Question

Let \(\phi: Z_{12} \rightarrow \mathbb{Z}_{3}\) be a homomorphism with Kern \(\phi=\\{0,3,6,9\\}\) and \(\phi(4)=2\). Find all the elcments \(x \in \mathbb{Z}_{12}\) such that \(\phi(x)=1,\) and show that they form a coset of Kern \(\phi\) in \(Z_{12}\).

Short answer

The elements \(\{5, 11\}\) in \(\mathbb{Z}_{12}\) have \(\phi(x) = 1\) and form a coset of the kernel \(\{0, 3, 6, 9\}\).

Step-by-step solution

Understanding the Problem

We have a homomorphism \(\phi: \mathbb{Z}_{12} \rightarrow \mathbb{Z}_{3}\) with a given kernel and property \(\phi(4)=2\). We need to find all \(x \in \mathbb{Z}_{12}\) such that \(\phi(x) = 1\) and show these elements form a coset of the kernel in \(\mathbb{Z}_{12}\).

Analyzing the Kernel

The kernel of a homomorphism \(\phi\) is the set of all elements in \(\mathbb{Z}_{12}\) that map to zero in \(\mathbb{Z}_{3}\). From the problem, we know the kernel is \(\{0, 3, 6, 9\}\). This tells us how the homomorphism maps other elements, especially when considering the full image group.

Using Given Properties

We have \(\phi(4) = 2\). So, elements that map to 1 can be found by adding another element that maps \(2\) to \(1\) in \(\mathbb{Z}_{3}\), as \(1 = 2 + (-1)\). This leads to checking \(\phi(x) = 1\) where \(\phi(x) = \phi(4) + \phi(y)\) where \(\phi(y) = -1 \equiv 2 \mod 3\) within this cyclic group.

Finding \(\phi(x) = 1\) Elements

We need elements \(x\) such that \(\phi(x) = 1\). As \(2 \equiv 2 \mod 3\) from \(\phi(4) = 2\), we require finding an element \(y\) such that \(\phi(y) = 2\). Therefore, \(x = 4 + k\) for elements not in the kernel that would transform 2 to 1: Check \(x = 4 + 1 \equiv 5 \mod 12\) and \(x = 4 - 1 \equiv 3 \mod 12\).

Solve Using Structure of \(\mathbb{Z}_{12}\)

For cosets: \(x ot\in \text{Kern} \phi\) is \(\{5, 8, 11\}\), calculated as \(5 = 4+1, 8=7+1, 11=10+1\) mapping from 2 to 1. They map to distinct elements by composition, forming \(1\). Thus, \(\{5, 8, 11\}\) in system \(\mathbb{Z}_{12}\) via reconstruction of mod30.

Verify if Forming Coset

The set \(\{5, 8, 11\}\) is a coset of the kernel \(\{0, 3, 6, 9\}\) as a translation. These elements are not in the kernel but map to 1 since each element differs from any in kernel by a base element to obtain new congruences representing distinct equivalence classes.

Key concepts

Kernel of a Homomorphism

In group theory, the kernel of a homomorphism is a fundamental concept that helps us understand how the homomorphism behaves. The kernel of a homomorphism \(\phi: G \rightarrow H\) is the set of elements in \(G\) that are mapped to the identity element in \(H\). In simpler terms, it reflects those elements that are essentially "annihilated" by the function because they collapse to the identity when the function is applied.

For the homomorphism \(\phi: \mathbb{Z}_{12} \rightarrow \mathbb{Z}_{3}\) given in the exercise, the kernel is \(\{0, 3, 6, 9\}\). Each of these elements maps to 0 in \(\mathbb{Z}_{3}\).

Understanding the kernel is crucial because it tells us how the homomorphism affects the structure of the group. It helps in identifying cosets, which are related to figuring out the quotient group. By analyzing the kernel, we can see which elements collectively have the same result under the homomorphism, offering insight into the deeper structure of the group itself.

Cosets

Cosets are a way to partition a group into equal-sized, non-overlapping subsets. When you have a subgroup \(H\) of a group \(G\), you create cosets by combining each element \(a\) of \(G\) with every element in \(H\). In essence, cosets are translations of subgroups across the group.

For the exercise, once you recognize the kernel of the homomorphism \(\phi\), you can identify the cosets of that kernel in \(\mathbb{Z}_{12}\). Each coset has the same size as the kernel, and importantly, they partition \(\mathbb{Z}_{12}\) without any overlap. The elements \(\{5, 8, 11\}\) form a coset of the kernel \(\{0, 3, 6, 9\}\) because if you add any element from the kernel to 5, 8, or 11, the result always stays within the coset pattern. This aligns with how cosets fundamentally reorganize the group while respecting its structure.

Cyclic Groups

A group is considered cyclic if there is an element in the group, known as a generator, from which every other element of the group can be obtained by repeatedly applying the group operation. In mathematical terms, a group \(G\) is cyclic if \(G = \langle g \rangle\), where \(g\) is the generator.

In our problem, both \(\mathbb{Z}_{12}\) and \(\mathbb{Z}_{3}\) are classic examples of cyclic groups. The group \(\mathbb{Z}_{12}\) is generated by the element 1, as every other element is a multiple of 1 added multiple times. Similarly, \(\mathbb{Z}_{3}\) is generated by 1, with three elements: 0, 1, and 2. These cyclic properties make calculations simpler, as operations in these groups repeatedly cycle through a set of elements.

Understanding cyclic groups helps to manipulate and understand the behavior of every element under operations, especially when working with homomorphisms and the transformations they apply.

Modular Arithmetic

Modular arithmetic is a system of arithmetic for integers, where numbers wrap around after reaching a certain value, called the modulus. It's like what happens with the hours on a clock.

In the given exercise, we work within \(\mathbb{Z}_{12}\) and \(\mathbb{Z}_{3}\). These are rings of integers under addition modulo 12 and 3, respectively. For example, if you add 10 and 3 in \(\mathbb{Z}_{12}\), you get 1, because 13 mod 12 equals 1.

Modular arithmetic is crucial in computing the image of a homomorphism. It helps us understand how adding or multiplying elements affects their equivalences. Every move within these groups can be systemized under congruence relations provided by their modulus. This lays the groundwork for deeper understanding in more advanced topics like number theory and cryptography.