Problem 4 - A Book of Abstract Algebra

Question

Let \(f\) be a homomorphism from \(G\) onto \(H\) with kernel \(K\) : $$ G \underset{K}{R} H $$ If \(S\) is any subgroup of \(H\), let \(S^{*}=\\{x \in G: f(x) \in S\\}\). Prove: $$ S \cong S^{*} / K $$

Step-by-Step Solution

Answer

The subgroups are isomorphic: \(S \cong S^{*} / K\) via a bijective homomorphism defined by the preimage and factor group mapping.

Step 1: Understand the Mapping

To prove that \(S \cong S^{*} / K\), we must first understand the nature of the homomorphism \(f: G \to H\). The kernel \(K\) is the set of elements in \(G\) that map to the identity element in \(H\). Given that \(f\) is onto, every element in \(H\) is the image of at least one element in \(G\).

Step 2: Analyze the Preimage Subgroup

The set \(S^{*} = \{x \in G: f(x) \in S\}\) is the preimage of subgroup \(S\) under homomorphism \(f\). Hence, \(S^{*}\) is a subgroup of \(G\). Since \(f\) is a homomorphism, preimages of subgroups in \(H\) remain subgroups in \(G\).

Step 3: Consider the Factor Group

We need to consider the factor group \(S^{*} / K\). The elements of \(S^{*} / K\) are the cosets of \(K\) in \(S^{*}\). The subgroup \(K\) is normal in \(G\), and consequently in \(S^{*}\), so \(S^{*} / K\) forms a group.

Step 4: Establish the Isomorphism

To show isomorphism \(S \cong S^{*} / K\), we define the mapping \(\phi: S^{*} / K \to S\) by \(\phi(xK) = f(x)\). This is well-defined because if \(xK = yK\), then \(x^{-1}y \in K\), implying \(f(x) = f(y)\). Thus, \(\phi\) is one-to-one and onto since every element of \(S\) is represented as some \(f(x)\) with \(x \in S^{*}\).

Step 5: Verify Homomorphism Property

The mapping \(\phi(xK) = f(x)\) preserves group operations. Consider two cosets \(xK\) and \(yK\), their product is \((xy)K\). Under \(\phi\), this maps to \(f(xy) = f(x)f(y)\), confirming \(\phi\) is a homomorphism.

Step 6: Conclude with Isomorphism Theorem

Since \(\phi: S^{*} / K \to S\) is a well-defined bijective homomorphism, by the definition of isomorphism, \(S \cong S^{*} / K\). Therefore, the structure of the subgroup in \(H\) and the corresponding quotient in \(G\) are isomorphic.

Key Concepts

Homomorphism

A homomorphism is a type of function between two groups that respects the group operation. In simpler terms, a homomorphism, denoted as \(f: G \to H\), maps elements from group \(G\) to group \(H\) such that for any two elements \(a\) and \(b\) in \(G\), their image under \(f\) satisfies \(f(a * b) = f(a) * f(b)\). This property ensures that the structure of \(G\) is somewhat preserved when mapped to \(H\).

The importance of a homomorphism lies in its ability to transform one group into another while keeping the group's fundamental properties intact. It's often used to relate different algebraic structures and study their relationships. An onto homomorphism means every element of \(H\) is the image of some element in \(G\). Such a relationship facilitates deeper insights into the structure of both groups.

Kernel

The kernel of a homomorphism \(f: G \to H\) is a subset of \(G\), capturing all elements that map to the identity element of \(H\). Mathematically, the kernel, \(K\), is defined as \(K = \{x \in G \mid f(x) = e_H\}\), where \(e_H\) is the identity in \(H\).

The kernel is a crucial part of the homomorphism because it tells us about the inherent symmetry or redundancy within the group \(G\). If the kernel only contains the identity of \(G\), the homomorphism is injective, implying a sort of one-to-one correspondence, without redundancies. More often, the kernel forms a normal subgroup of \(G\), allowing for the construction of factor groups by partitioning \(G\) into cosets of \(K\).

Factor Group

A factor group, also known as a quotient group, arises when you partition a group \(G\) by a normal subgroup \(K\). The factor group is denoted as \(G / K\) and its elements are the cosets of \(K\) in \(G\). Each coset is a set of elements formed by multiplying all elements of \(K\) by a particular element from \(G\).

In the context of the isomorphism theorem, the factor group \(S^{*}/K\) involves using the subgroup \(S^{*}\), the preimage of \(S\) under the homomorphism, and the kernel \(K\). The structure of this factor group provides insights into the overall structure of the transformation imposed by the homomorphism \(f\), connecting it to the subgroup \(S\) in \(H\). Factor groups are pivotal in simplifying complex group structures and enabling easier analysis.

Subgroup

A subgroup \(H\) is a "smaller" group within a larger group \(G\). Subgroups must themselves satisfy the basic properties of a group: closure, associativity, identity, and inverses. This means when you take any two elements from a subgroup and perform the group operation, the result is still in that subgroup.

Subgroups in this context play a role whereby they allow us to study smaller sections or components of larger groups. The subgroup \(S\) within the group \(H\), and its preimage \(S^{*}\) within \(G\), are at the heart of the given problem. They help us understand how properties and structures are mirrored across different levels of the homomorphic mapping. Exploring subgroups gives clarity on how complex systems can be broken down into simpler, yet still meaningful parts.

Preimage

The concept of a preimage in the context of a homomorphism refers to the collection of elements in the domain \(G\) that map to a particular set in the codomain \(H\). For the homomorphism \(f: G \to H\) and a subset \(S \subseteq H\), the preimage is \(S^{*} = \{ x \in G \mid f(x) \in S \}\).

Understanding preimages is crucial when studying mappings because they allow us to comprehend how subsets of \(H\) relate back to their counterparts in \(G\). In this specific problem, \(S^{*}\) represents all elements of \(G\) that, when transformed by \(f\), land within the subgroup \(S\) of \(H\). Analyzing preimages helps to uncover the alignment between structures of groups before and after the application of the homomorphism.