Group Isomorphism

As we did in the case of rings. we are interested in finding groups with the same algebraic properties i.e. isomorphic groups.

7.8.1 Definition. Let (G, *) and (G', ⋅) two groups. We say that G is isomorphic to G' and write G ≃ G' if there is a bijective function φ: G ⟶ G', such that

φ(a * b) = φ(a) ⋅ φ(b), ∀a,b ∈ G

such a mapping is known as isomorphism. The isomorphism is said to "preserve the operation", i.e the result is the same whether the group operation is performed before or after the mapping. An isomoprhism from a group G to G itself is called an automorphism.

We've already studied isomorphic groups, e.g D₃ ≃ 𝓢₃. In this case an isomorphism is given associating to every rigid motion of the plane the correspondent permutation of the three vertices of the triangle.

Klein's group is isomorphic to the group of invertible elements of ℤ₈.
(ℝ, +) is isomorphic to the multiplicative group of real positive numbers through the exponential function

f (x+y) = e^x+y = e^x+y = e^x ⋅ e^y = f(x) ⋅ f(t)

7.8.2 Example. Let c be a fixed element of a group G. Define f: G ⟶ G by f(g) = c⁻¹gc, Then

f(b)f(b) = (c⁻¹ac)(c⁻¹bc) = c⁻¹a (cc⁻¹)be = c⁻¹abc = f(ab)

If g ∈ G

f(cgc⁻¹) = c⁻¹(cgc⁻¹)c = (c⁻¹c)g(c⁻¹c) = ege = g

f is surjective. To show that f is injective, suppose f(a) = f(b. The c⁻¹ac = c⁻¹bc. Canceling c on the right side and c⁻¹ on the left side by Propositions 7.1.2, 7.1.3, and corollary Corollary 7.1.4, we have a = b. Hence, f is injective. Therefore f is an isomorphism, called the inner automorphism of G induced by c. ■

When two groups are isomorphic all algebraic properties are shared: cardinality, the order of the elements, the fact to cyclic and abelian, to possess subgroups of a given order, etc. When they different for one of these properties they are not isomorphic.

(ℤ, +) is not isomorphic to (ℚ \ {0}, ⋅), since every element in ℤ different from zero has infinite order, while in ℚ \ {0} there exist two element 1 and −1 with finite order.
(ℚ, +) is not isomorphic to (ℚ \ {0}, ⋅), since every element different form the neutral element in (ℚ, +) has infinite order whilst in ℚ \ {0} the element −1 has order 2.
(U(ℤ₈), +) and (ℤ₄, +) are both groups with order four, but they are not isomorphic since in ℤ₄ there exists an element of order 4, while in U(ℤ₈) there is not such an element.
(ℚ \ {0}, ⋅) and (ℝ \ {0}) are not isomorphic, since in (ℝ \ {0}) the orders of the elements are &infty;, 1 and 2 while in (ℂ\{0},⋅) there exist element of any order (as n^th roots of unity).

7.8.3 Cayley’s Theorem. Every group is isomorphic to a group of permutations.

Proof. Let G be a given group. The permutations that we use in the proof will be mappings defined on the set of all elements in G.
For each element a in G, we define a mapping

T_a: G ⟶ G
x ⟼ ax ∀x ∈ G

That is, the image of each x in G is obtained by multiplying x on the left by a. Now f a is one-to-one since

T_a(x) = T_a(y) ⇒ ax = ay
⇒ x = y

To see that T_a is onto, let b be arbitrary in G. Then x = i⁻¹b is in G, and for this particular x we have

T_a(x) = ax
⇒ a(a⁻¹b) = b

Thus T_a is a permutation on the set of elements of G i.e. T_a ∈ 𝓢(G).
We define the following mapping

Ψ (G, ⋅) ⟶ 𝓢(G, ∘)
a ⟼ T_a

Ψ preserves the operations:

Ψ(ab) = Ψ(a) ∘ Ψ(b)

Indeed

T_ab(x) = ab ⋅ x = a(bx) = T_a(T_b(x)) ∀ x∈G

hence T_ab = T_a ∘ T_b

Ψ is injective. Indeed Ψ(a) = Ψ(b) that is T_b = T_a.

In particular

a = T_a(e) = T_b(e) = b

This mapping is not surjective: compare the cardinality of G and 𝓢(G). Instead the image of Ψ

Ψ(G) := {T_a | a ∈ G}

is a subgroup of 𝓢(G) isomorphic to G. □

Exercise

Determine four distinct subgroups of 𝓢₄ isomorphic to 𝓢₃ and nine isomorphic to 𝓢₂.
Verify Cayley's theorem for the group G = ⟨g | g⁴ = 1⟩ and for the Klein group.
Let G the multiplicative group of invertible elements ℤ₈ and let G' the multiplicative group of invertible elements of ℤ₁₂. Prove that they are isomorphic.
Prove that the set
$S = {(\begin{array}{cc} x & y \\ - y & x \end{array}) | x, y \in R, x, y \neq 0}$
is a subgroup of (GL₂(ℝ), ⋅) isomorphic to (ℂ*,⋅), where ℂ* = ℂ \{0}.

Solutions

Consider the symmetric group the subgroups 𝓢₄. Let H = {σ ∈ 𝓢₄| σ₄(4) = 4}. By definition H consists of those permutations in 𝓢₄ that leave 4 fixed. Hence elements of H exactly the permutations of the elements {1,2,3}. This means that H is a permutation group isomorphic to 𝓢₃. We can identify other subgroups H_i of permutation of 𝓢₄ tha leave the element i fixed:
1. H₁ = {id,(23),(24),(34),(234),(243)}
2. H₂ = {id,(13),(14),(34),(134),(143)}
3. H₃ = {id,(12),(14),(24),(124),(142)}
4. H₄ = {id,(12),(13),(23),(123),(132)}
Order 2 groups are cyclic so for {e,s} with s ∈ S₄ we need s² = e. Thus we have the subgroups made by transpositions H₁ = {id, (1 2)}, H = {id, (1 2)(3 4)}
Lagrange theorem avoid using S₃ in the case. We choose the group S₄ and find an order 4 element of S₄ we've that ⟨1234⟩ = {(1234),(13)(24),(1432), id}. For V we have the following isomorphic subgroup of S₄ = {id,(12)(34), (14)(23), (13) (24)}
G = U(ℤ₈) = {[1]₈, [3]₈, [5]₈, [7]₈}; G = U(ℤ₁₂) = {[1]₁₂, [5]₁₂, [7]₁₂, [11]₁₂}. An isomorphism between G and G' is the one that maps [3]₈ to [5]₁₂, [5]₈ to [7]₁₂. [7]₈ to [11]₁₂.

They are both isomorphic to the Klein group,(in V every non-trivial element is order 2), we can take the isomorphism [1]₈ ↦ id, [3]₈ ↦ r, [5]₈↦ s₁, [7]₈ ↦ s₂.
The product of elements of S is an element of S, (1 0; 0 1) ∈ S, the inverse of (x y; −y x) is 1/(x² + y²) (x −y; y x) ∈ S.

An isomorphism with (ℂ*,⋅) is obtained considering the map z = x + iy ↦ (x y; −y x); Multiply out (x₁ + i y₁)(x₂ + i y₂) and (x₁ y₁; −y₁ x₁)(x₂ y₂; −y₂ x₂) to verify.

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