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

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

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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(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
Transcribed Image Text:**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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