金声玉亮2.0模型评测报告

问题

There is a single choice question about abstract algebra. Answer the question by replying A, B, C or D.
Question: Statement 1 | If H is a subgroup of G and a belongs to G then aH is a subgroup of G if and only if a is in H. Statement 2 | If H is a subgroup of G and a and b belong to G then aH = bH if and only if ab is in H.
A. True, True
B. False, False
C. True, False
D. False, True
Answer:

答案

模型	模型答案	正确答案
金声玉亮2.0	A. True, True. Reasoning Skill: Consequence Evaluation This question requires the test-taker to evaluate the consequences of certain conditions in abstract algebra. Statement 1 implies that if aH is a subgroup of G, then a must be in H, and vice versa. Statement 2 implies that if aH = bH, then ab must be in H, and vice versa. The test-taker needs to analyze the conditions and determine the consequences of each statement, which requires Consequence Evaluation reasoning skill. Note: The question is designed to test the understanding of the properties of subgroups and cosets in abstract algebra, and the ability to evaluate the consequences of certain conditions. The correct answer is A, True, True, as both statements are correct.	C
deepseekr1	The correct answer is C. True, False. Statement 1 Analysis: If \( aH \) is a subgroup of \( G \), it must contain the identity element \( e \). This implies \( a \in H \) (since \( aH \) can only contain \( e \) if \( a \in H \)). Conversely, if \( a \in H \), then \( aH = H \), which is a subgroup. Thus, Statement 1 is true. Statement 2 Analysis: The standard criterion for \( aH = bH \) is \( a^{-1}b \in H \), not \( ab \in H \). A counterexample: Let \( G = S_3 \), \( H = \{e, (12)\} \), \( a = (13) \), and \( b = (132) \). Here, \( ab = e \in H \), but \( aH = \{(13), (123)\} \) and \( bH = \{(132), (23)\} \), which are distinct. Thus, Statement 2 is false. Answer: C	C
gpto1	C	C
GPT-4o	D. False, True	C
llama3.1-70B-Instruct	A	C
llama3.1-405B-Instruct	A	C
DeepSeek-Coder-V2	B	C
Qwen2-72B-Instruct	B. False, False	C
baichuan4	D. False, True	C
doubao	B For Statement 1, aH is a left coset of H in G, not necessarily a subgroup. For Statement 2, aH = bH if and only if a^{-1}b is in H, not ab.	C
minimax	B. False, False Explanation: Statement 1 is false because aH being a subgroup of G does not imply that a must be in H. It is true that if a is in H, then aH is a subgroup of G, but the converse is not necessarily true. The set aH is a left coset of H in G, and it is a subgroup of G if and only if H is a normal subgroup of G. If H is not normal, then aH is not necessarily a subgroup. Statement 2 is also false. The correct statement is that aH = bH if and only if ab^(-1) is in H, where b^(-1) is the inverse of b in G. This is because aH = bH implies that for every element h in H, ah is in bH, and thus there exists some h' in H such that ah = bh'. This means that a = bh'b^(-1), and since h' is in H, bh'b^(-1) is in H if and only if b^(-1) is in H, which is true if and only if ab^(-1) is in H.	C
kimi	D. False, True Explanation: Statement 1: If H is a subgroup of G and a belongs to G, then aH is a subgroup of G if and only if a is in H. This statement is false. The set aH is a left coset of H in G, and it is a subgroup of G if and only if aH = H, which happens if and only if a is in H. However, the statement as it is written implies that aH is a subgroup of G for any a in G, which is not true. Statement 2: If H is a subgroup of G and a and b belong to G, then aH = bH if and only if ab^(-1) is in H. This statement is true. The left cosets aH and bH are equal if and only if there exists an element h in H such that a = bh, which means that ab^(-1) is in H.	C
wenxinyiyan	B Explanation: Statement 1: * If $a \in H$, then $aH = H$ since multiplying every element of $H$ by $a$ (which is also in $H$) will still give an element of $H$. So $aH$ is a subgroup of $G$ in this case. * However, the converse is not true. It is possible that $aH$ is a subgroup of $G$ even if $a \notin H$. For example, consider the group $G = \mathbb{Z}/4\mathbb{Z} = \{0, 1, 2, 3\}$ under addition modulo 4. Let $H = \{0, 2\}$ be a subgroup of $G$. Then $1 + H = \{1, 3\}$ is also a subgroup of $G$ (isomorphic to $H$ via the mapping $x \mapsto x + 1$) but $1 \notin H$. Therefore, Statement 1 is false. Statement 2: * If $aH = bH$, then there exists $h_1, h_2 \in H$ such that $ah_1 = bh_2$. Then $a = bh_2h_1^{-1}$ and $b^{-1}a = h_2h_1^{-1} \in H$. However, this does not imply that $ab \in H$. It only tells us about $b^{-1}a$ being in $H$. * For example, consider the same group $G = \mathbb{Z}/4\mathbb{Z}$ and subgroup $H = \{0, 2\}$. Let $a = 1$ and $b = 3$. Then $aH = \{1, 3\} = bH$ but $ab = 1 \times 3 = 3 \notin H$. Therefore, Statement 2 is also false.	C
zhipu	The correct answer is D. False, True. Here's the reasoning: Statement 1: "If H is a subgroup of G and a belongs to G then aH is a subgroup of G if and only if a is in H." This statement is false. For a left coset \( aH \) to be a subgroup of \( G \), it must satisfy the subgroup criteria, which include closure under the group operation and the existence of an identity element within the coset. However, \( aH \) is generally not a subgroup unless \( a \) is in \( H \) and \( H \) is a normal subgroup of \( G \). The statement as given is too restrictive and does not account for the normality condition. Statement 2: "If H is a subgroup of G and a and b belong to G then aH = bH if and only if ab is in H." This statement is true. The condition \( aH = bH \) is equivalent to saying that \( a^{-1}b \in H \). This can be rewritten as \( b = ah \) for some \( h \in H \), implying \( ab = a(ah) = (aa)h \in H \) if \( a^{-1}b \in H \). Conversely, if \( ab \in H \), then \( b = a(a^{-1}b) \in aH \), showing \( bH \subseteq aH \) and similarly \( aH \subseteq bH \), thus \( aH = bH \). So, the correct choice is D. False, True.	C
gpto1	C	C
deepseekv3	The correct answer is: D. False, True ### Explanation: #### Statement 1: "If \( H \) is a subgroup of \( G \) and \( a \) belongs to \( G \), then \( aH \) is a subgroup of \( G \) if and only if \( a \) is in \( H \)." This statement is false. \( aH \) is a subgroup of \( G \) if and only if \( aH = H \), which happens if and only if \( a \in H \). However, \( aH \) is not a subgroup in general unless \( a \) is in \( H \). Therefore, the statement is false because it claims that \( aH \) is always a subgroup under the given condition, which is not the case. #### Statement 2: "If \( H \) is a subgroup of \( G \) and \( a \) and \( b \) belong to \( G \), then \( aH = bH \) if and only if \( ab^{-1} \) is in \( H \)." This statement is true. The cosets \( aH \) and \( bH \) are equal if and only if \( a \) and \( b \) are in the same coset of \( H \), which is equivalent to \( ab^{-1} \in H \). Thus, the correct answer is D. False, True.	C

模型

模型答案

正确答案

金声玉亮2.0

A. True, True.

Reasoning Skill: Consequence Evaluation

This question requires the test-taker to evaluate the consequences of certain conditions in abstract algebra. Statement 1 implies that if aH is a subgroup of G, then a must be in H, and vice versa. Statement 2 implies that if aH = bH, then ab must be in H, and vice versa. The test-taker needs to analyze the conditions and determine the consequences of each statement, which requires Consequence Evaluation reasoning skill.

Note: The question is designed to test the understanding of the properties of subgroups and cosets in abstract algebra, and the ability to evaluate the consequences of certain conditions. The correct answer is A, True, True, as both statements are correct.

deepseekr1

The correct answer is **C. True, False**.

**Statement 1 Analysis**:
If $ aH $ is a subgroup of $ G $, it must contain the identity element $ e $. This implies $ a \in H $ (since $ aH $ can only contain $ e $ if $ a \in H $). Conversely, if $ a \in H $, then $ aH = H $, which is a subgroup. Thus, Statement 1 is **true**.

**Statement 2 Analysis**:
The standard criterion for $ aH = bH $ is $ a^{-1}b \in H $, not $ ab \in H $. A counterexample: Let $ G = S_3 $, $ H = \{e, (12)\} $, $ a = (13) $, and $ b = (132) $. Here, $ ab = e \in H $, but $ aH = \{(13), (123)\} $ and $ bH = \{(132), (23)\} $, which are distinct. Thus, Statement 2 is **false**.

**Answer**: C

gpto1

GPT-4o

D. False, True

llama3.1-70B-Instruct

llama3.1-405B-Instruct

DeepSeek-Coder-V2

Qwen2-72B-Instruct

B. False, False

baichuan4

D. False, True

doubao

B

For Statement 1, aH is a left coset of H in G, not necessarily a subgroup.

For Statement 2, aH = bH if and only if a^{-1}b is in H, not ab.

minimax

B. False, False

Explanation:

Statement 1 is false because aH being a subgroup of G does not imply that a must be in H. It is true that if a is in H, then aH is a subgroup of G, but the converse is not necessarily true. The set aH is a left coset of H in G, and it is a subgroup of G if and only if H is a normal subgroup of G. If H is not normal, then aH is not necessarily a subgroup.

Statement 2 is also false. The correct statement is that aH = bH if and only if ab^(-1) is in H, where b^(-1) is the inverse of b in G. This is because aH = bH implies that for every element h in H, ah is in bH, and thus there exists some h' in H such that ah = bh'. This means that a = bh'b^(-1), and since h' is in H, bh'b^(-1) is in H if and only if b^(-1) is in H, which is true if and only if ab^(-1) is in H.

kimi

D. False, True

Explanation:
Statement 1: If H is a subgroup of G and a belongs to G, then aH is a subgroup of G if and only if a is in H. This statement is false. The set aH is a left coset of H in G, and it is a subgroup of G if and only if aH = H, which happens if and only if a is in H. However, the statement as it is written implies that aH is a subgroup of G for any a in G, which is not true.

Statement 2: If H is a subgroup of G and a and b belong to G, then aH = bH if and only if ab^(-1) is in H. This statement is true. The left cosets aH and bH are equal if and only if there exists an element h in H such that a = bh, which means that ab^(-1) is in H.

wenxinyiyan

B

Explanation:

Statement 1:

* If $a \in H$, then $aH = H$ since multiplying every element of $H$ by $a$ (which is also in $H$) will still give an element of $H$. So $aH$ is a subgroup of $G$ in this case.
* However, the converse is not true. It is possible that $aH$ is a subgroup of $G$ even if $a \notin H$. For example, consider the group $G = \mathbb{Z}/4\mathbb{Z} = \{0, 1, 2, 3\}$ under addition modulo 4. Let $H = \{0, 2\}$ be a subgroup of $G$. Then $1 + H = \{1, 3\}$ is also a subgroup of $G$ (isomorphic to $H$ via the mapping $x \mapsto x + 1$) but $1 \notin H$.

Therefore, Statement 1 is false.

Statement 2:

* If $aH = bH$, then there exists $h_1, h_2 \in H$ such that $ah_1 = bh_2$. Then $a = bh_2h_1^{-1}$ and $b^{-1}a = h_2h_1^{-1} \in H$. However, this does not imply that $ab \in H$. It only tells us about $b^{-1}a$ being in $H$.
* For example, consider the same group $G = \mathbb{Z}/4\mathbb{Z}$ and subgroup $H = \{0, 2\}$. Let $a = 1$ and $b = 3$. Then $aH = \{1, 3\} = bH$ but $ab = 1 \times 3 = 3 \notin H$.

Therefore, Statement 2 is also false.

zhipu

The correct answer is **D. False, True**.

Here's the reasoning:

**Statement 1:**
"If H is a subgroup of G and a belongs to G then aH is a subgroup of G if and only if a is in H."

This statement is **false**. For a left coset $ aH $ to be a subgroup of $ G $, it must satisfy the subgroup criteria, which include closure under the group operation and the existence of an identity element within the coset. However, $ aH $ is generally not a subgroup unless $ a $ is in $ H $ and $ H $ is a normal subgroup of $ G $. The statement as given is too restrictive and does not account for the normality condition.

**Statement 2:**
"If H is a subgroup of G and a and b belong to G then aH = bH if and only if ab is in H."

This statement is **true**. The condition $ aH = bH $ is equivalent to saying that $ a^{-1}b \in H $. This can be rewritten as $ b = ah $ for some $ h \in H $, implying $ ab = a(ah) = (aa)h \in H $ if $ a^{-1}b \in H $. Conversely, if $ ab \in H $, then $ b = a(a^{-1}b) \in aH $, showing $ bH \subseteq aH $ and similarly $ aH \subseteq bH $, thus $ aH = bH $.

So, the correct choice is **D. False, True**.

gpto1

deepseekv3

The correct answer is:

**D. False, True**

### Explanation:

#### Statement 1:
"If $ H $ is a subgroup of $ G $ and $ a $ belongs to $ G $, then $ aH $ is a subgroup of $ G $ if and only if $ a $ is in $ H $."

This statement is **false**. $ aH $ is a subgroup of $ G $ if and only if $ aH = H $, which happens if and only if $ a \in H $. However, $ aH $ is not a subgroup in general unless $ a $ is in $ H $. Therefore, the statement is false because it claims that $ aH $ is always a subgroup under the given condition, which is not the case.

#### Statement 2:
"If $ H $ is a subgroup of $ G $ and $ a $ and $ b $ belong to $ G $, then $ aH = bH $ if and only if $ ab^{-1} $ is in $ H $."

This statement is **true**. The cosets $ aH $ and $ bH $ are equal if and only if $ a $ and $ b $ are in the same coset of $ H $, which is equivalent to $ ab^{-1} \in H $.

Thus, the correct answer is **D. False, True**.