    # 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

• S2 is isomorphic to C2, the cyclic group of order 2, i.e., the multiplicative group comprising 1 and -1 (1.a). • The Klein 4-group K4 and the cyclic group of order 4 C4 are not isomorphic. because the Klein four-group is a group with four elements, in which each element is self-inverse (composing it with itself produces the identity). However, σ·σ = σ2 ≠ identity. -1.b.-
• (ℤ, +) is isomorphic to (2ℤ, +) where 2ℤ = {2n | n ∈ ℤ}.
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).
• (ℝ, +) is isomorphic to (ℝ+, ·). Fig. 1.c.
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).
• Conversely, we could have demonstrated that (ℝ+, ·) is isomorphic to (ℝ, +).
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).
• The circle group, $\Tau$ = {z ∈ ℂ: |z|=1} is a proper subgroup of ℂ*. It is the multiplicate group of all complex numbers with absolute value 1, that is, the unit circle in the complex plane. It has infinite order. 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.

• 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 (ℤn, +n) As a result, U10 ≃ ℤ4, U7 ≃ ℤ6, and U5 ≃ ℤ4 because U10 and U5 are both cyclic of order 4, and U7 is cyclic of order 6. Therefore, for every n, there is really only one cyclic group of order n.

U7 = ⟨3⟩ = {1, 3, 2, 6, 4, 5}, Φ: ℤ6 → U7, Φ(x) = 3x ∀x ∈ ℤ6.

• 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

• There is no isomorphism from U12 to U10.

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, ⊥

• Let G = SL(2, ℝ) the group of two by two real matrices with determinant one. Let M be any two by two matrix with determinant one and define a mapping ΦM: SL(2, ℝ) → SL(2, ℝ), ΦM(A) = MAM-1.
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.

• If g ∈ G, consider the function fg: G → G, fg(x) = g · x. By the existence of the inverses, fg has an inverse fg-1, fg-1(x) = g-1·x

Theorem. A function is bijective if and only if has an inverse. Therefore, fg is bijective ⇒ fg is a permutation of G.

• The set K = {fg: g ∈ G} is a group under function composition.
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.
• We state that Φ between G and K is an isomorphism.
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.

• 2, +mod2 = {0, 1}. 0 corresponds to the identity permutation. 1 → (12).
• 3, +mod3. 0 corresponds to the identity permutation, 1 → (123), and 2 → (132)
• 4, +mod4. 0 corresponds to the identity permutation, 1 → (1234), 2 → (13)(24), and 3 → (1432)
• Let’s calculate a group of permutations that is isomorphic to U8 = {1, 3, 5, 7}, U’8 = {T1, T3, T5, T7} • Klein four-group {e, a, b, c} corresponds to e, (12)(34), (13)(24), and (14)(23)
• D6 = {id (e), rotations (ρ1 -90° around the center anticlockwise-, ρ2 -180°-, and ρ3 -270°-), reflections or mirror images in the x -horizontal- and y -diagonal- axes (μ1, μ2), and reflections in the diagonal δ1 and δ2} = {e, s, rs, r2s, r3s, r, r2} → e, (12)(35)(46), (13)(26)(45), (14)(25)(36), (156)(243), (165)(234) # Properties of Isomorphisms Acting on Elements

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

• Φ carries the identity of G to G’.

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.

• ∀n ∈ ℤ, ∀a ∈ G, Φ(an) = (Φ(a))n.

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)

• 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, 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 ∎.

• Isomorphism preserves order, i.e., ∀a ∈ G, |a| = |Φ(a)|.

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 ⊥

• If two groups are isomorphic, then corresponding equations have the same number of solutions. For a fixed k ∈ ℤ and a fixed b ∈ G, the equation xk = b has the same number of solutions in G as does the equation xk = Φ(b) in G'.

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,

• Φ-1 is an isomorphism from G’ onto G.
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’).
• |G| = |G’|
• G is Abelian if and only if G’ is Abelian
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
• G is cyclic if and only if G’ is cyclic
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⟩
• If G has a subgroup H, then Φ(H) = {Φ(h) | h ∈ H} is a subgroup of 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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