Datasets:

agatha-duzan
/

advanced_algebra_af

name stringlengths 14 14	formal_statement stringlengths 114 355	informal_statement stringlengths 104 577	tags stringclasses 3 values	header stringclasses 7 values	split stringclasses 1 value
putnam_1985_b6	theorem putnam_1985_b6 (n : ℕ) (npos : n > 0) (G : Finset (Matrix (Fin n) (Fin n) ℝ)) (groupG : (∀ g ∈ G, ∀ h ∈ G, g * h ∈ G) ∧ 1 ∈ G ∧ (∀ g ∈ G, ∃ h ∈ G, g * h = 1)) (hG : ∑ M in G, Matrix.trace M = 0) : (∑ M in G, M = 0) := sorry	Let $G$ be a finite set of real $n\times n$ matrices $\{M_i\}$, $1 \leq i \leq r$, which form a group under matrix multiplication. Suppose that $\sum_{i=1}^r \mathrm{tr}(M_i)=0$, where $\mathrm{tr}(A)$ denotes the trace of the matrix $A$. Prove that $\sum_{i=1}^r M_i$ is the $n \times n$ zero matrix.	['abstract_algebra', 'linear_algebra']		valid
putnam_2009_a5	theorem putnam_2009_a5 : (∃ (G : Type*) (_ : CommGroup G) (_ : Fintype G), ∏ g : G, orderOf g = 2^2009) ↔ putnam_2009_a5_solution := sorry	Is there a finite abelian group $G$ such that the product of the orders of all its elements is 2^{2009}?	['abstract_algebra']	abbrev putnam_2009_a5_solution : Prop := sorry -- False	valid
putnam_1969_b2	theorem putnam_1969_b2 (G : Type*) [Group G] [Finite G] (h : ℕ → Prop := fun n => ∃ H : Fin n → Subgroup G, (∀ i : Fin n, (H i) < ⊤) ∧ ((⊤ : Set G) = ⋃ i : Fin n, (H i))) : ¬(h 2) ∧ ((¬(h 3)) ↔ putnam_1969_b2_solution) := sorry	Show that a finite group can not be the union of two of its proper subgroups. Does the statement remain true if 'two' is replaced by 'three'?	['abstract_algebra']	abbrev putnam_1969_b2_solution : Prop := sorry -- False	valid
putnam_1977_b6	theorem putnam_1977_b6 [Group G] (H : Subgroup G) (h : ℕ := Nat.card H) (a : G) (ha : ∀ x : H, (xa)^3 = 1) (P : Set G := {g : G \| ∃ xs : List H, (xs.length ≥ 1) ∧ g = (List.map (fun h : H => ha) xs).prod}) : (Finite P) ∧ (P.ncard ≤ 3*h^2) := sorry	Let $G$ be a group and $H$ be a subgroup of $G$ with $h$ elements. Suppose that $G$ contains some element $a$ such that $(xa)^3 = 1$ for all $x \in H$ (here $1$ represents the identity element of $G$). Let $P$ be the subset of $G$ containing all products of the form $x_1 a x_2 a \cdots x_n a$ with $n \ge 1$ and $x_i \in H$ for all $i \in \{1, 2, \dots, n\}$. Prove that $P$ is a finite set and contains no more than $3h^2$ elements.	['abstract_algebra']		valid
putnam_2012_a2	theorem putnam_2012_a2 (S : Type) [CommSemigroup S] (a b c : S) (hS : ∀ x y : S, ∃ z : S, x z = y) (habc : a * c = b * c) : a = b := sorry	Let $$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$ and $y$ in $S$, there exists $z$ in $S$ such that $xz=y$. (This $z$ may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and $ac=bc$, then $a=b$.	['abstract_algebra']		valid
putnam_1984_b3	theorem putnam_1984_b3 : (∀ (F : Type*) (_ : Fintype F), Fintype.card F ≥ 2 → (∃ mul : F → F → F, ∀ x y z : F, (mul x z = mul y z → x = y) ∧ (mul x (mul y z) ≠ mul (mul x y) z))) ↔ putnam_1984_b3_solution := sorry	Prove or disprove the following statement: If $F$ is a finite set with two or more elements, then there exists a binary operation $$ on F such that for all $x,y,z$ in $F$, \begin{enumerate} \item[(i)] $xz=yz$ implies $x=y$ (right cancellation holds), and \item[(ii)] $x(yz) \neq (xy)*z$ (\emph{no} case of associativity holds). \end{enumerate}	['abstract_algebra']	abbrev putnam_1984_b3_solution : Prop := sorry -- True	valid
putnam_2019_b6	theorem putnam_2019_b6 (n : ℕ) (neighbors : (Fin n → ℤ) → (Fin n → ℤ) → Prop) (hneighbors : ∀ p q : Fin n → ℤ, neighbors p q = (∃ i : Fin n, abs (p i - q i) = 1 ∧ ∀ j ≠ i, p j = q j)) : (n ≥ 1 ∧ ∃ S : Set (Fin n → ℤ), (∀ p ∈ S, ∀ q : Fin n → ℤ, neighbors p q → q ∉ S) ∧ (∀ p ∉ S, {q ∈ S \| neighbors p q}.encard = 1)) ↔ n ∈ putnam_2019_b6_solution := sorry	Let $ \mathbb{Z}^n $ be the integer lattice in $ \mathbb{R}^n $. Two points in $ \mathbb{Z}^n $ are called neighbors if they differ by exactly 1 in one coordinate and are equal in all other coordinates. For which integers $ n \geq 1 $ does there exist a set of points $ S \subset \mathbb{Z}^n $ satisfying the following two conditions? \begin{enumerate} \item If $ p $ is in $ S $, then none of the neighbors of $ p $ is in $ S $. \item If $ p \in \mathbb{Z}^n $ is not in $ S $, then exactly one of the neighbors of $ p $ is in $ S $. \end{enumerate}	['abstract_algebra']	abbrev putnam_2019_b6_solution : Set ℕ := sorry -- Set.Ici 1	valid
putnam_1979_b3	theorem putnam_1979_b3 (F : Type) [Field F] [Fintype F] (n : ℕ := Fintype.card F) (nodd : Odd n) (b c : F) (p : Polynomial F := X ^ 2 + (C b) X + (C c)) (hp : Irreducible p) : ({d : F \| Irreducible (p + (C d))}.ncard = putnam_1979_b3_solution n) := sorry	Let $F$ be a finite field with $n$ elements, and assume $n$ is odd. Suppose $x^2 + bx + c$ is an irreducible polynomial over $F$. For how many elements $d \in F$ is $x^2 + bx + c + d$ irreducible?	['abstract_algebra']	abbrev putnam_1979_b3_solution : ℕ → ℤ := sorry -- fun n ↦ (n - 1) / 2	valid
putnam_1972_a2	theorem putnam_1972_a2 : (∀ (S : Type) (_ : Mul S), (∀ x y : S, x (x * y) = y ∧ ((y * x) * x) = y) → (∀ x y : S, x * y = y * x)) ∧ ∃ (S : Type) (_ : Mul S), (∀ x y : S, x (x * y) = y ∧ ((y * x) * x) = y) ∧ ¬(∀ x y z : S, x * (y * z) = (x * y) * z) := sorry	Let $S$ be a set and $\cdot$ be a binary operation on $S$ satisfying: (1) for all $x,y$ in $S$, $x \cdot (x \cdot y) = y$ (2) for all $x,y$ in $S$, $(y \cdot x) \cdot x = y$. Show that $\cdot$ is commutative but not necessarily associative.	['abstract_algebra']		valid
putnam_2001_a1	theorem putnam_2001_a1 (S : Type) [Mul S] (hS : ∀ a b : S, (a b) * a = b) : ∀ a b : S, a * (b * a) = b := sorry	Consider a set $S$ and a binary operation $$, i.e., for each $a,b\in S$, $ab\in S$. Assume $(ab)a=b$ for all $a,b\in S$. Prove that $a(ba)=b$ for all $a,b\in S$.	['abstract_algebra']		valid
putnam_2018_a4	theorem putnam_2018_a4 (m n : ℕ) (a : ℕ → ℤ) (G : Type) [Group G] (g h : G) (mnpos : m > 0 ∧ n > 0) (mngcd : Nat.gcd m n = 1) (ha : ∀ k : Set.Icc 1 n, a k = Int.floor (m k / (n : ℝ)) - Int.floor (m * ((k : ℤ) - 1) / (n : ℝ))) (ghprod : ((List.Ico 1 (n + 1)).map (fun k : ℕ => g * h ^ (a k))).prod = 1) : g * h = h * g := sorry	Let $m$ and $n$ be positive integers with $\gcd(m,n)=1$, and let $a_k=\left\lfloor \frac{mk}{n} \right\rfloor - \left\lfloor \frac{m(k-1)}{n} \right\rfloor$ for $k=1,2,\dots,n$. Suppose that $g$ and $h$ are elements in a group $G$ and that $gh^{a_1}gh^{a_2} \cdots gh^{a_n}=e$, where $e$ is the identity element. Show that $gh=hg$. (As usual, $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.)	['abstract_algebra', 'number_theory']		valid
putnam_1990_b4	theorem putnam_1990_b4 : (∀ (G : Type) (_ : Fintype G) (_ : Group G) (n : ℕ) (a b : G), (n = Fintype.card G ∧ a ≠ b ∧ G = Subgroup.closure {a, b}) → (∃ g : ℕ → G, (∀ x : G, {i : Fin (2 n) \| g i = x}.encard = 2) ∧ (∀ i : Fin (2 * n), (g ((i + 1) % (2 * n)) = g i * a) ∨ (g ((i + 1) % (2 * n)) = g i * b))) ↔ putnam_1990_b4_solution) := sorry	Let $G$ be a finite group of order $n$ generated by $a$ and $b$. Prove or disprove: there is a sequence $g_1,g_2,g_3,\dots,g_{2n}$ such that \begin{itemize} \item[(1)] every element of $G$ occurs exactly twice, and \item[(2)] $g_{i+1}$ equals $g_ia$ or $g_ib$ for $i=1,2,\dots,2n$. (Interpret $g_{2n+1}$ as $g_1$.) \end{itemize}	['abstract_algebra']	abbrev putnam_1990_b4_solution : Prop := sorry -- True	valid
putnam_1972_b3	theorem putnam_1972_b3 (G : Type) [Group G] (A B : G) (hab : A B * A = B * A^2 * B ∧ A^3 = 1 ∧ (∃ n : ℤ, n > 0 ∧ B^(2*n - 1) = 1)) : B = 1 := sorry	Let $A$ and $B$ be two elements in a group such that $ABA = BA^2B$, $A^3 = 1$, and $B^{2n-1} = 1$ for some positive integer $n$. Prove that $B = 1$.	['abstract_algebra']		valid

Abstract Algebra Autoformalization Dataset

Dataset Summary

This dataset consists of abstract algebra problems, with informal (natural language) statements and the corresponding formal statements written in Lean 4. It is designed for autoformalization tasks in the domain of abstract algebra. The dataset is a subset of the benchmark Putnam Bench.

Dataset Structure

The final dataset includes 25 abstract algebra problems from PutnamBench with a balanced test/validation split (50/50).

Data Fields

name: A unique identifier for the problem.
formal_statement: The formal statement of the mathematical problem written in Lean 4.
informal_statement: The informal, natural language description of the problem, usually expressed in LaTeX.
tags: A list of tags related to the problem, which typically includes mathematical areas or subfields (e.g., "number theory", "algebra").
header: Contextual metadata necessary for type-checking the formal statements. The header contains information such as packages to import or additional definitions required for the formalization to be complete. Without the header, the formal statement may not "type-check" correctly.
split: Indicates whether the problem belongs to the test or valid set.

Dataset Splits

Split	Number of Problems
Test	12
Valid	13
Total	25

Data Example

Here’s an example of a problem from the dataset:

{
  "name":"putnam_2001_a1",
  "formal_statement":"theorem putnam_2001_a1\n(S : Type*)\n[Mul S]\n(hS : \u2200 a b : S, (a * b) * a = b)\n: \u2200 a b : S, a * (b * a) = b :=\nsorry",
  "informal_statement":"Consider a set $S$ and a binary operation $*$, i.e., for each $a,b\\in S$, $a*b\\in S$.  Assume $(a*b)*a=b$ for all $a,b\\in S$.  Prove that $a*(b*a)=b$ for all $a,b\\in S$.",
  "tags":"['abstract_algebra']",
  "header":"",
  "split":"valid"
}

Usage

The dataset can be used directly in your code as follows:

Loading the Dataset

You can load the dataset using the datasets library from Hugging Face:

from datasets import load_dataset

# Load the entire dataset
dataset = load_dataset('agatha-duzan/advanced_algebra_af')

# Load a specific split (test or validation)
test_set = dataset['test']
valid_set = dataset['valid']

Downloads last month: 8

Size of downloaded dataset files:

21.2 kB

Size of the auto-converted Parquet files:

24.7 kB

Number of rows: