Isomorphisms II. Properties.

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Recall that two groups (G, *) and (G', ⋄) are said to be 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).

Properties of Isomorphisms Acting on Elements

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

• Isomorphism carries or preserves the identity.

Proof: Φ(e) =[G group, ∃e identity e·e = e] Φ(e·e) =[Φ is isomorphism, the group operation is preserved] Φ(e)·Φ(e) ⇒ Φ(e) = Φ(e)·Φ(e) ⇒ [By cancellation property, ab = ac ⇒ b = c Λ Φ(e) = Φ(e)·Φ(e)] e’ = Φ(e).

• ∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ.

Proof:
n = 0, Φ(a⁰) = Φ(e) = [Previous result] e’ = Φ(a)⁰.
n = 1, Φ(a¹) = Φ(a) = Φ(a)¹. Trivial
n = 2, Φ(a²) = Φ(a·a) = [isomorphism, the group operation is preserved] 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 ≤ 0, -n ≥ 0, e’ = Φ(e) = Φ(aⁿa^-n) = Φ(aⁿ)Φ(a^-n) = [We have already demonstrated for all positive integers Λ -n ≥ 0] Φ(aⁿ)(Φ(a))^-n ⇒ [e’ = Φ(aⁿ)(Φ(a))^-n and multiplying both sides by (Φ(a))ⁿ] (Φ(a))ⁿ = Φ(aⁿ)∎

• Isomorphism preserves commutativity.

Proof:

⇒) Suppose (G, ·) is Abelian, ∀a,b: a·b = b·a. Besides, ∀a’, b’ ∈ G’: ∃a, b ∈ G: Φ(a) = a’, Φ(b) = b'.

a’·b’ = [Φ is onto] Φ(a)·Φ(b) = [Φ is isomorphism ⇒ homomorphism] Φ(a·b) = [By assumption, (G, ·) is Abelian, a·b = b·a] Φ(b·a) = [Φ is homomorphism] Φ(b)·Φ(a) = b’·a’ ⇒ (G’, ·) is Abelian.

⇐) Suppose G’ is Abelian and for the sake of contradiction, assume G is not Abelian ⇒ ∃a, b ∈ G: a·b ≠ b·a ⇒ [Φ is injective] Φ(a·b) ≠ Φ(b·a) ⇒ [Φ is homomorphism] Φ(a)·Φ(b) ≠ Φ(b)·Φ(a) ⊥ (G’ is not Abelian).

• Isomorphism between cyclic groups carries the generator, i.e., G = ⟨a⟩ if and only if G’ = ⟨Φ(a)⟩.

Proof: Suppose G = ⟨a⟩ ⇒ [Φ(a) ∈ G’ and (G’, ·) group, so by closure, Φ(a)·Φ(a)∈ G’. In general, Φ(a)ⁿ ∈ G’] ⟨Φ(a)⟩ ⊆ G’. Since Φ is onto, ∀b ∈ G’, ∃k ∈ℤ such that a^k ∈ G Λ Φ(a^k) = b ⇒ b = Φ(a^k) = [∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] (Φ(a))^k ∈ ⟨Φ(a)⟩ ⇒ ∀b ∈ G’, ∃k ∈ℤ such that b = (Φ(a))^k ∈ ⟨Φ(a)⟩ ⇒ G’ ⊆ ⟨Φ(a)⟩ ⇒ ⟨Φ(a)⟩ = G'.

Suppose G’ = ⟨Φ(a)⟩ ⇒ [a ∈ G, (G, ·) group, and by closure] ⟨a⟩ ⊆ G. ∀b ∈ G, Φ(b) ∈ G’ = ⟨Φ(a)⟩ ⇒ ∃k ∈ℤ, Φ(b) = (Φ(a))^k = [∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] Φ(a^k), so Φ(b) = Φ(a^k) ⇒ [Φ is one-to-one] b = a^k ∈ G ⇒ ∀b ∈ G ∃k∈ℤ such that b = a^k ⇒ G ⊆ ⟨a⟩ ⇒ ⟨a⟩ = G ∎

• Isomorphism preserves the order of an element, 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) = [∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] Φ(a)ⁿ = e’ ⇒ [By assumption, |Φ(a)| = m] m ≤ n.

|Φ(a)| = m ⇒ (Φ(a))^m = [∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] Φ(a^m) = e’. Therefore, Φ(a^m) = Φ(aⁿ) = e’ ⇒ [Φ is one-to-one] aⁿ = a^m = e ⇒ [By assumption, |a| = n] 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 carries the identity] 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 x^k = b has the same number of solutions in G as does the equation x^k = Φ(b) in G'.

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^k = Φ(b).

Let q’ ∈ G’ be a solution to x^k= Φ(b) in G’: q’^k = Φ(b). Since Φ is onto, ∃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 ⇒ q ∈ A ⇒ q’=Φ(q) ∈ Φ(A) ⇒[q’ ∈ G’ be an arbitrary solution to x^k= Φ(b) in G’] 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 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’ ⇒[Φ is isomorphism] ∃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 group homomorphism] Φ(ab) = [By assumption, G is Abelian] Φ(ba) =[Φ is a group homomorphism] Φ(b)Φ(a) = b’a'
2. Suppose G’ is Abelian, ∀a, b ∈ G, Φ(ab) =[Φ is a group homomorphism] Φ(a)Φ(b) = [G’ is Abelian] Φ(b)Φ(a) = [Φ is a group homomorphism] = 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 ⇒ a’ = Φ(gⁿ) =[∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] (Φ(g))ⁿ ⇒ ∀a’∈ G’, ∃n ∈ ℤ, such that a’ = (Φ(g))ⁿ ⇒ G’ = ⟨Φ(g)⟩.
2. Suppose G’ is cyclic. G’ = ⟨g’⟩ ⇒ ∃ g ∈ G: g’ = Φ(g). Let a ∈ G ⇒ [G’ = ⟨g’⟩] Φ(a) = g’ⁿ for some n ∈ ℤ = (Φ(g))ⁿ =[∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] Φ(gⁿ) ⇒ Φ(a) = Φ(gⁿ) ⇒ [Φ is one-to-one] a = gⁿ ⇒ ∀a∈G, a = gⁿ for some n ∈ ℤ ⇒ G = ⟨g⟩∎

• If G has a subgroup H, then Φ(H) = {Φ(h) | h ∈ H} is a subgroup of G', i.e., H ≤ G ⇒ Φ(H) ≤ 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 is a group homomorphism] Φ(a)Φ(b^-1) =[∀n ∈ ℤ, ∀a ∈ G, Φ(aⁿ) = (Φ(a))ⁿ] Φ(a)(Φ(b))^-1 = a’b’^-1 ∈ Φ(H) ⇒ ∀a’, b’ ∈ Φ(H), a’b’^-1 ∈ Φ(H) ⇒ Φ(H) ≤ G’ ∎

• Isomorphism of groups is an equivalence relation.

Proof.

1. Reflexive. G ≋ G because a group is isomorphic to itself, Φ: G → G, ∀a ∈ G, Φ(a) = a, i.e., Φ is the identity.
2. Symmetric Property. If G ≋ G’ via Φ, then G’ ≋ G via Φ^-1.
3. Transitivity. If G ≋ H via Φ and H ≋ K via φ, then Φφ: G → K is an isomorphism, i.e., G ≋ K.

Proof: The composition of two bijections is a bijection, too. Besides, (Φφ)(g₁g₂) = Φ(φ(g₁g₂)) = [φ is a group homomorphism] Φ(φ(g₁)φ(g₂)) = [Φ is a group homomorphism] Φ(φ(g₁))Φ(φ(g₂)) = (Φφ)(g₁)(Φφ)(g₂)∎

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