| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B | B |
| deepseekr1 | The analysis of both statements is as follows: **Statement 1** claims that for any subgroup \( H \) of \( G \) and \( a \in G \), the left coset \( aH \) equals the right coset \( Ha \). This is only true if \( H \) is a **normal subgroup** of \( G \). However, the statement does not require \( H \) to be normal. A counterexample in \( S_3 \), where a non-normal subgroup (e.g., \( \{(1), (12)\} \)) shows \( aH \neq Ha \), proves the statement **false**. **Statement 2** asserts that if \( H \) is normal in \( G \), then \( ah = ha \) for all \( h \in H \) and \( a \in G \). Normality ensures \( aH = Ha \), meaning for every \( h \in H \), there exists \( h' \in H \) such that \( ah = h'a \). However, this does **not** imply \( h = h' \) for all \( h \). A counterexample using \( H = A_3 \) (normal in \( S_3 \)) and \( a = (12) \in S_3 \) demonstrates \( ah \neq ha \), making the statement **false**. Both statements are false. **Answer:** B. False, False |
B |
| gpto1 | B | B |
| GPT-4o | D. False, True | B |
| llama3.1-70B-Instruct | Statement 1 is false. Consider the dihedral group D4. Let H = {e, h} and a = r. Then aH = {r, rh} and Ha = {r, hr}. Since r ≠ hr, aH ≠ Ha. Statement 2 is false. Consider the dihedral group D4. Let H = {e, r2} and a = r. Then H is normal in G. But r.r2 = r3 and r2.r = r. Since r3 ≠ r, we have ar ≠ ra. |
B |
| llama3.1-405B-Instruct | A |
B |
| DeepSeek-Coder-V2 | B | B |
| Qwen2-72B-Instruct | D. False, True | B |
| baichuan4 | D. False, True | B |
| doubao | D. False, True. The first statement is not generally true for all subgroups. But the second statement is a property of a normal subgroup. |
B |
| minimax | A. True, True Explanation: Statement 1 is true because if H is a subgroup of G, then for any element a in G, the left coset aH and the right coset Ha are related by the equation aH = Ha if and only if H is a normal subgroup of G. However, in general, without H being normal, the statement does not hold true. Statement 2 is true because it is the definition of a normal subgroup. A subgroup H of a group G is said to be normal if for every element a in G and every element h in H, the equation ah = ha holds. This is exactly what Statement 2 is saying. |
B |
| kimi | D. False, True Explanation: Statement 1: If H is a subgroup of a group G and a belongs to G, then aH = Ha. This statement is false. In general, the left coset aH and the right coset Ha are not necessarily equal unless H is a normal subgroup of G. A coset aH is the set of all elements of the form ah where h is in H, and a coset Ha is the set of all elements of the form ha where h is in H. If H is not a normal subgroup, then aH and Ha can be different. Statement 2: If H is normal of G and a belongs to G, then ah = ha for all h in H. This statement is true. If H is a normal subgroup of G, then for any element a in G and any element h in H, the element ah is in H and the element ha is also in H. This means that the operation of multiplying an element from G with an element from H on either side results in an element that is still in H, which is the definition of a normal subgroup. |
B |
| wenxinyiyan | D | B |
| zhipu | D. False, True Explanation: - Statement 1: If \( H \) is a subgroup of a group \( G \) and \( a \) belongs to \( G \), then \( aH \) (left coset of \( H \) by \( a \)) does not necessarily equal \( Ha \) (right coset of \( H \) by \( a \)) unless \( H \) is a normal subgroup. Thus, Statement 1 is False. - Statement 2: If \( H \) is normal in \( G \) and \( a \) belongs to \( G \), then for all \( h \) in \( H \), \( ah = ha \). This is a defining property of normal subgroups. Thus, Statement 2 is True. Therefore, the correct answer is D. False, True. |
B |
| gpto1 | B | B |
| deepseekv3 | The correct answer is **D. False, True**. ### Explanation: **Statement 1**: "If \( H \) is a subgroup of a group \( G \) and \( a \) belongs to \( G \), then \( aH = Ha \)." This statement is **False**. The equality \( aH = Ha \) holds if and only if \( H \) is a **normal subgroup** of \( G \). For an arbitrary subgroup \( H \), this is not necessarily true. **Statement 2**: "If \( H \) is normal in \( G \) and \( a \) belongs to \( G \), then \( ah = ha \) for all \( h \) in \( H \)." This statement is **True**. If \( H \) is normal in \( G \), then for any \( a \in G \) and \( h \in H \), we have \( ah = ha \). This is the definition of a normal subgroup. Thus, the correct answer is **D. False, True**. |
B |