Problem 3 - 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: Let \(g\) be the restriction of \(f\) to \(S^{*}\). [That is, \(g(x)=f(x)\) for every \(x \in S^{*}\), and \(S^{*}\) is the domain of \(g\). ] Then \(g\) is a homomorphism from \(S^{*}\) onto \(S\), and \(K=\operatorname{ker} g\).

Step-by-Step Solution

Answer

The function \( g \) is a homomorphism from \( S^{*} \) onto \( S \) with kernel \( K \).

Step 1: Understand the given properties

We have a homomorphism \( f: G \to H \) where \( K = \ker(f) \) and \( f \) is onto. We need to look at a subgroup \( S \subseteq H \) and define \( S^{*} = \{x \in G \mid f(x) \in S\} \). Note that \( S^{*} \) is the preimage of \( S \) under \( f \).

Step 2: Define the restriction homomorphism

Define a new function \( g: S^{*} \to S \) by \( g(x) = f(x) \) for all \( x \in S^{*} \). Since \( f(x) \in S \), \( g \) maps each element appropriately into \( S \).

Step 3: Prove that g is a homomorphism

To show \( g \) is a homomorphism, we need to prove that \( g(xy) = g(x)g(y) \) for all \( x, y \in S^{*} \). We've defined \( g(x) = f(x) \), hence \( g(xy) = f(xy) = f(x)f(y) = g(x)g(y) \), using the property of \( f \) as a homomorphism.

Step 4: Prove that g is onto S

For \( g \) to be onto, for every \( s \in S \), there is an \( x \in S^{*} \) such that \( g(x) = s \). Given \( f \) is onto, for each \( s \), there exists some \( x \in G \) with \( f(x) = s \). Hence, \( x \in S^{*} \) and \( g(x) = s \), proving \( g \) is onto.

Step 5: Find the kernel of g

Determine \( \ker(g) = \{x \in S^{*} \mid g(x) = e_S \}\), where \( e_S \) is the identity in \( S \). Since \( g(x) = f(x) \), this set is precisely \( \ker(f) \cap S^{*} = K \), since \( K = \ker(f) \). Hence, \( \ker(g) = K \).

Key Concepts

Subgroup

In group theory, a **subgroup** is a subset of a group that is itself a group under the same operation. When discussing subgroups, it's vital to ensure that the subset fulfills two main criteria:

• Closure: If you take two elements from the subgroup and perform the group's operation on them, the result should also be in the subgroup.


• Identity and Inverse: The subgroup must contain the identity element of the group, and for every element in the subgroup, its inverse should also be present in the subgroup.

Understanding these properties helps to comprehend the structure of the group and how subgroups interact within it. For example, in our exercise, we consider a subgroup \( S \) of a group \( H \), which signifies that \( S \) satisfies these subgroup criteria within the group \( H \). This plays a crucial role in defining the preimage, \( S^{*} \), which involves grouping elements from the original group \( G \) that map into \( S \) under the homomorphism \( f \).

Kernel of a Homomorphism

The **kernel of a homomorphism** is an important concept in understanding the structure of group homomorphisms. For a homomorphism \( f: G \to H \), the kernel, denoted as \( \ker(f) \), is the set of elements in \( G \) that map to the identity element in \( H \). Mathematically, it is expressed as:\[\ker(f) = \{ g \in G \mid f(g) = e_H \}\]where \( e_H \) represents the identity element in \( H \). The kernel is always a normal subgroup of \( G \), meaning it commutes well with every element of \( G \). In the context of our exercise, the kernel \( K \) reflects those elements of \( G \) that become the identity in \( H \) when the homomorphism \( f \) is applied. Understanding the kernel can provide insight into the structure of the group and the nature of the homomorphism, particularly regarding its injectivity. A kernel that consists solely of the identity element indicates \( f \) is injective.

Onto Homomorphism

An **onto homomorphism**, also known as a surjective homomorphism, is where every element in the target group \( H \) is the image of at least one element from the source group \( G \). Simply put, a function \( f: G \to H \) is onto if \[\forall h \in H, \exists g \in G \text{ such that } f(g) = h.\]This relationship ensures the entire group \( H \) is covered by the function \( f \), and no elements in \( H \) are left unmapped. In the exercise, the function \( f \) is an onto homomorphism, meaning it maps \( G \) onto \( H \) completely. This characteristic is significant when defining \( g \), the restriction of \( f \) on \( S^{*} \), because it assures that \( g \) will also be onto the subgroup \( S \). It means every element in \( S \) has a corresponding preimage in \( S^{*} \).

Preimage

The concept of a **preimage** arises frequently in the study of functions and homomorphisms. Given a function \( f: G \to H \), the preimage of a subset \( S \subseteq H \) refers to all elements in \( G \) that map into \( S \). The preimage is formally written as:\[S^{*} = \{ x \in G \mid f(x) \in S \}\]This set \( S^{*} \) assembles all the elements of the original group \( G \) that, through the homomorphism \( f \), find their image in the group \( S \). In the exercise, this concept is employed to define the preimage \( S^{*} \), allowing us to create a new homomorphism \( g \) restricted to \( S^{*} \). By examining preimages, we can understand how different elements and substructures of a group relate to each other through the function \( f \). This also aids in studying how certain characteristics of \( G \) reflect in \( H \), further illuminating the relationship established by the homomorphism \( f \).