Problem 3 - A Book of Abstract Algebra

Question

The following three facts about isomorphism are true for all groups: 1\. Every group is isomorphic to itself. 2\. If \(G_{1} \cong G_{2}\), then \(G_{2} \cong G_{1}\). 3\. If \(G_{1} \cong G_{2}\) and \(G_{2} \cong G_{3}\), then \(G_{1} \cong G_{3}\). Fact 1 asserts that for any group \(G\), there exists an isomorphism from \(G\) to \(G\). Fact 2 asserts that, if there is an isomorphism \(f\) from \(G_{1}\) to \(G_{2}\), there must be some isomorphism from \(G_{2}\) to \(G_{1}\). Well, the inverse of \(f\) is such an isomorphism. Fact 3 asserts that, if there are isomorphisms \(f: G_{1} \rightarrow G_{2}\) and \(g: G_{2} \rightarrow G_{3}\), there must be an isomorphism from \(G_{1}\) to \(G_{3}\). One can easily guess that \(g \circ f\) is such an isomorphism. The details of facts 1,2 , and 3 are left as exercises. Let \(G_{1}, G_{2}\), and \(G_{3}\) be groups, and let \(f: G_{1} \rightarrow G_{2}\) and \(g: G_{2} \rightarrow G_{3}\) be isomorphisms. Prove that \(g \circ f: G_{1} \rightarrow G_{3}\) is an isomorphism.

Step-by-Step Solution

Answer

The composition \(g \circ f: G_{1} \rightarrow G_{3}\) is an isomorphism because it is bijective and homomorphic.

Step 1: Define isomorphism

An isomorphism between two groups \(G\) and \(H\) is a bijective homomorphism \(f: G \rightarrow H\). This means it is a function that is both injective (one-to-one) and surjective (onto), and it respects the group operation.

Step 2: Prove Composition Property

If \(f: G_{1} \rightarrow G_{2}\) is an isomorphism and \(g: G_{2} \rightarrow G_{3}\) is an isomorphism, then the composition \(g \circ f: G_{1} \rightarrow G_{3}\) is a bijective homomorphism. This is because the composition of two bijective and homomorphic functions is bijective and homomorphic.

Step 3: Show Bijectivity of g \circ f

Since \(f\) and \(g\) are bijective, \(g \circ f\) is injective (no two elements map to the same element) and surjective (every element in the target group \(G_3\) is mapped to by an element of \(G_1\)). Therefore, \(g \circ f\) is bijective.

Step 4: Show Homomorphism Property

To be a homomorphism, the function \(g \circ f\) must satisfy \((g \circ f)(ab) = (g \circ f)(a)(g \circ f)(b)\) for all \(a, b \in G_1\). Since \(f\) and \(g\) are homomorphisms, this property is preserved through the composition.

Step 5: Conclude g \circ f is an Isomorphism

Since \(g \circ f\) is both bijective and homomorphic, \(g \circ f: G_{1} \rightarrow G_{3}\) is an isomorphism.

Key Concepts

Isomorphism Properties

Group isomorphism is a foundational concept in group theory where you have a structural match between two groups. This matching, known as an isomorphism, allows you to treat two different groups as identical regarding their algebraic structure. Here are the three fundamental properties of isomorphisms:

• Reflexivity: Every group is isomorphic to itself. This means that for any group \(G\), there exists an isomorphism from \(G\) to itself, often the identity mapping (sending each element to itself).
• Symmetry: If a group \(G_1\) is isomorphic to \(G_2\), then \(G_2\) is isomorphic to \(G_1\). This entails that if you can map \(G_1\) to \(G_2\) through an isomorphism, you can map back from \(G_2\) to \(G_1\) using the inverse function.
• Transitivity: If group \(G_1\) is isomorphic to \(G_2\), and \(G_2\) is isomorphic to \(G_3\), then \(G_1\) is isomorphic to \(G_3\). Such a chain of isomorphisms lets you bridge isomorphic connections, implying a consistent structural equivalence across multiple groups.

Understanding these properties gives you a clearer insight into how groups relate to one another under isomorphism and highlight how broad the concept of isomorphism is in group theory.

Group Homomorphism

Homomorphisms are a crucial component in understanding isomorphisms. A homomorphism is a function that respects group operations. Specifically, if \(f: G \rightarrow H\) is a homomorphism, it satisfies \(f(ab) = f(a)f(b)\) for all elements \(a, b\) in the group \(G\). This ensures the operation in the domain is mirrored in the codomain.
Homomorphisms can be injective, surjective, or bijective, and when it achieves the latter, it is much more significant. A bijective homomorphism is what we call an isomorphism.

• Injective: One-to-one mapping or distinct mapping ensures different elements in \(G\) map to different elements in \(H\).
• Surjective: Onto mapping guarantees every element in \(H\) has a corresponding element in \(G\) that maps to it.
• Bijective: Both injective and surjective properties hold, making the function both one-to-one and onto. This is the perfect condition for an isomorphism.

Understanding the nature of homomorphisms helps in constructing functions between groups that accurately reflect their structural properties, paving the way to identify and establish isomorphisms.

Composition of Functions

Function composition plays an essential role in connecting isomorphic structures across multiple groups. When you have two isomorphisms, say \(f: G_1 \rightarrow G_2\) and \(g: G_2 \rightarrow G_3\), it's possible to create a new function by composing these two functions, written as \(g \circ f\). This new function \(g \circ f: G_1 \rightarrow G_3\) carries over the properties of each component.

• Preservation of Properties: If both \(f\) and \(g\) are bijective and homomorphic, \(g \circ f\) is also bijective and homomorphic. This ensures \(g \circ f\) is an isomorphism.
• Associativity: Function composition is associative, meaning \((h \circ g) \circ f = h \circ (g \circ f)\) for any three functions, ensuring consistent results regardless of how you group the compositions.
• Simplicity: The composition process creates a straightforward path from the domain of the first function to the codomain of the second, simplifying the often complex construction of direct maps.

Composing functions is a powerful method that simplifies the verification of transitive properties among group isomorphisms, providing the structural framework needed to analyze and compare complex systems within group theory.