Mathwords logoMathwords

Isomorphic — Definition, Formula & Examples

Isomorphic describes two algebraic structures (like groups, rings, or vector spaces) that are structurally identical — they have the same form even if their elements look different. An isomorphism is a bijective map between them that preserves the operation.

Two algebraic structures (G,∗)(G, *) and (H,∘)(H, \circ) are isomorphic, written G≅HG \cong H, if there exists a bijection φ:G→H\varphi: G \to H such that φ(a∗b)=φ(a)∘φ(b)\varphi(a * b) = \varphi(a) \circ \varphi(b) for all a,b∈Ga, b \in G. The map φ\varphi is called an isomorphism.

Key Formula

φ(a∗b)=φ(a)∘φ(b)\varphi(a * b) = \varphi(a) \circ \varphi(b)
Where:
  • φ\varphi = A bijective map from structure G to structure H
  • ∗* = The operation in G
  • ∘\circ = The operation in H

How It Works

To show two structures are isomorphic, you must (1) define a function between them, (2) prove it is a bijection (one-to-one and onto), and (3) prove it preserves the operation. If any step fails, the structures are not isomorphic. To show two structures are *not* isomorphic, find a structural property one has that the other lacks — for example, one is cyclic and the other is not, or they have different orders of elements.

Worked Example

Problem: Show that the group (Z3,+)(\mathbb{Z}_3, +) is isomorphic to the group ({1,ω,ω2},⋅)(\{1, \omega, \omega^2\}, \cdot), where ω=e2πi/3\omega = e^{2\pi i/3} is a primitive cube root of unity.
Define the map: Let φ:Z3→{1,ω,ω2}\varphi: \mathbb{Z}_3 \to \{1, \omega, \omega^2\} be defined by φ(k)=ωk\varphi(k) = \omega^k.
φ(0)=1,φ(1)=ω,φ(2)=ω2\varphi(0) = 1, \quad \varphi(1) = \omega, \quad \varphi(2) = \omega^2
Verify bijection: The map sends 0, 1, 2 to three distinct elements, so it is both injective and surjective.
Verify operation preservation: For any a,b∈Z3a, b \in \mathbb{Z}_3, check that φ(a+b)=φ(a)⋅φ(b)\varphi(a + b) = \varphi(a) \cdot \varphi(b). Since φ(a+b)=ωa+b=ωa⋅ωb=φ(a)⋅φ(b)\varphi(a+b) = \omega^{a+b} = \omega^a \cdot \omega^b = \varphi(a) \cdot \varphi(b), the operation is preserved.
φ(a+b)=ωa+b=ωa⋅ωb=φ(a)⋅φ(b)\varphi(a + b) = \omega^{a+b} = \omega^a \cdot \omega^b = \varphi(a) \cdot \varphi(b)
Answer: Since φ\varphi is a bijection that preserves the group operation, (Z3,+)≅({1,ω,ω2},⋅)(\mathbb{Z}_3, +) \cong (\{1, \omega, \omega^2\}, \cdot).

Why It Matters

Isomorphism is the central notion of "sameness" in abstract algebra. Recognizing that two structures are isomorphic lets you transfer known theorems from one to the other, which is essential throughout group theory, ring theory, and in applications like cryptography and coding theory.

Common Mistakes

Mistake: Confusing a homomorphism with an isomorphism.
Correction: A homomorphism only needs to preserve the operation. An isomorphism must also be bijective (one-to-one and onto). Every isomorphism is a homomorphism, but not every homomorphism is an isomorphism.