| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B | B |
| deepseekr1 | The correct answer is **B. False, False**. **Analysis:** 1. **Statement 1:** A factor group of a non-Abelian group is non-Abelian. - **Counterexample:** The symmetric group \( S_3 \) is non-Abelian. Taking the normal subgroup \( A_3 \) (alternating group of order 3), the factor group \( S_3/A_3 \) has order 2 and is cyclic (hence Abelian). This shows a non-Abelian group can have an Abelian factor group, making Statement 1 **false**. 2. **Statement 2:** If \( K \trianglelefteq H \) and \( H \trianglelefteq G \), then \( K \trianglelefteq G \). - **Counterexample:** Let \( G = A_4 \), \( H \) be the Klein 4-group (normal in \( A_4 \)), and \( K \) a subgroup of \( H \) of order 2 (normal in \( H \)). Conjugating \( K \) by elements of \( A_4 \) (e.g., conjugation by \( (123) \)) results in elements outside \( K \), proving \( K \not\trianglelefteq G \). Thus, Statement 2 is **false**. Both statements are false, so the answer is **B**. |
B |
| gpto1 | B | B |
| GPT-4o | D. False, True | B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | A factor group of a non-Abelian group can be Abelian. For example, consider the dihedral group D4 = {e, a, a2, a3, b, ab, a2b, a3b} where a4 = e, b2 = e and ab = ba-1. The subgroup H = {e, a2} is normal in D4 and the factor group D4/H is Abelian. Therefore, statement 1 is false. If K is a normal subgroup of H and H is a normal subgroup of G, then K is a normal subgroup of G. This is because if k is an element of K, h is an element of H, and g is an element of G, then ghk = (ghg-1)(gkg-1) is an element of K since H is normal in G and K is normal in H. Therefore, statement 2 is true. The correct answer is D. False, True. |
B |
| DeepSeek-Coder-V2 | B. False, False | B |
| Qwen2-72B-Instruct | D. False, True | B |
| baichuan4 | D | B |
| doubao | D. False, True Counterexample for Statement 1: The factor group of the non-Abelian group S_3 (symmetric group on 3 elements) by the normal subgroup A_3 (alternating group on 3 elements) is Abelian. Statement 2 is a well-known result in abstract algebra and is true. |
B |
| minimax | D. False, True Explanation: Statement 1 is false because a factor group (also known as a quotient group) of a non-Abelian group can be Abelian. For example, consider the symmetric group S3, which is non-Abelian, and its normal subgroup A3, which is the alternating group consisting of even permutations. The factor group S3/A3 is isomorphic to Z2, which is Abelian. Statement 2 is true because if K is a normal subgroup of H and H is a normal subgroup of G, then for any g in G and k in K, we have gkg^(-1) in H (since H is normal in G) and also in K (since K is normal in H). Therefore, gkg^(-1) is in K, which shows that K is normal in G. |
B |
| kimi | C. True, False Statement 1 is true because a factor group (also known as a quotient group) of a non-Abelian group can be non-Abelian. The commutativity of a group is not necessarily preserved in its factor groups. Statement 2 is false because the normality of a subgroup is not transitive. If K is a normal subgroup of H and H is a normal subgroup of G, it does not necessarily mean that K is a normal subgroup of G. Normality is a property that depends on the specific group and subgroup in question, and it is not inherited in the way that the statement suggests. |
B |
| wenxinyiyan | D | B |
| zhipu | The correct answer is B. False, False. Explanation: **Statement 1: A factor group of a non-Abelian group is non-Abelian.** This statement is false. It is possible for a non-Abelian group to have an Abelian factor group. For example, consider the symmetric group \( S_3 \), which is non-Abelian. The subgroup \( A_3 \) (the alternating group of order 3) is normal in \( S_3 \), and the factor group \( S_3 / A_3 \) is isomorphic to \( \mathbb{Z}_2 \), which is Abelian. **Statement 2: If K is a normal subgroup of H and H is a normal subgroup of G, then K is a normal subgroup of G.** This statement is also false. The property of being a normal subgroup is not transitive. A counterexample is the dihedral group \( D_8 \) (the group of symmetries of a square). Let \( K \) be the subgroup of rotations by 90 degrees, \( H \) be the subgroup of rotations (which includes \( K \)), and \( G \) be \( D_8 \). \( K \) is normal in \( H \) and \( H \) is normal in \( G \), but \( K \) is not normal in \( G \). Therefore, both statements are false, making the correct answer B. False, False. |
B |
| gpto1 | B | B |
| deepseekv3 | The correct answer is **D. False, True**. ### Explanation: 1. **Statement 1**: A factor group of a non-Abelian group is non-Abelian. This is **False**. A factor group \( G/N \) of a non-Abelian group \( G \) can be Abelian. For example, the symmetric group \( S_3 \) is non-Abelian, but its factor group by the alternating group \( A_3 \) is Abelian (since \( S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z} \), which is Abelian). 2. **Statement 2**: If \( K \) is a normal subgroup of \( H \) and \( H \) is a normal subgroup of \( G \), then \( K \) is a normal subgroup of \( G \). This is **True**. Normality is transitive in this case: if \( K \triangleleft H \) and \( H \triangleleft G \), then \( K \triangleleft G \). Thus, **D. False, True** is the correct answer. |
B |