Math Problem Statement

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.

Solution

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

The four options are:

  • A: Z5\mathbb{Z}_5 (cyclic group of order 5)
  • B: A3A_3 (alternating group on 3 elements)
  • C: S3S_3 (symmetric group on 3 elements)
  • D: The subgroup 4\langle 4 \rangle of Z\mathbb{Z}, which is the set 4Z4\mathbb{Z} (multiples of 4).

Key Concept: First Isomorphism Theorem

The First Isomorphism Theorem states: G/ker(Φ)Φ(G)G/\ker(\Phi) \cong \Phi(G) Where GG is the original group, ker(Φ)\ker(\Phi) is the kernel of the homomorphism, and Φ(G)\Phi(G) is the image of GG under Φ\Phi.

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

Analyzing the Options:

  1. Option A: Z5\mathbb{Z}_5

    • Z5\mathbb{Z}_5 is cyclic of order 5, so it could be a homomorphic image of Z\mathbb{Z}.
  2. Option B: A3A_3

    • A3A_3 is the alternating group on 3 elements, which is isomorphic to Z3\mathbb{Z}_3 (cyclic group of order 3). Therefore, it is cyclic and could be a homomorphic image of Z\mathbb{Z}.
  3. Option C: S3S_3

    • S3S_3 is the symmetric group on 3 elements, which is not cyclic (it's generated by two elements and is not abelian). Since S3S_3 is not cyclic, it cannot be isomorphic to a homomorphic image of Z\mathbb{Z}.
  4. Option D: The subgroup 4\langle 4 \rangle of Z\mathbb{Z}

    • 4=4Z\langle 4 \rangle = 4\mathbb{Z} is cyclic (since any subgroup of Z\mathbb{Z} is cyclic), so it could be a homomorphic image of Z\mathbb{Z}.

Conclusion:

The group that cannot be isomorphic to Φ(G)\Phi(G) is Option C: S3S_3, because it is not cyclic.

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


Here are five related questions you might consider:

  1. Can every homomorphism from Z\mathbb{Z} to any group be described by a single integer?
  2. What are the properties of the kernel of a homomorphism from Z\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 S3S_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).

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Math Problem Analysis

Mathematical Concepts

Group Theory
Isomorphisms
Homomorphisms
Cyclic Groups
First Isomorphism Theorem

Formulas

G/ker(Φ) ≅ Φ(G)

Theorems

First Isomorphism Theorem

Suitable Grade Level

Undergraduate (Abstract Algebra)