Question

In Exercises 1 through \(10,\) determine whether or not the indicated map \(\phi\) is a homomorphism, and in the cases where \(\phi\) is a homomorphism, determine Kem \(\phi\). $$ \phi: S_{3} \rightarrow \mathbb{Z}_{2}, \text { where } $$ $$ \phi(\sigma)=\left\\{\begin{array}{l} 0 \text { if } \sigma \text { is an even permutation } \\ 1 \text { if } \sigma \text { is an odd permutation } \end{array}\right. $$

Short answer

Yes, \( \phi \) is a homomorphism. The kernel \( \ker\phi = \{e, (123), (132)\} \).

Step-by-step solution

Understand the Structure of S3 and Z2

The group \( S_3 \) is the symmetric group of all permutations of 3 elements, with 6 elements: the identity \( e \), 2-cycles (\( (12), (13), (23) \)), and the 3-cycles (\( (123), (132) \)). Meanwhile, \( \mathbb{Z}_2 \) is the group with 2 elements: \( \{0, 1\} \), with addition modulo 2 as its operation.

Define a Group Homomorphism

A map \( \phi: G \rightarrow H \) between two groups is a homomorphism if it preserves the group operation, i.e., \( \phi(g_1g_2) = \phi(g_1) + \phi(g_2) \) for all \( g_1, g_2 \in G \). In this context, it would mean: \( \phi(\sigma_1 \circ \sigma_2) = \phi(\sigma_1) + \phi(\sigma_2) \pmod{2} \).

Check Homomorphism Property of φ

To verify if \( \phi \) is a homomorphism, consider the cases of permutation pairs from \( S_3 \). For example, a cycle composition such as \( (12) \circ (23) = (123) \). Evaluate \( \phi((12)) + \phi((23)) = 1 + 1 = 0 \pmod{2} \) and \( \phi((123)) = 0 \), which matches the left side. Check all other pairs similarly. The consistency shows \( \phi \) preserves the group operation.

Determine the Kernel of φ

The kernel of a homomorphism is the set of elements in the domain that map to the identity element of the codomain. Here, \( \ker\phi = \{ \sigma \in S_3 \mid \phi(\sigma) = 0 \} \). Since \( \phi(\sigma) = 0 \) for even permutations, the kernel consists of the identity \( e \), and the 3-cycles: \( (123), (132) \).

Key concepts

Understanding the Symmetric Group S3

The symmetric group, denoted as \( S_3 \), is a fascinating mathematical structure that consists of all possible permutations of three distinct objects. Imagine you have three items, labeled 1, 2, and 3. The group \( S_3 \) includes any arrangement of these items.
It contains six elements: the identity permutation, which leaves the objects in their original order, three 2-cycles like \((12), (13), (23)\), where two items are swapped, and two 3-cycles like \((123), (132)\), where all three items rotate in a cycle.
These permutations together form a group under the operation of composition, where the result of applying one permutation and then another is also a permutation in the group. This showcases the rich interplay between arithmetic, geometry, and function composition within group theory.

Exploring the Homomorphism Property

A homomorphism in group theory is a map between two groups that respects their structure. To determine if a function \( \phi \) between groups \( G \) and \( H \) is a homomorphism, it must satisfy the property: \( \phi(g_1g_2) = \phi(g_1) + \phi(g_2) \) for all elements \( g_1, g_2 \) in \( G \). This ensures that the operation in the domain \( G \) is aligned with the operation in the codomain \( H \).
In the context of the exercise, we explore the homomorphism \( \phi: S_3 \rightarrow \mathbb{Z}_2 \), where \( \phi(\sigma) = 0 \) for even permutations and \( \phi(\sigma) = 1 \) for odd permutations. Checking examples such as \( (12) \circ (23) = (123) \) verifies that \( \phi((12)) + \phi((23)) = 1 + 1 = 0 \pmod{2} \) aligns with \( \phi((123)) = 0 \). This confirms that \( \phi \) respects the group operation, hence is a valid homomorphism.

Unraveling the Kernel of a Homomorphism

The kernel of a homomorphism, represented as \( \ker\phi \), is the set of elements in the domain that map to the identity element in the codomain. It plays a crucial role in understanding the structure of the group and the nature of the homomorphism.
For the homomorphism \( \phi: S_3 \rightarrow \mathbb{Z}_2 \), we identify the kernel as consisting of elements in \( S_3 \) that map to 0 in \( \mathbb{Z}_2 \). These are the even permutations. In \( S_3 \), the identity \( e \) and the 3-cycles \((123), (132)\) are even permutations and therefore form the kernel: \( \ker\phi = \{ e, (123), (132) \} \).
Understanding the kernel helps in analyzing whether \( \phi \) is injective, and it provides insights into the internal structure of a group homomorphism.

Understanding Permutations

Permutations are an essential concept in group theory, representing all possible arrangements of a set of objects. They can be thought of as bijective functions that rearrange elements of a particular set.
In the symmetric group \( S_3 \), permutations are crucial because they form the basic elements of the group. Each permutation can be decomposed into cycles, which represent independent circular shifts of a subset of elements. For instance, a 2-cycle like \((12)\) swaps elements 1 and 2, while a 3-cycle like \((123)\) rotates all three elements in sequence.
Permutations are fundamental not only in group theory but also in many fields such as combinatorics, cryptography, and even quantum mechanics. They highlight the intrinsic properties of symmetry and arrangement in mathematics.