for any k. Any automorphism α is determined by α(1), α(1) ∈ Un.

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

  1. is Φ one-to-one? Let α, β ∈ 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 described 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+···+1) [β(1) terms] = α(1)+α(1)+···+α(1) [β(1) terms] = α(1)β(1) = Φ(α)Φ(β)

Theorem. G/ℤ(G) ≃ Inn(G).

Proof: (Partial) ℤ(G) = {g ∈ G | gx =xg ∀x ∈ G}

Φ: G → Inn(G), Φ(a) = θg where θg(x) = gxg-1

  1. Φ is an homomorphism, ∀x ∈ G, Φ(ab)(x) = θab(x) = (ab)x(ab)-1 = [The shoe-sock theorem] (ab)x(b-1a-1) = a(bxb-1)a-1 = aθb(x)a-1 = θab)(x) = Φ(a)Φ(b)(x) ⇒ ∀x ∈ G, Φ(ab)(x) = Φ(a)Φ(b)(x) ⇒ Φ(ab) = Φ(a)Φ(b)
  2. Φ is Surjective. Almost trivial.
  3. KerΦ = {g ∈G }
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