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4. Let p: G → H be an onto homomorphism. Show that if G is cyclic, so is H.

4. Let p: G → H be an onto homomorphism. Show that if G is cyclic, so is H.

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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**Transcription for Educational Website:** --- **Problem 4:** Let \( \varphi: G \rightarrow H \) be an onto homomorphism. Show that if \( G \) is cyclic, so is \( H \). --- In this problem, we're examining a homomorphism between two groups and exploring the implications for \( H \) when \( G \) is known to be cyclic. ### Explanation: - **Cyclic Group:** A group is cyclic if there exists an element (called a generator) such that every element of the group can be expressed as a power of this generator. - **Homomorphism:** A function between two groups that respects the group operation. - **Onto (Surjective) Homomorphism:** A homomorphism where every element in the target group \( H \) has a pre-image in the source group \( G \). ### Key Concepts: To solve this problem, consider: 1. Identifying a generator \( g \) of \( G \) since it's cyclic. 2. Using the property of the homomorphism to show that \( \varphi(g) \) generates \( H \). Conclusion: You will demonstrate that a generator of \( G \) is mapped to a generator of \( H \), proving that \( H \) is cyclic. *Note: There are no graphs or diagrams in this text.*
Transcribed Image Text:**Transcription for Educational Website:** --- **Problem 4:** Let \( \varphi: G \rightarrow H \) be an onto homomorphism. Show that if \( G \) is cyclic, so is \( H \). --- In this problem, we're examining a homomorphism between two groups and exploring the implications for \( H \) when \( G \) is known to be cyclic. ### Explanation: - **Cyclic Group:** A group is cyclic if there exists an element (called a generator) such that every element of the group can be expressed as a power of this generator. - **Homomorphism:** A function between two groups that respects the group operation. - **Onto (Surjective) Homomorphism:** A homomorphism where every element in the target group \( H \) has a pre-image in the source group \( G \). ### Key Concepts: To solve this problem, consider: 1. Identifying a generator \( g \) of \( G \) since it's cyclic. 2. Using the property of the homomorphism to show that \( \varphi(g) \) generates \( H \). Conclusion: You will demonstrate that a generator of \( G \) is mapped to a generator of \( H \), proving that \( H \) is cyclic. *Note: There are no graphs or diagrams in this text.*
Expert Solution
Step 1

Given that ϕ:G→H is an onto homomorphism.

Suppose that G is cyclic, then,

G=x , for some x

Let y=ϕx and h∈H.

Since ϕ is onto, there exists a g∈G with ϕg=h.

Since G is cyclic, there exists an integer n with xn=g.

But then , 

h=ϕg   =ϕxn   =ϕxn   =yn

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