"dataset_name": "abstract_algebra"
"description": "The following are multiple choice questions (with answers) about abstract\
  \ algebra.\n\nQ: Statement 1 | Every element of a group generates a cyclic subgroup\
  \ of the group. Statement 2 | The symmetric group S_10 has 10 elements.\n(A) True,\
  \ True (B) False, False (C) True, False (D) False, True\nA: Let's think step by\
  \ step. A cyclic group is a group that is generated by a single element. Hence a\
  \ subgroup generated by a single element of a group is cyclic and Statement 1 is\
  \ True. The answer is (C).\n\nQ: The symmetric group $S_n$ has $\nactorial{n}$ elements,\
  \ hence it is not true that $S_{10}$ has 10 elements.\nFind the characteristic of\
  \ the ring 2Z.\n(A) 0 (B) 3 (C) 12 (D) 30\nA: Let's think step by step. A characteristic\
  \ of a ring is R is $n$ if the statement $ka = 0$ for all $a\\in 2Z$ implies that\
  \ $k$ is a multiple of $n$. Assume that $ka = 0$ for all $a\\in 2Z$ for some $k$.\
  \ In particular $2k = 0$. Hence $k=0$ and $n=0$. The answer is (A).\n\nQ: Statement\
  \ 1| Every function from a finite set onto itself must be one to one. Statement\
  \ 2 | Every subgroup of an abelian group is abelian.\n(A) True, True (B) False,\
  \ False (C) True, False (D) False, True\nA: Let's think step by step. Statement\
  \ 1 is true. Let $S$ be a finite set. If $f:S \nightarrow S$ is a onto function,\
  \ then $|S| = |f(S)|$. If $f$ was not one to one, then for finite domain $S$ the\
  \ image would have less than $S$ elements, a contradiction.\nStatement 2 is true.\
  \ Let $G$ be an abelian group and $H$ be a subgroup of $G$. We need to show that\
  \ $H$ is abelian. Let $a,b \\in H$. Then $a,b \\in G$ and $ab=ba$. Since $G$ is\
  \ abelian, $ab=ba$. Since $H$ is a subgroup of $G$, $ab \\in H$. Therefore, $ab=ba$\
  \ and $H$ is abelian. The answer is (A).\n\nQ: Statement 1 | If aH is an element\
  \ of a factor group, then |aH| divides |a|. Statement 2 | If H and K are subgroups\
  \ of G then HK is a subgroup of G.\n(A) True, True (B) False, False (C) True, False\
  \ (D) False, True\nA: Let's think step by step. Statement 2 is false. Let $H$ be\
  \ a subgroup of $S_3$ generated by the cycle $(1,2)$ and $K$ be a subgroup of $S_3$\
  \ generated by the cycle $(1,3)$. Both $H$ and $K$ have two elements, the generators\
  \ and the identity. However $HK$ contains cycles (1,2), (1,3) and (2,3,1), but the\
  \ inverse of (2,3,1) is (2,1,3) and it does not belong to HK, hence HK is not a\
  \ subgroup. The answer is (B).\n\nQ: Find all c in Z_3 such that Z_3[x]/(x^2 + c)\
  \ is a field.\n(A) 0 (B) 1 (C) 2 (D) 3\nA: Let's think step by step. Z_3[x]/(x^2\
  \ + c) is a field if and only if x^2 + c does not have roots in Z_3. That is x^2\
  \ + c != 0 for every x in Z_3. If c = 0, then x^2 + c = x^2 has root 0. If c = 1\
  \ then x^2 + c = x^2 + 1 = 0 + 1 for x = 0, 1 + 1 = 2 for x = 1 and 1 + 1 = 2 for\
  \ x = 2, hence x^2 + 1 does not have any roots. For c = 2 the polynomial x^2 + 2\
  \ has two roots at x = 1 and x = 2. Hence Z_3[x]/(x^2 + c) is a field if and only\
  \ if c = 1. The answer is (B).\n\n"
"group": "mmlu_flan_cot_fewshot_stem"
"include": "_mmlu_flan_cot_fewshot_template_yaml"
"task": "mmlu_flan_cot_fewshot_abstract_algebra"
