**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.*

Answered: 4. Let p: G → H be an onto…

Math

Advanced Math

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

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.*

Expert Solution

Step 1

Given that $ϕ : G \to H$ is an onto homomorphism.

Suppose that G is cyclic, then,

$G = <x>$ , for some x

Let $y = ϕ (x)$ and $h \in H$ .

Since $ϕ$ is onto, there exists a $g \in G$ with $ϕ (g) = h$ .

Since G is cyclic, there exists an integer n with $x^{n} = g$ .

But then ,

$h = ϕ (g) = ϕ (x^{n}) = ϕ {(x)}^{n} = y^{n}$

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