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10. If : G→ H is a group homomorphism and G is cyclic, prove that (G) is also cyclic.

10. If : G→ H is a group homomorphism and G is cyclic, prove that (G) is also cyclic.

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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**Problem 10:** If \( \phi : G \rightarrow H \) is a group homomorphism and \( G \) is cyclic, prove that \( \phi(G) \) is also cyclic. **Explanation:** The problem prompts you to demonstrate that if you have a group homomorphism \( \phi \) from group \( G \) to group \( H \) and \( G \) is a cyclic group, then the image of \( G \) under \( \phi \), denoted as \( \phi(G) \), is also a cyclic group. **Key Concepts:** - **Group Homomorphism:** A function between two groups that respects the group operation. - **Cyclic Group:** A group that can be generated by a single element. - **Image of a Homomorphism:** The set of all outputs \( \phi(g) \) where \( g \in G \). **Approach to the Solution:** 1. **Understand Cyclic Groups:** Since \( G \) is cyclic, there exists an element \( g \in G \) such that every element in \( G \) can be expressed as \( g^n \) for some integer \( n \). 2. **Apply Homomorphism to Generators:** Consider \( \phi(g) \). Since \( \phi \) is a homomorphism, it respects the group operation: \( \phi(g^n) = \phi(g)^n \). 3. **Show Image is Cyclic:** Demonstrate that every element in \( \phi(G) \) can be expressed as \( \phi(g)^n \), showing that \( \phi(g) \) is a generator for \( \phi(G) \). By following these steps, you can conclude that \( \phi(G) \) is indeed cyclic.
Transcribed Image Text:**Problem 10:** If \( \phi : G \rightarrow H \) is a group homomorphism and \( G \) is cyclic, prove that \( \phi(G) \) is also cyclic. **Explanation:** The problem prompts you to demonstrate that if you have a group homomorphism \( \phi \) from group \( G \) to group \( H \) and \( G \) is a cyclic group, then the image of \( G \) under \( \phi \), denoted as \( \phi(G) \), is also a cyclic group. **Key Concepts:** - **Group Homomorphism:** A function between two groups that respects the group operation. - **Cyclic Group:** A group that can be generated by a single element. - **Image of a Homomorphism:** The set of all outputs \( \phi(g) \) where \( g \in G \). **Approach to the Solution:** 1. **Understand Cyclic Groups:** Since \( G \) is cyclic, there exists an element \( g \in G \) such that every element in \( G \) can be expressed as \( g^n \) for some integer \( n \). 2. **Apply Homomorphism to Generators:** Consider \( \phi(g) \). Since \( \phi \) is a homomorphism, it respects the group operation: \( \phi(g^n) = \phi(g)^n \). 3. **Show Image is Cyclic:** Demonstrate that every element in \( \phi(G) \) can be expressed as \( \phi(g)^n \), showing that \( \phi(g) \) is a generator for \( \phi(G) \). By following these steps, you can conclude that \( \phi(G) \) is indeed cyclic.
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