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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#4. Thanks.
![### Problem Statement
4. Suppose that \( G \) is a finite group with an element \( g \) of order 5 and an element \( h \) of order 7. Why must \( |G| \geq 35 \)? (1 from 6.5)
### Explanation
Given:
- \( G \) is a finite group.
- \( g \) is an element in \( G \) with order 5.
- \( h \) is an element in \( G \) with order 7.
To show:
- The order of \( G \), \( |G| \), must be at least 35.
### Solution Sketch
1. **Basic Concepts**:
- The **order of an element** in a group is the smallest positive integer \( n \) such that \( g^n = e \), where \( e \) is the identity element.
- By **Lagrange's Theorem**, the order of any element in a finite group divides the order of the group.
2. **Orders and Multiples**:
- Given \( g \) has an order of 5, \( 5 \) divides \( |G| \).
- Given \( h \) has an order of 7, \( 7 \) divides \( |G| \).
3. **Relatively Prime Orders**:
- The greatest common divisor (gcd) of 5 and 7 is 1, since they are relatively prime.
4. **Conclusion**:
- Since 5 and 7 are both divisors of \( |G| \) and are relatively prime, their least common multiple must divide \( |G| \).
- The least common multiple (LCM) of 5 and 7 is \( 35 \).
Thus,
\[ |G| \text { must be a multiple of 35, and therefore } |G| \geq 35 \]
This completes the proof that \( |G| \geq 35 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbb66fa7-7c22-4982-a22f-aaed542f65b3%2F8fb26300-2783-4520-a282-22c7cee69912%2Fzipfwg_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
4. Suppose that \( G \) is a finite group with an element \( g \) of order 5 and an element \( h \) of order 7. Why must \( |G| \geq 35 \)? (1 from 6.5)
### Explanation
Given:
- \( G \) is a finite group.
- \( g \) is an element in \( G \) with order 5.
- \( h \) is an element in \( G \) with order 7.
To show:
- The order of \( G \), \( |G| \), must be at least 35.
### Solution Sketch
1. **Basic Concepts**:
- The **order of an element** in a group is the smallest positive integer \( n \) such that \( g^n = e \), where \( e \) is the identity element.
- By **Lagrange's Theorem**, the order of any element in a finite group divides the order of the group.
2. **Orders and Multiples**:
- Given \( g \) has an order of 5, \( 5 \) divides \( |G| \).
- Given \( h \) has an order of 7, \( 7 \) divides \( |G| \).
3. **Relatively Prime Orders**:
- The greatest common divisor (gcd) of 5 and 7 is 1, since they are relatively prime.
4. **Conclusion**:
- Since 5 and 7 are both divisors of \( |G| \) and are relatively prime, their least common multiple must divide \( |G| \).
- The least common multiple (LCM) of 5 and 7 is \( 35 \).
Thus,
\[ |G| \text { must be a multiple of 35, and therefore } |G| \geq 35 \]
This completes the proof that \( |G| \geq 35 \).
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