Question

Define homomorphism with an example

Answer 1

A **homomorphism** is a structure-preserving map between two algebraic structures (such as groups, rings, or vector spaces) that respects the operations defined on those structures. Specifically, if you have two algebraic structures \( (A, \cdot) \) and \( (B, * ) \), a function \( f: A \to B \) is called a homomorphism if for all elements \( x, y \in A \):

\[
f(x \cdot y) = f(x) * f(y)
\]

This means that the image of the product (or operation) of two elements in the first structure corresponds to the product of their images in the second structure.

### Example: Group Homomorphism

Consider the following example involving groups:

Let \( G = (\mathbb{Z}, +) \), the group of integers under addition, and \( H = (\mathbb{Z}/2\mathbb{Z}, +) \), the group of integers modulo 2 under addition.

Define the function \( f: G \to H \) by the rule:

\[
f(x) = x \mod 2
\]

We will check that \( f \) is a homomorphism.

1. Take any two integers \( a, b \in \mathbb{Z} \).
2. Compute \( f(a + b) \):

\[
f(a + b) = (a + b) \mod 2
\]

3. Compute \( f(a) + f(b) \) in \( H \):

\[
f(a) + f(b) = (a \mod 2) + (b \mod 2)
\]

In modulo 2 arithmetic, this is equivalent to:

\[
f(a) + f(b) = (a + b) \mod 2
\]

4. Thus, we see that:

\[
f(a + b) = f(a) + f(b)
\]

Since this holds for all \( a, b \in \mathbb{Z} \), we conclude that \( f \) is indeed a homomorphism.

### Summary
In this example, \( f \) preserves the operation of addition from the integers to the integers modulo 2, demonstrating that it is a homomorphism.