Could you show me how to do this in detail? Could you state any theorems and definition you used for this problem?(Q is quaternion group) **Problem 3:**Show that $ \mathbb{Q} $ has no subgroup isomorphic to $ \mathbb{Z}_2 \times \mathbb{Z}_2 $.### Explanation:- $ \mathbb{Q} $ represents the group of rational numbers under addition.- $ \mathbb{Z}_2 $ denotes the cyclic group of order 2.- $ \mathbb{Z}_2 \times \mathbb{Z}_2 $ is the direct product of $\mathbb{Z}_2$ with itself, which is known to be an abelian group of order 4 where every element has order 2.**Objective:**Prove that within the group $ \mathbb{Q} $, there is no subgroup that matches the structure of $ \mathbb{Z}_2 \times \mathbb{Z}_2 $.### Solution Outline:1. **Basic Properties:** - $ \mathbb{Q} $ as an additive group is divisible; for any positive integer $ n $ and any element $ q \in \mathbb{Q} $, there exists an element $ r \in \mathbb{Q} $ such that $ nr = q $.2. **Subgroup Structure of $ \mathbb{Z}_2 \times \mathbb{Z}_2 $:** - Elements of $ \mathbb{Z}_2 \times \mathbb{Z}_2 $ are $(0,0), (0,1), (1,0), (1,1)$ with the property that $ 2x = 0$ for all $x$.3. **Approach:** - Assume for contradiction that there exists a subgroup $ H \subset \mathbb{Q} $ isomorphic to $ \mathbb{Z}_2 \times \mathbb{Z}_2 $. - Identify elements $ a, b \in \mathbb{Q} $ such that every non-identity element in $ H $ has order 2. Thus, $ 2a = 0 $ and $ 2b = 0 $, implying $ a = 0 $ and $ b = 0 $. - Conclude that the only solution is the trivial subgroup, which contradicts the assumed structure

The objective is to show that ℚ has no subgroup isomorphic to ℤ2×ℤ2

Answered: 3. Show that Q has no subgroup…

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3. Show that Q has no subgroup isomorphic to Z2 × Z2.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter4: More On Groups

Section4.6: Quotient Groups

Problem 11E: Find all homomorphic images of the quaternion group.

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Could you show me how to do this in detail? Could you state any theorems and definition you used for this problem?

(Q is quaternion group)

**Problem 3:**

Show that $ \mathbb{Q} $ has no subgroup isomorphic to $ \mathbb{Z}_2 \times \mathbb{Z}_2 $.

### Explanation:
- $ \mathbb{Q} $ represents the group of rational numbers under addition.
- $ \mathbb{Z}_2 $ denotes the cyclic group of order 2.
- $ \mathbb{Z}_2 \times \mathbb{Z}_2 $ is the direct product of $\mathbb{Z}_2$ with itself, which is known to be an abelian group of order 4 where every element has order 2.

**Objective:**
Prove that within the group $ \mathbb{Q} $, there is no subgroup that matches the structure of $ \mathbb{Z}_2 \times \mathbb{Z}_2 $.

### Solution Outline:

1. **Basic Properties:**
- $ \mathbb{Q} $ as an additive group is divisible; for any positive integer $ n $ and any element $ q \in \mathbb{Q} $, there exists an element $ r \in \mathbb{Q} $ such that $ nr = q $.

2. **Subgroup Structure of $ \mathbb{Z}_2 \times \mathbb{Z}_2 $:**
- Elements of $ \mathbb{Z}_2 \times \mathbb{Z}_2 $ are $(0,0), (0,1), (1,0), (1,1)$ with the property that $ 2x = 0$ for all $x$.

3. **Approach:**
- Assume for contradiction that there exists a subgroup $ H \subset \mathbb{Q} $ isomorphic to $ \mathbb{Z}_2 \times \mathbb{Z}_2 $.
- Identify elements $ a, b \in \mathbb{Q} $ such that every non-identity element in $ H $ has order 2. Thus, $ 2a = 0 $ and $ 2b = 0 $, implying $ a = 0 $ and $ b = 0 $.
- Conclude that the only solution is the trivial subgroup, which contradicts the assumed structure

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