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

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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