| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B. 2 Explanation: A homomorphism from Z to Z_2 is a function f: Z → Z_2 that satisfies f(a + b) = f(a) + f(b) for all a, b in Z. Since Z_2 has only two elements, 0 and 1, there are only two possible homomorphisms: 1. The trivial homomorphism, which maps every element of Z to 0 in Z_2. 2. The homomorphism that maps every even integer to 0 and every odd integer to 1. Therefore, there are exactly 2 homomorphisms from Z to Z_2. |
B |
| deepseekr1 | The question asks for the number of group homomorphisms from ℤ (the integers under addition) to ℤ₂ (the integers modulo 2). 1. **Key Insight**: A group homomorphism from ℤ is determined entirely by where the generator (1 ∈ ℤ) is mapped, as ℤ is cyclic and generated by 1. 2. **Possible Images**: The codomain ℤ₂ has two elements: 0 and 1. Thus, there are two choices for the image of 1: - **Case 1**: If 1 ↦ 0, the homomorphism maps every integer to 0 (trivial homomorphism). - **Case 2**: If 1 ↦ 1, the homomorphism maps even integers to 0 and odd integers to 1 (reduction modulo 2). 3. **Verification**: Both choices satisfy the homomorphism property \( f(a + b) = f(a) + f(b) \). No other mappings exist because ℤ₂ only has two elements, and ℤ's structure is cyclic. Thus, there are exactly **2 homomorphisms**. Answer: B |
B |
| gpto1 | B | B |
| GPT-4o | A. 1 | B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | B. 2 Explanation: A homomorphism from Z to Z_2 is a function f: Z → Z_2 that satisfies f(a + b) = f(a) + f(b) for all a, b in Z. Since Z_2 has only two elements, 0 and 1, there are only two possible homomorphisms: 1. The trivial homomorphism, which maps every element of Z to 0 in Z_2. 2. The homomorphism that maps every even integer to 0 and every odd integer to 1. Therefore, there are exactly 2 homomorphisms from Z to Z_2. |
B |
| DeepSeek-Coder-V2 | B | B |
| Qwen2-72B-Instruct | B. 2 | B |
| baichuan4 | A. 1 | B |
| doubao | B. 2. A homomorphism from Z to Z_2 is completely determined by the image of 1. There are two possible images of 1 in Z_2, namely 0 and 1, so there are two homomorphisms. |
B |
| minimax | B. 2 Explanation: A homomorphism from the integers Z into the cyclic group Z_2 must map even integers to the identity element of Z_2 (which we can denote as 0) and odd integers to the non-identity element of Z_2 (which we can denote as 1). There are only two possible mappings that satisfy this condition: 1. The trivial homomorphism that maps every integer to 0 in Z_2. 2. The homomorphism that maps even integers to 0 and odd integers to 1 in Z_2. These are the only two homomorphisms possible because any other mapping would not preserve the group operation. For example, if you map an odd integer to 0, then the image of its sum with any other integer would not be consistent with the group operation in Z_2. |
B |
| kimi | B. 2 | B |
| wenxinyiyan | B | B |
| zhipu | B. 2 Explanation: A homomorphism from \(\mathbb{Z}\) (the integers) to \(\mathbb{Z}_2\) (the integers modulo 2) must map the generator of \(\mathbb{Z}\) (which is 1) to an element of \(\mathbb{Z}_2\). The elements of \(\mathbb{Z}_2\) are 0 and 1. Therefore, there are two possible homomorphisms: one that maps 1 to 0 and one that maps 1 to 1. These correspond to the zero homomorphism and the identity homomorphism, respectively. Hence, there are 2 homomorphisms. |
B |
| gpto1 | B | B |
| deepseekv3 | The question asks for the number of homomorphisms from the group of integers \(\mathbb{Z}\) to the group \(\mathbb{Z}_2\). 1. **Understanding the groups**: - \(\mathbb{Z}\) is the group of integers under addition. It is an infinite cyclic group generated by \(1\). - \(\mathbb{Z}_2\) is the group of integers modulo 2 under addition. It has two elements: \([0]\) and \([1]\). 2. **Homomorphism definition**: A homomorphism \(f: \mathbb{Z} \to \mathbb{Z}_2\) must satisfy \(f(a + b) = f(a) + f(b)\) for all \(a, b \in \mathbb{Z}\). 3. **Determining the homomorphisms**: Since \(\mathbb{Z}\) is cyclic, any homomorphism is completely determined by the image of the generator \(1\). Let \(f(1) = x\) where \(x \in \mathbb{Z}_2\). - If \(x = [0]\), then \(f(n) = n \cdot [0] = [0]\) for all \(n \in \mathbb{Z}\). This is the zero homomorphism. - If \(x = [1]\), then \(f(n) = n \cdot [1] = [n \mod 2]\). This is a nontrivial homomorphism. These are the only two possibilities because \(\mathbb{Z}_2\) has only two elements. 4. **Conclusion**: There are exactly two homomorphisms from \(\mathbb{Z}\) to \(\mathbb{Z}_2\). **Answer: B. 2** |
B |