Datasets:

agatha-duzan

/

advanced_algebra_af

like 0

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, (x*a)^3 = 1) (P : Set G := {g : G | ∃ xs : List H, (xs.length ≥ 1) ∧ g = (List.map (fun h : H => h*a) 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 $x*z=y$. (This $z$ may depend on $x$ and $y$.) Show that if $a,b,c$ are in $S$ and $a*c=b*c$, 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)] $x*z=y*z$ implies $x=y$ (right cancellation holds), and \item[(ii)] $x*(y*z) \neq (x*y)*z$ (\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 $a,b\in S$, $a*b\in S$. Assume $(a*b)*a=b$ for all $a,b\in S$. Prove that $a*(b*a)=b$ for all $a,b\in S$.

['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