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The problem you uploaded is related to the First Isomorphism Theorem and homomorphisms between groups. Let's analyze the question step by step:

### Problem Summary:
- \( G = \mathbb{Z} \) (the group of integers under addition), and we are considering a homomorphism \( \Phi: G \to \overline{G} \), where \( G \) and \( \overline{G} \) are groups.
- The question asks which of the following groups **cannot** be isomorphic to the image of \( G \) under \( \Phi \), i.e., \( \Phi(G) \).

The four options are:
- A: \( \mathbb{Z}_5 \) (cyclic group of order 5)
- B: \( A_3 \) (alternating group on 3 elements)
- C: \( S_3 \) (symmetric group on 3 elements)
- D: The subgroup \( \langle 4 \rangle \) of \( \mathbb{Z} \), which is the set \( 4\mathbb{Z} \) (multiples of 4).

### Key Concept: First Isomorphism Theorem
The First Isomorphism Theorem states:
\[
G/\ker(\Phi) \cong \Phi(G)
\]
Where \( G \) is the original group, \( \ker(\Phi) \) is the kernel of the homomorphism, and \( \Phi(G) \) is the image of \( G \) under \( \Phi \).

If \( G = \mathbb{Z} \), then any homomorphic image of \( \mathbb{Z} \) is **cyclic** (since \( \mathbb{Z} \) is cyclic itself). This means that \( \Phi(G) \) must be isomorphic to a cyclic group.

### Analyzing the Options:
1. **Option A: \( \mathbb{Z}_5 \)** 
   - \( \mathbb{Z}_5 \) is cyclic of order 5, so it **could** be a homomorphic image of \( \mathbb{Z} \).

2. **Option B: \( A_3 \)** 
   - \( A_3 \) is the alternating group on 3 elements, which is isomorphic to \( \mathbb{Z}_3 \) (cyclic group of order 3). Therefore, it is cyclic and **could** be a homomorphic image of \( \mathbb{Z} \).

3. **Option C: \( S_3 \)**
   - \( S_3 \) is the symmetric group on 3 elements, which is **not cyclic** (it's generated by two elements and is not abelian). Since \( S_3 \) is not cyclic, it **cannot** be isomorphic to a homomorphic image of \( \mathbb{Z} \).

4. **Option D: The subgroup \( \langle 4 \rangle \) of \( \mathbb{Z} \)**
   - \( \langle 4 \rangle = 4\mathbb{Z} \) is cyclic (since any subgroup of \( \mathbb{Z} \) is cyclic), so it **could** be a homomorphic image of \( \mathbb{Z} \).

### Conclusion:
The group that **cannot** be isomorphic to \( \Phi(G) \) is **Option C: \( S_3 \)**, because it is not cyclic.

Would you like more details or have any questions about this solution?

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Here are five related questions you might consider:
1. Can every homomorphism from \( \mathbb{Z} \) to any group be described by a single integer?
2. What are the properties of the kernel of a homomorphism from \( \mathbb{Z} \)?
3. How does the structure of the image of a homomorphism reflect the structure of the original group?
4. What is the significance of cyclic groups in group homomorphisms?
5. Can non-abelian groups like \( S_3 \) ever be homomorphic images of abelian groups?

**Tip:** When analyzing potential homomorphic images, always check whether the target group shares key structural properties (e.g., cyclic vs non-cyclic, abelian vs non-abelian).

Let G and G̅ be groups and Φ: G → G̅ be a homomorphism. If G = Z, which of the following groups cannot be isomorphic to Φ(G)? A: Z₅, B: A₃, C: S₃, D: The subgroup ⟨4⟩ of Z.

This problem addresses the First Isomorphism Theorem and group homomorphisms. It explores which group from Z₅, A₃, S₃, or ⟨4⟩ of Z cannot be isomorphic to the homomorphic image of Z under a given homomorphism. Ideal for undergraduate students studying Abstract Algebra.

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