| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | D |
D |
| deepseekr1 | The first statement claims all solvable groups have prime-power order. However, counterexamples like the symmetric group \( S_3 \) (order 6, solvable but not a prime power) show this is false. Thus, Statement 1 is **False**. For Statement 2, all prime-power order groups (p-groups) are solvable. This follows because p-groups have nontrivial centers, allowing induction on their order to demonstrate solvability via normal subgroups and solvable extensions. Hence, Statement 2 is **True**. **Answer:** D. False, True |
D |
| gpto1 | D | D |
| GPT-4o | D. False, True | D |
| llama3.1-70B-Instruct | Statement 1 is false. For example, the symmetric group $S_{3}$ is solvable but not of prime-power order. Statement 2 is true. In fact, every group of prime-power order is nilpotent, and hence solvable. Final Answer: The final answer is D. I hope it is correct. |
D |
| llama3.1-405B-Instruct | D |
D |
| DeepSeek-Coder-V2 | D | D |
| Qwen2-72B-Instruct | D. False, True | D |
| baichuan4 | D | D |
| doubao | D. False, True. Not every solvable group is of prime-power order (so Statement 1 is false), but every group of prime-power order is solvable (Statement 2 is true). |
D |
| minimax | D. False, True Statement 1 is false because not every solvable group is of prime-power order. A solvable group is one that has a normal series whose factor groups (quotient groups) are all abelian, and there are solvable groups that are not of prime power order. Statement 2 is true because every group of prime-power order is solvable. This is a consequence of the fact that groups of prime power order have non-trivial center, and by using the class equation and induction on the order of the group, one can show that such groups have a normal series with abelian factor groups. |
D |
| kimi | D. False, True Statement 1 is false because a solvable group is one where there exists a subnormal series with abelian factors, and it does not necessarily have to be of prime-power order. Statement 2 is true because a group of prime-power order is indeed solvable. This is a result of the fact that a group of prime-power order is a p-group, and p-groups are solvable. |
D |
| wenxinyiyan | D | D |
| zhipu | D. False, True Explanation: - Statement 1: "Every solvable group is of prime-power order." This is false. A solvable group does not necessarily have to be of prime-power order. For example, the symmetric group \( S_3 \) is solvable but not of prime-power order. - Statement 2: "Every group of prime-power order is solvable." This is true. Groups of prime-power order are known to be solvable by the structure theorem for finite abelian groups and the fact that p-groups (groups of order \( p^n \) for some prime \( p \) and integer \( n \)) are solvable. |
D |
| gpto1 | D | D |
| deepseekv3 | The correct answer is: **D. False, True** ### Explanation: 1. **Statement 1**: "Every solvable group is of prime-power order." This is **false**. Not all solvable groups have prime-power order. For example, the symmetric group \( S_3 \) is solvable but its order is 6, which is not a prime power. 2. **Statement 2**: "Every group of prime-power order is solvable." This is **true**. By Burnside's theorem, all groups of prime-power order (i.e., \( p \)-groups) are solvable. Thus, Statement 1 is false, and Statement 2 is true. The correct choice is **D**. |
D |