Question
define homomorphism and provide one example with a solution
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Answered by
GPT-4o mini
In mathematics, a **homomorphism** is a structure-preserving map between two algebraic structures. Specifically, it is a function \( f: A \to B \) between two algebraic structures \( (A, \ast) \) and \( (B, \odot) \) that respects the operations defined on those structures. This means that for all elements \( x, y \) in \( A \):
\[
f(x \ast y) = f(x) \odot f(y)
\]
This definition indicates that the homomorphism preserves the operation; that is, the result of operating on two elements in the first structure and then applying the homomorphism is the same as first applying the homomorphism to each element and then operating on the results in the second structure.
### Example: Homomorphism in Groups
Let's consider an example involving groups. Let \( G = (\mathbb{Z}, +) \) be the group of integers under addition, and let \( H = (\mathbb{Z}/2\mathbb{Z}, +) \) be the group of integers modulo 2, also under addition.
We can define a function \( f: G \to H \) as follows:
\[
f(x) = x \mod 2
\]
### Verifying Homomorphism
To check if \( f \) is a homomorphism, we need to show that \( f \) preserves the group operation. Specifically, for all \( a, b \in \mathbb{Z} \), we need to prove:
\[
f(a + b) = f(a) + f(b)
\]
Let's calculate both sides:
1. **Left-hand side:**
\[
f(a + b) = (a + b) \mod 2
\]
2. **Right-hand side:**
\[
f(a) + f(b) = (a \mod 2) + (b \mod 2)
\]
We consider cases for \( a \mod 2 \) and \( b \mod 2 \):
- If both \( a \) and \( b \) are even, then \( f(a) = 0 \) and \( f(b) = 0 \), so \( f(a) + f(b) = 0 + 0 = 0 \).
- If both \( a \) and \( b \) are odd, then \( f(a) = 1 \) and \( f(b) = 1 \), so \( f(a) + f(b) = 1 + 1 \equiv 0 \mod 2\).
- If one is even and the other is odd, \( f(a) + f(b) = 0 + 1 \equiv 1 \mod 2\) or \( 1 + 0 \equiv 1 \mod 2\).
In all cases:
\[
(a + b) \mod 2 = f(a + b)
\]
Thus, we have established that:
\[
f(a + b) = f(a) + f(b)
\]
Since the operation is preserved, \( f \) is indeed a homomorphism from the group of integers under addition to the group of integers modulo 2 under addition.
\[
f(x \ast y) = f(x) \odot f(y)
\]
This definition indicates that the homomorphism preserves the operation; that is, the result of operating on two elements in the first structure and then applying the homomorphism is the same as first applying the homomorphism to each element and then operating on the results in the second structure.
### Example: Homomorphism in Groups
Let's consider an example involving groups. Let \( G = (\mathbb{Z}, +) \) be the group of integers under addition, and let \( H = (\mathbb{Z}/2\mathbb{Z}, +) \) be the group of integers modulo 2, also under addition.
We can define a function \( f: G \to H \) as follows:
\[
f(x) = x \mod 2
\]
### Verifying Homomorphism
To check if \( f \) is a homomorphism, we need to show that \( f \) preserves the group operation. Specifically, for all \( a, b \in \mathbb{Z} \), we need to prove:
\[
f(a + b) = f(a) + f(b)
\]
Let's calculate both sides:
1. **Left-hand side:**
\[
f(a + b) = (a + b) \mod 2
\]
2. **Right-hand side:**
\[
f(a) + f(b) = (a \mod 2) + (b \mod 2)
\]
We consider cases for \( a \mod 2 \) and \( b \mod 2 \):
- If both \( a \) and \( b \) are even, then \( f(a) = 0 \) and \( f(b) = 0 \), so \( f(a) + f(b) = 0 + 0 = 0 \).
- If both \( a \) and \( b \) are odd, then \( f(a) = 1 \) and \( f(b) = 1 \), so \( f(a) + f(b) = 1 + 1 \equiv 0 \mod 2\).
- If one is even and the other is odd, \( f(a) + f(b) = 0 + 1 \equiv 1 \mod 2\) or \( 1 + 0 \equiv 1 \mod 2\).
In all cases:
\[
(a + b) \mod 2 = f(a + b)
\]
Thus, we have established that:
\[
f(a + b) = f(a) + f(b)
\]
Since the operation is preserved, \( f \) is indeed a homomorphism from the group of integers under addition to the group of integers modulo 2 under addition.