**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 $\mathbb{Z}_2 \times \mathbb{Z}_2$.

We know that if two groups are isomorphic than they have same number of elements i.e. their…

Answered: Show that the multiplicative group Z is…

Math

Advanced Math

Show that the multiplicative group Z is isomorphic to the group Z2 X Z2 8,

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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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 $\mathbb{Z}_2 \times \mathbb{Z}_2$.

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