Show that the multiplicative group Z is isomorphic to the group Z2 X Z2 8,
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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![**Title: Isomorphism of Multiplicative and Cartesian Groups**
**Objective:**
Show that the multiplicative group \(\mathbb{Z}_8^\times\) is isomorphic to the group \(\mathbb{Z}_2 \times \mathbb{Z}_2\).
**Problem Statement:**
The task is to demonstrate that the multiplicative group of units modulo 8, denoted \(\mathbb{Z}_8^\times\), is structurally identical to the direct product of the cyclic group of order 2, denoted \(\mathbb{Z}_2\), with itself.
This means we need to find a bijective homomorphism:
\[ f: \mathbb{Z}_8^\times \to \mathbb{Z}_2 \times \mathbb{Z}_2 \]
which is a function that preserves the group operation, ensuring that the structure of \(\mathbb{Z}_8^\times\) matches that of \(\mathbb{Z}_2 \times \mathbb{Z}_2\).
To begin, let's examine the elements of each group.
### Elements:
- \(\mathbb{Z}_8^\times\): The group of units modulo 8 consists of those integers less than 8 that are coprime to 8. These elements are {1, 3, 5, 7}.
- \(\mathbb{Z}_2\): The cyclic group of order 2 consists of the elements {0, 1}.
- \(\mathbb{Z}_2 \times \mathbb{Z}_2\): This direct product consists of the ordered pairs {(0,0), (0,1), (1,0), (1,1)}.
### Verification:
Now, we will establish the isomorphism by verifying the group structure.
1. **Group Operation Preservation:**
- For \(\mathbb{Z}_8^\times\), the operation is multiplication modulo 8.
- For \(\mathbb{Z}_2 \times \mathbb{Z}_2\), the operation is component-wise addition modulo 2.
2. **Bijection and Homomorphism:**
- Define a mapping \( f \) from \(\mathbb{Z}_8^\times\) to \(\mathbb{Z}_2 \times \mathbb{Z}_2\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd95b1c5e-3673-40a0-a87b-179a3e73672c%2Fb92db3c1-cbe9-44f0-afd4-1929e7c3fcee%2F5f2gc7x_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Isomorphism of Multiplicative and Cartesian Groups**
**Objective:**
Show that the multiplicative group \(\mathbb{Z}_8^\times\) is isomorphic to the group \(\mathbb{Z}_2 \times \mathbb{Z}_2\).
**Problem Statement:**
The task is to demonstrate that the multiplicative group of units modulo 8, denoted \(\mathbb{Z}_8^\times\), is structurally identical to the direct product of the cyclic group of order 2, denoted \(\mathbb{Z}_2\), with itself.
This means we need to find a bijective homomorphism:
\[ f: \mathbb{Z}_8^\times \to \mathbb{Z}_2 \times \mathbb{Z}_2 \]
which is a function that preserves the group operation, ensuring that the structure of \(\mathbb{Z}_8^\times\) matches that of \(\mathbb{Z}_2 \times \mathbb{Z}_2\).
To begin, let's examine the elements of each group.
### Elements:
- \(\mathbb{Z}_8^\times\): The group of units modulo 8 consists of those integers less than 8 that are coprime to 8. These elements are {1, 3, 5, 7}.
- \(\mathbb{Z}_2\): The cyclic group of order 2 consists of the elements {0, 1}.
- \(\mathbb{Z}_2 \times \mathbb{Z}_2\): This direct product consists of the ordered pairs {(0,0), (0,1), (1,0), (1,1)}.
### Verification:
Now, we will establish the isomorphism by verifying the group structure.
1. **Group Operation Preservation:**
- For \(\mathbb{Z}_8^\times\), the operation is multiplication modulo 8.
- For \(\mathbb{Z}_2 \times \mathbb{Z}_2\), the operation is component-wise addition modulo 2.
2. **Bijection and Homomorphism:**
- Define a mapping \( f \) from \(\mathbb{Z}_8^\times\) to \(\mathbb{Z}_2 \times \mathbb{Z}_2\).
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