Problem 25 - Algebra: Pure and Applied

Question

In Exercises 23 through 30 construct an example of a nontrivial homomorphism between the two indicated groups, if this is possible, or explain why this is not possible. $$ \phi: \mathbb{Z}_{4} \rightarrow \mathbb{Z}_{12} $$

Step-by-Step Solution

Answer

A homomorphism \( \varphi: \mathbb{Z}_4 \rightarrow \mathbb{Z}_{12} \) is given by \( \varphi(1) = 3 \), resulting in \( \varphi(2) = 6 \), \( \varphi(3) = 9 \), and \( \varphi(0) = 0 \).

Step 1: Understanding Homomorphism Conditions

A group homomorphism from a group \( G \) to a group \( H \) is a function \( \varphi: G \rightarrow H \) that respects the group operation: \( \varphi(a \cdot b) = \varphi(a) \cdot \varphi(b) \). For \( \mathbb{Z}_4 \) and \( \mathbb{Z}_{12} \), this means \( \varphi((a+b) \mod 4) = (\varphi(a) + \varphi(b)) \mod 12 \).

Step 2: Checking Identity Element Mapping

The identity element in \( \mathbb{Z}_4 \) is 0, and any homomorphism must map 0 to the identity element in \( \mathbb{Z}_{12} \), which is also 0. Thus, \( \varphi(0) = 0 \).

Step 3: Determining Generator Mapping

\( \mathbb{Z}_4 \) is a cyclic group with generator 1. We need to determine \( \varphi(1) \). We choose \( \varphi(1) = 3 \) (or any number in \( \mathbb{Z}_{12} \) that respects homomorphism properties). Check for consistency across group operation.

Step 4: Checking Consistency with Generator

Since \( \mathbb{Z}_4 \) is generated by 1, we compute the other mappings: \( \varphi(2) = 6 \), \( \varphi(3) = 9 \). This scaling by 3 works since it maintains the operation: \( (2 + 2) \mod 4 = 0 \rightarrow (6+6) \mod 12 = 0 \).

Step 5: Verifying Homomorphism

Verify \( \varphi(a+b \mod 4) = (\varphi(a) + \varphi(b)) \mod 12 \). Check each combination of group elements, such as: \( \varphi(1+2) = \varphi(3) = 9 \) and \( \varphi(1) + \varphi(2) = 3 + 6 = 9 \). The function is consistent.

Key Concepts

Cyclic Groups

Cyclic groups are fundamental in group theory. They are groups generated by a single element, meaning every element of the group can be expressed as some power (or multiple, in the case of additive groups) of this generator.
In the context of the exercise, \( \mathbb{Z}_4 \) is a cyclic group. It means there is an element in \( \mathbb{Z}_4 \) that, through repeated addition, can generate every other element in the group. This element is 1.
Therefore, through the addition operation, the cyclic group \( \mathbb{Z}_4 \) can be represented as:

• The addition of 1 with itself enough times to cover the group, like:
  • 1 + 1 = 2
  • 1 + 1 + 1 = 3
  • 1 + 1 + 1 + 1 = 0 (mod 4)

Understanding cyclic groups is crucial when constructing homomorphisms, as seeing how elements map between groups requires knowing the generators and their properties.

Modular Arithmetic

Modular arithmetic deals with numbers wrapped around after reaching a certain value—the modulus. It's like a clock, where after 12, it resets to 1.
In the exercise, both \( \mathbb{Z}_4 \) and \( \mathbb{Z}_{12} \) operate under modular arithmetic, where elements reset or wrap around when they reach the modulus. This means:

• For \( \mathbb{Z}_4 \), any addition results in a number taken modulo 4.
• For \( \mathbb{Z}_{12} \), it's the same but with a modulus of 12.

When constructing a homomorphism like \( \varphi: \mathbb{Z}_4 \rightarrow \mathbb{Z}_{12} \), understanding how numbers behave with modular arithmetic is crucial.
The homomorphism needs to respect this structure, ensuring that operations map as expected within their respective groups.

Group Theory Concepts

In group theory, understanding the nature of groups and functions like homomorphisms is vital for solving related problems. Each group, like \( \mathbb{Z}_4 \) and \( \mathbb{Z}_{12} \), has an identity element, which is the neutral element for the group operation. Here, it's 0 in both groups.
A homomorphism is a function between two groups that preserves the group operation. Meaning, if you perform an operation on the elements in the first group and then map them, it should yield the same result as mapping the individual elements first and then performing the operation in the second group.
Thus, a homomorphism like \( \varphi: \mathbb{Z}_4 \rightarrow \mathbb{Z}_{12} \) must:

• Map the identity element of \( \mathbb{Z}_4 \), which is 0, to the identity element of \( \mathbb{Z}_{12} \), which is also 0.
• Preserve operations, i.e., \( \varphi((a+b) \mod 4) = (\varphi(a) + \varphi(b)) \mod 12 \). This ensures the function respects both group structures.

Effective homomorphisms maintain this property across all elements, ensuring consistent and orderly transformation from one group to another.