Problem 2 - 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 \(G_{1}\) and \(G_{2}\) be groups, and \(f: G_{1} \rightarrow G_{2}\) an isomorphism. Show that \(f^{-1}\) : \(G_{2} \rightarrow G_{1}\) is an isomorphism. [HINT: Review the discussion of inverse functions at the end of Chapter 6. Then, for arbitrary elements \(a, b \in G_{1}\), let \(c=f(a)\) and \(d=f(b)\). Note that \(a=f^{-1}(c)\) and \(b=f^{-1}(d)\). Show that \(\left.f^{-1}(c d)=f^{-1}(c) f^{-1}(d) .\right]\)

Step-by-Step Solution

Answer

The inverse function \(f^{-1}\) is an isomorphism because it maintains bijection and the homomorphism property \(f^{-1}(cd) = f^{-1}(c)f^{-1}(d)\).

Step 1: Understanding Isomorphism

Recall that an isomorphism between two groups \(G_1\) and \(G_2\) is a bijective homomorphism. This means \(f: G_1 \rightarrow G_2\) is a function such that for all elements \(a, b \in G_1\), the function \(f(ab) = f(a)f(b)\) holds.

Step 2: Understanding the Inverse Function

The inverse function \(f^{-1}: G_2 \rightarrow G_1\) reverses the mapping done by \(f\). Since \(f\) is bijective, \(f^{-1}\) is well-defined for all elements in \(G_2\). Our goal is to show that \(f^{-1}\) is also a homomorphism.

Step 3: Assign Variables for Group Elements

Let arbitrary elements \(a, b\) in \(G_1\), and let \(c = f(a)\) and \(d = f(b)\) be their images in \(G_2\). Consequently, we have \(a = f^{-1}(c)\) and \(b = f^{-1}(d)\) using the properties of the inverse function.

Step 4: Prove Homomorphism Property for \(f^{-1}\)

To prove that \(f^{-1}\) is a homomorphism, consider the product \(cd\) in \(G_2\). Since \(f\) is a homomorphism, \(f(ab) = f(a)f(b) = cd\). We want \(f^{-1}(cd) = f^{-1}(f(ab)) = ab\). Since \(a = f^{-1}(c)\) and \(b = f^{-1}(d)\), it follows that \(ab = f^{-1}(c)f^{-1}(d)\). Therefore, \(f^{-1}(cd) = f^{-1}(c)f^{-1}(d)\), proving \(f^{-1}\) is a homomorphism.

Step 5: Conclusion

Since \(f^{-1}\) is bijective (inherited from \(f\)) and satisfies the homomorphism property, \(f^{-1}\) is an isomorphism from \(G_2\) to \(G_1\).

Key Concepts

Bijective Homomorphism

A bijective homomorphism is a fundamental concept in group theory, particularly when understanding group isomorphisms. At its core, a homomorphism is a function that preserves the group operation. This means if you have a function from one group to another, the image of a product of elements must equal the product of their images. In mathematical terms, for any elements \(a\) and \(b\) in a group \(G_1\), if \(f: G_1 \rightarrow G_2\) is a homomorphism, then \(f(ab) = f(a)f(b)\).

A homomorphism becomes bijective when it is both injective (one-to-one) and surjective (onto). Injective means that no two different elements in \(G_1\) have the same image in \(G_2\), ensuring each element in the codomain has its own unique pre-image. Meanwhile, surjective means every element of \(G_2\) is mapped from some element in \(G_1\).

• **Injective (one-to-one):** Different elements in the domain map to different elements in the codomain.
• **Surjective (onto):** Every element in the codomain is associated with an element from the domain.

Thus, a bijective homomorphism not only maps structures accurately but covers the entirety, establishing a perfect one-to-one correspondence between two groups.

Inverse Function

An inverse function is a concept that appears frequently in mathematics, notably in the context of group theory and functions. Specifically, in group isomorphism, where we deal with the existence of isomorphisms back and forth between groups.

For a function \(f\) to have an inverse, it must be bijective. This means it needs to be both injective and surjective, allowing the existence of a one-to-one correspondence between the elements of the involved groups.

In practical terms, if \(f: G_1 \rightarrow G_2\) is a bijective function, then its inverse function \(f^{-1}: G_2 \rightarrow G_1\) maps every element of \(G_2\) back to \(G_1\). This reverses the process undertaken by \(f\), maintaining the structure of group operations backward. For elements \(c\) and \(d\) in \(G_2\), if \(f(a) = c\) and \(f(b) = d\), the inverse satisfies \(a = f^{-1}(c)\) and \(b = f^{-1}(d)\).

• **Reverse Mapping:** The function \(f^{-1}\) undoes the mapping done by \(f\).
• **Group Structure Preservation:** It follows that \(f^{-1}(cd) = f^{-1}(c)f^{-1}(d)\), ensuring that multiplicative properties are retained.

The inverse function concept is pivotal in proving that if one group maps onto another bijectively, the reverse can also happen.

Group Homomorphism

In the study of group theory within abstract algebra, a group homomorphism is a special kind of function that reflects the group's algebraic structure from one group to another. Essentially, it is a function \(f: G_1 \rightarrow G_2\) that respects the operation defined on the groups. This means that for any two elements \(a\), \(b\) in \(G_1\), \(f(ab) = f(a)f(b)\).

This foundational property ensures that the operation between any two elements in the source group \(G_1\), when mapped by a homomorphism, translates identically in the target group \(G_2\). However, not all homomorphisms are isomorphisms. For a homomorphism \(f\) to be considered an isomorphism, it must also be bijective.

• **Homomorphism Property:** The operation inside the domain maps directly to the operation inside the codomain.
• **Preserves Group Structure:** The mappings keep the group operations consistent between \(G_1\) and \(G_2\).

Understanding homomorphisms is crucial for comprehending more complex concepts like kernels and quotient groups.

Abstract Algebra

Abstract Algebra is a wide branch of mathematics that provides the language and structure for understanding various algebraic systems, including groups, rings, and fields. It explores the properties of algebraic operations and the structures on which they are defined.

One of the key components of abstract algebra is group theory, which studies groups as an abstraction to number sets. It focuses on the algebraic structures formed by sets equipped with an operation that combines any two elements to form a third element, complying with rules such as associativity, identity, and invertibility.

Abstract algebra breaks down into:

• **Groups:** Sets coupled with a single associative operation possessing an identity element and inverses.
• **Rings:** Algebraic structures combining addition and multiplication.
• **Fields:** Like rings but with division operations (excluding division by zero).

These structures, combined with the study of homomorphisms and isomorphisms, form the foundation for many mathematical arguments and proofs, underpinning concepts that are foundational in advanced mathematical, scientific, and engineering disciplines.