Improving Autoformalization using Type Checking
Paper • 2406.07222 • Published
problem_name large_stringlengths 17 29 | formal_statement large_stringlengths 54 343 | header large_stringclasses 20
values | informal_statement large_stringlengths 43 377 | informal_solution large_stringlengths 0 3.81k | tags listlengths 1 1 | split large_stringclasses 1
value | name large_stringlengths 17 29 | uuid large_stringlengths 16 16 | goal_state large_stringlengths 18 313 |
|---|---|---|---|---|---|---|---|---|---|
Artin|exercise_2_2_9 | theorem exercise_2_2_9 {G : Type*} [Group G] {a b : G}
(h : a * b = b * a) :
∀ x y : closure {x | x = a ∨ x = b}, x*y = y*x := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Let $H$ be the subgroup generated by two elements $a, b$ of a group $G$. Prove that if $a b=b a$, then $H$ is an abelian group. | \begin{proof}
Since $a$ and $b$ commute, for any $g, h\in H$ we can write $g=a^ib^j$ and $h = a^kb^l$. Then $gh = a^ib^ja^kb^l = a^kb^la^ib^j = hg$. Thus $H$ is abelian.
\end{proof} | [
"Artin"
] | validation | Artin_exercise_2_2_9 | 0ed67d40c9366608 | G : Type u_1
inst✝ : Group G
a b : G
h : a * b = b * a
⊢ ∀ (x y : ↥(Subgroup.closure {x | x = a ∨ x = b})), x * y = y * x |
Artin|exercise_2_4_19 | theorem exercise_2_4_19 {G : Type*} [Group G] {x : G}
(hx : orderOf x = 2) (hx1 : ∀ y, orderOf y = 2 → y = x) :
x ∈ center G := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that if a group contains exactly one element of order 2 , then that element is in the center of the group. | \begin{proof}
Let $x$ be the element of order two. Consider the element $z=y^{-1} x y$, we have: $z^2=\left(y^{-1} x y\right)^2=\left(y^{-1} x y\right)\left(y^{-1} x y\right)=e$. So: $z=x$, and $y^{-1} x y=x$. So: $x y=y x$. So: $x$ is in the center of $G$.
\end{proof} | [
"Artin"
] | validation | Artin_exercise_2_4_19 | 7a4b1375a2dbf338 | G : Type u_1
inst✝ : Group G
x : G
hx : orderOf x = 2
hx1 : ∀ (y : G), orderOf y = 2 → y = x
⊢ x ∈ center G |
Artin|exercise_2_11_3 | theorem exercise_2_11_3 {G : Type*} [Group G] [Fintype G]
(hG : Even (card G)) : ∃ x : G, orderOf x = 2 := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that a group of even order contains an element of order $2 .$ | \begin{proof}
Pair up if possible each element of $G$ with its inverse, and observe that
$$
g^2 \neq e \Longleftrightarrow g \neq g^{-1} \Longleftrightarrow \text { there exists the pair }\left(g, g^{-1}\right)
$$
Now, there is one element that has no pairing: the unit $e$ (since indeed $e=e^{-1} \Longleftrigh... | [
"Artin"
] | validation | Artin_exercise_2_11_3 | 6495bea222ec5c3c | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
hG : Even (card G)
⊢ ∃ x, orderOf x = 2 |
Artin|exercise_3_5_6 | theorem exercise_3_5_6 {K V : Type*} [Field K] [AddCommGroup V]
[Module K V] {S : Set V} (hS : Set.Countable S)
(hS1 : span K S = ⊤) {ι : Type*} (R : ι → V)
(hR : LinearIndependent K R) : Countable ι := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Let $V$ be a vector space which is spanned by a countably infinite set. Prove that every linearly independent subset of $V$ is finite or countably infinite. | \begin{proof}
Let $A$ be the countable generating set, and let $U$ be an uncountable linearly independent set. It can be extended to a basis $B$ of the whole space. Now consider the subset $C$ of elements of $B$ that appear in the $B$-decompositions of elements of $A$.
Since only finitely many elements are involv... | [
"Artin"
] | validation | Artin_exercise_3_5_6 | c12a792567d45f6a | K : Type u_1
V : Type u_2
inst✝² : Field K
inst✝¹ : AddCommGroup V
inst✝ : Module K V
S : Set V
hS : S.Countable
hS1 : Submodule.span K S = ⊤
ι : Type u_3
R : ι → V
hR : LinearIndependent K R
⊢ Countable ι |
Artin|exercise_6_1_14 | theorem exercise_6_1_14 (G : Type*) [Group G]
(hG : IsCyclic $ G ⧸ (center G)) :
center G = ⊤ := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Let $Z$ be the center of a group $G$. Prove that if $G / Z$ is a cyclic group, then $G$ is abelian and hence $G=Z$. | \begin{proof}
We have that $G / Z(G)$ is cyclic, and so there is an element $x \in G$ such that $G / Z(G)=\langle x Z(G)\rangle$, where $x Z(G)$ is the coset with representative $x$. Now let $g \in G$
We know that $g Z(G)=(x Z(G))^m$ for some $m$, and by definition $(x Z(G))^m=x^m Z(G)$.
Now, in general, if $H \... | [
"Artin"
] | validation | Artin_exercise_6_1_14 | 536644866eec384b | G : Type u_1
inst✝ : Group G
hG : IsCyclic (G ⧸ center G)
⊢ center G = ⊤ |
Artin|exercise_6_4_3 | theorem exercise_6_4_3 {G : Type*} [Group G] [Fintype G] {p q : ℕ}
(hp : Prime p) (hq : Prime q) (hG : card G = p^2 *q) :
IsSimpleGroup G → false := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that no group of order $p^2 q$, where $p$ and $q$ are prime, is simple. | \begin{proof}
We may as well assume $p<q$. The number of Sylow $q$-subgroups is $1 \bmod q$ and divides $p^2$. So it is $1, p$, or $p^2$. We win if it's 1 and it can't be $p$, so suppose it's $p^2$. But now $q \mid p^2-1$, so $q \mid p+1$ or $q \mid p-1$.
Thus $p=2$ and $q=3$. But we know no group of order 36 is ... | [
"Artin"
] | validation | Artin_exercise_6_4_3 | f2b4df68ef33291c | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
p q : ℕ
hp : Prime p
hq : Prime q
hG : card G = p ^ 2 * q
⊢ IsSimpleGroup G → false = true |
Artin|exercise_6_8_1 | theorem exercise_6_8_1 {G : Type*} [Group G]
(a b : G) : closure ({a, b} : Set G) = Subgroup.closure {b*a*b^2, b*a*b^3} := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that two elements $a, b$ of a group generate the same subgroup as $b a b^2, b a b^3$. | \begin{proof}
Let $H = \langle bab^2, bab^3\rangle$. It is clear that $H\subset \langle a, b\rangle$. Note that $(bab^2)^{-1}(bab^3)=b$, therefore $b\in H$. This then implies that $b^{-1}(bab^2)b^{-2}=a\in H$. Thus $\langle a, b\rangle\subset H$.
\end{proof} | [
"Artin"
] | validation | Artin_exercise_6_8_1 | d1810fa1d0f5f64f | G : Type u_1
inst✝ : Group G
a b : G
⊢ Subgroup.closure {a, b} = Subgroup.closure {b * a * b ^ 2, b * a * b ^ 3} |
Artin|exercise_10_2_4 | theorem exercise_10_2_4 :
span ({2} : Set $ Polynomial ℤ) ⊓ (span {X}) =
span ({2 * X} : Set $ Polynomial ℤ) := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that in the ring $\mathbb{Z}[x],(2) \cap(x)=(2 x)$. | \begin{proof}
Let $f(x) \in(2 x)$. Then there exists some polynomial $g(x) \in \mathbb{Z}$ such that
$$
f(x)=2 x g(x)
$$
But this means that $f(x) \in(2)$ (because $x g(x)$ is a polynomial), and $f(x) \in$ $(x)$ (because $2 g(x)$ is a polynomial). Thus, $f(x) \in(2) \cap(x)$, and
$$
(2 x) \subseteq(2) \cap(x... | [
"Artin"
] | validation | Artin_exercise_10_2_4 | cc288d1293efbdcb | ⊢ Ideal.span {2} ⊓ Ideal.span {X} = Ideal.span {2 * X} |
Artin|exercise_10_4_6 | theorem exercise_10_4_6 {R : Type*} [CommRing R]
(I J : Ideal R) (x : ↑(I ⊓ J)) :
IsNilpotent ((Ideal.Quotient.mk (I*J)) x) := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Let $I, J$ be ideals in a ring $R$. Prove that the residue of any element of $I \cap J$ in $R / I J$ is nilpotent. | \begin{proof}
If $x$ is in $I \cap J, x \in I$ and $x \in J . R / I J=\{r+a b: a \in I, b \in J, r \in R\}$. Then $x \in I \cap J \Rightarrow x \in I$ and $x \in J$, and so $x^2 \in I J$. Thus
$$
[x]^2=\left[x^2\right]=[0] \text { in } R / I J
$$
\end{proof} | [
"Artin"
] | validation | Artin_exercise_10_4_6 | 08443dd3fdc8035d | R : Type u_1
inst✝ : CommRing R
I J : Ideal R
x : ↥(I ⊓ J)
⊢ IsNilpotent ((Ideal.Quotient.mk (I * J)) ↑x) |
Artin|exercise_10_7_10 | theorem exercise_10_7_10 {R : Type*} [Ring R]
(M : Ideal R) (hM : ∀ (x : R), x ∉ M → IsUnit x)
(hProper : ∃ x : R, x ∉ M) :
IsMaximal M ∧ ∀ (N : Ideal R), IsMaximal N → N = M := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Let $R$ be a ring, with $M$ an ideal of $R$. Suppose that every element of $R$ which is not in $M$ is a unit of $R$. Prove that $M$ is a maximal ideal and that moreover it is the only maximal ideal of $R$. | \begin{proof}
Suppose there is an ideal $M\subset I\subset R$. If $I\neq M$, then $I$ contains a unit, thus $I=R$. Therefore $M$ is a maximal ideal.
Suppose we have an arbitrary maximal ideal $M^\prime$ of $R$. The ideal $M^\prime$ cannot contain a unit, otherwise $M^\prime =R$. Therefore $M^\prime \subset M$. But... | [
"Artin"
] | validation | Artin_exercise_10_7_10 | 8e0a008b9dd99d7a | R : Type u_1
inst✝ : Ring R
M : Ideal R
hM : ∀ x ∉ M, IsUnit x
hProper : ∃ x, x ∉ M
⊢ M.IsMaximal ∧ ∀ (N : Ideal R), N.IsMaximal → N = M |
Artin|exercise_11_4_1b | theorem exercise_11_4_1b :
Irreducible (12 + 6 * X + X ^ 3 : Polynomial ℚ) := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that $x^3 + 6x + 12$ is irreducible in $\mathbb{Q}$. | \begin{proof}
Apply Eisenstein's criterion with $p=3$.
\end{proof} | [
"Artin"
] | validation | Artin_exercise_11_4_1b | 5467acee71ff5948 | ⊢ Irreducible (12 + 6 * X + X ^ 3) |
Artin|exercise_11_4_6b | theorem exercise_11_4_6b {F : Type*} [Field F] [Fintype F] (hF : card F = 7) :
Irreducible (X ^ 2 + 1 : Polynomial F) := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that $x^2+1$ is irreducible in $\mathbb{F}_7$ | \begin{proof}
If $p(x)=x^2+1$ were reducible, its factors must be linear. But no $p(a)$ for $a\in\mathbb{F}_7$ evaluates to 0, therefore $x^2+1$ is irreducible.
\end{proof} | [
"Artin"
] | validation | Artin_exercise_11_4_6b | 6bb05edcee13ad0c | F : Type u_1
inst✝¹ : Field F
inst✝ : Fintype F
hF : card F = 7
⊢ Irreducible (X ^ 2 + 1) |
Artin|exercise_11_4_8 | theorem exercise_11_4_8 (p : ℕ) (hp : Prime p) (n : ℕ) (hn : n > 0) :
Irreducible (X ^ n - (p : Polynomial ℚ) : Polynomial ℚ) := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Let $p$ be a prime integer. Prove that the polynomial $x^n-p$ is irreducible in $\mathbb{Q}[x]$. | \begin{proof}
Straightforward application of Eisenstein's criterion with $p$.
\end{proof} | [
"Artin"
] | validation | Artin_exercise_11_4_8 | bb2d01f40dd6d261 | p : ℕ
hp : Prime p
n : ℕ
hn : n > 0
⊢ Irreducible (X ^ n - ↑p) |
Artin|exercise_13_4_10 | theorem exercise_13_4_10
{p : ℕ} {hp : Nat.Prime p} (h : ∃ r : ℕ, p = 2 ^ r + 1) :
∃ (k : ℕ), p = 2 ^ (2 ^ k) + 1 := by
| import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| Prove that if a prime integer $p$ has the form $2^r+1$, then it actually has the form $2^{2^k}+1$. | \begin{proof}
In particular, we have
$$
\frac{x^a+1}{x+1}=\frac{(-x)^a-1}{(-x)-1}=1-x+x^2-\cdots+(-x)^{a-1}
$$
by the geometric sum formula. In this case, specialize to $x=2^{2^m}$ and we have a nontrivial divisor.
\end{proof} | [
"Artin"
] | validation | Artin_exercise_13_4_10 | 41d5d658c10d1eb3 | p : ℕ
hp : Nat.Prime p
h : ∃ r, p = 2 ^ r + 1
⊢ ∃ k, p = 2 ^ 2 ^ k + 1 |
Axler|exercise_1_3 | theorem exercise_1_3 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] {v : V} : -(-v) = v := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Prove that $-(-v) = v$ for every $v \in V$. | \begin{proof}
By definition, we have
$$
(-v)+(-(-v))=0 \quad \text { and } \quad v+(-v)=0 .
$$
This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$.
\end{proof} | [
"Axler"
] | validation | Axler_exercise_1_3 | c3e856e76beb871b | F : Type u_1
V : Type u_2
inst✝² : AddCommGroup V
inst✝¹ : Field F
inst✝ : Module F V
v : V
⊢ - -v = v |
Axler|exercise_1_6 | theorem exercise_1_6 : ∃ U : Set (ℝ × ℝ),
(U ≠ ∅) ∧
(∀ (u v : ℝ × ℝ), u ∈ U ∧ v ∈ U → u + v ∈ U) ∧
(∀ (u : ℝ × ℝ), u ∈ U → -u ∈ U) ∧
(∀ U' : Submodule ℝ (ℝ × ℝ), U ≠ ↑U') := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Give an example of a nonempty subset $U$ of $\mathbf{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbf{R}^2$. | \begin{proof}
\[U=\mathbb{Z}^2=\left\{(x, y) \in \mathbf{R}^2: x, y \text { are integers }\right\}\]
$U=\mathbb{Z}^2$ satisfies the desired properties. To come up with this, note by assumption, $U$ must be closed under addition and subtraction, so in particular, it must contain 0 . We need to find a set which fai... | [
"Axler"
] | validation | Axler_exercise_1_6 | 18efa72abec50571 | ⊢ ∃ U, U ≠ ∅ ∧ (∀ (u v : ℝ × ℝ), u ∈ U ∧ v ∈ U → u + v ∈ U) ∧ (∀ u ∈ U, -u ∈ U) ∧ ∀ (U' : Submodule ℝ (ℝ × ℝ)), U ≠ ↑U' |
Axler|exercise_1_8 | theorem exercise_1_8 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] {ι : Type*} (u : ι → Submodule F V) :
∃ U : Submodule F V, (⋂ (i : ι), (u i).carrier) = ↑U := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Prove that the intersection of any collection of subspaces of $V$ is a subspace of $V$. | \begin{proof}
Let $V_1, V_2, \ldots, V_n$ be subspaces of the vector space $V$ over the field $F$. We must show that their intersection $V_1 \cap V_2 \cap \ldots \cap V_n$ is also a subspace of $V$.
To begin, we observe that the additive identity $0$ of $V$ is in $V_1 \cap V_2 \cap \ldots \cap V_n$. This is because... | [
"Axler"
] | validation | Axler_exercise_1_8 | 9a8e0fcb73422df3 | F : Type u_1
V : Type u_2
inst✝² : AddCommGroup V
inst✝¹ : Field F
inst✝ : Module F V
ι : Type u_3
u : ι → Submodule F V
⊢ ∃ U, ⋂ i, (u i).carrier = ↑U |
Axler|exercise_3_1 | theorem exercise_3_1 {F V : Type*}
[AddCommGroup V] [Field F] [Module F V] [FiniteDimensional F V]
(T : V →ₗ[F] V) (hT : finrank F V = 1) :
∃ c : F, ∀ v : V, T v = c • v:= by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\operatorname{dim} V=1$ and $T \in \mathcal{L}(V, V)$, then there exists $a \in \mathbf{F}$ such that $T v=a v$ for all $v \in V$. | \begin{proof}
If $\operatorname{dim} V=1$, then in fact, $V=\mathbf{F}$ and it is spanned by $1 \in \mathbf{F}$.
Let $T$ be a linear map from $V$ to itself. Let $T(1)=\lambda \in V(=\mathbf{F})$.
Step 2
2 of 3
Every $v \in V$ is a scalar. Therefore,
$$
\begin{aligned}
T(v) & =T(v \cdot 1) \\
& =v T(1) \ldo... | [
"Axler"
] | validation | Axler_exercise_3_1 | 39a9685a19cae4bc | F : Type u_1
V : Type u_2
inst✝³ : AddCommGroup V
inst✝² : Field F
inst✝¹ : Module F V
inst✝ : FiniteDimensional F V
T : V →ₗ[F] V
hT : finrank F V = 1
⊢ ∃ c, ∀ (v : V), T v = c • v |
Axler|exercise_4_4 | theorem exercise_4_4 (p : Polynomial ℂ) :
p.degree = @card (rootSet p ℂ) (rootSetFintype p ℂ) ↔
Disjoint
(@card (rootSet (derivative p) ℂ) (rootSetFintype (derivative p) ℂ))
(@card (rootSet p ℂ) (rootSetFintype p ℂ)) := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Suppose $p \in \mathcal{P}(\mathbf{C})$ has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p^{\prime}$ have no roots in common. | \begin{proof}
First, let $p$ have $m$ distinct roots. Since $p$ has the degree of $m$, then this could imply that $p$ can be actually written in the form of $p(z)=c\left(z-\lambda_1\right) \ldots\left(z-\lambda_m\right)$, which you have $\lambda_1, \ldots, \lambda_m$ being distinct.
To prove that both $p$ and $p^... | [
"Axler"
] | validation | Axler_exercise_4_4 | e966e0f70bbef197 | p : ℂ[X]
⊢ p.degree = ↑(card ↑(p.rootSet ℂ)) ↔ Disjoint (card ↑((derivative p).rootSet ℂ)) (card ↑(p.rootSet ℂ)) |
Axler|exercise_5_4 | theorem exercise_5_4 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] (S T : V →ₗ[F] V) (hST : S ∘ T = T ∘ S) (c : F):
Submodule.map S (ker (T - c • LinearMap.id)) = ker (T - c • LinearMap.id) := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Suppose that $S, T \in \mathcal{L}(V)$ are such that $S T=T S$. Prove that $\operatorname{null} (T-\lambda I)$ is invariant under $S$ for every $\lambda \in \mathbf{F}$. | \begin{proof}
First off, fix $\lambda \in F$. Secondly, let $v \in \operatorname{null}(T-\lambda I)$. If so, then $(T-\lambda I)(S v)=T S v-\lambda S v=$ $S T v-\lambda S v=S(T v-\lambda v)=0$. Therefore, $S v \in \operatorname{null}(T-\lambda I)$ since $n u l l(T-\lambda I)$ is actually invariant under $S$.
\end... | [
"Axler"
] | validation | Axler_exercise_5_4 | 32c8e114676bf986 | F : Type u_1
V : Type u_2
inst✝² : AddCommGroup V
inst✝¹ : Field F
inst✝ : Module F V
S T : V →ₗ[F] V
hST : ⇑S ∘ ⇑T = ⇑T ∘ ⇑S
c : F
⊢ Submodule.map S (T - c • LinearMap.id).ker = (T - c • LinearMap.id).ker |
Axler|exercise_5_12 | theorem exercise_5_12 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] {S : End F V}
(hS : ∀ v : V, ∃ c : F, v ∈ eigenspace S c) :
∃ c : F, S = c • LinearMap.id := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Suppose $T \in \mathcal{L}(V)$ is such that every vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator. | \begin{proof}
For every single $v \in V$, there does exist $a_v \in F$ such that $T v=a_v v$. Since $T 0=0$, then we have to make $a_0$ be the any number in F. However, for every single $v \in V\{0\}$, then the value of $a_V$ is uniquely determined by the previous equation of $T v=a_v v$.
Now, to show that $T$ ... | [
"Axler"
] | validation | Axler_exercise_5_12 | ad6e5aaa76c11823 | F : Type u_1
V : Type u_2
inst✝² : AddCommGroup V
inst✝¹ : Field F
inst✝ : Module F V
S : End F V
hS : ∀ (v : V), ∃ c, v ∈ S.eigenspace c
⊢ ∃ c, S = c • LinearMap.id |
Axler|exercise_5_20 | theorem exercise_5_20 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] [FiniteDimensional F V] {S T : End F V}
(h1 : card (T.Eigenvalues) = finrank F V)
(h2 : ∀ v : V, (∃ c : F, v ∈ eigenspace S c) ↔ (∃ c : F, v ∈ eigenspace T c)) :
S * T = T * S := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Suppose that $T \in \mathcal{L}(V)$ has $\operatorname{dim} V$ distinct eigenvalues and that $S \in \mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $S T=T S$. | \begin{proof}
First off, let $n=\operatorname{dim} V$. so, there is a basis of $\left(v_1, \ldots, v_j\right)$ of $V$ that consist of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues, then we would have $T v_j=\lambda_1 v_j$ for every single $j$.
Now, for every $v_j$... | [
"Axler"
] | validation | Axler_exercise_5_20 | 405ad5199154fcec | F : Type u_1
V : Type u_2
inst✝³ : AddCommGroup V
inst✝² : Field F
inst✝¹ : Module F V
inst✝ : FiniteDimensional F V
S T : End F V
h1 : card T.Eigenvalues = finrank F V
h2 : ∀ (v : V), (∃ c, v ∈ S.eigenspace c) ↔ ∃ c, v ∈ T.eigenspace c
⊢ S * T = T * S |
Axler|exercise_6_2 | theorem exercise_6_2 {V : Type*} [NormedAddCommGroup V] [NormedField F] [RCLike F]
[Module F V] [InnerProductSpace F V] (u v : V) :
⟪u, v⟫_F = 0 ↔ ∀ (a : F), ‖u‖ ≤ ‖u + a • v‖ := by
| import Mathlib
open InnerProductSpace RCLike ContinuousLinearMap Complex
open scoped BigOperators
| Suppose $u, v \in V$. Prove that $\langle u, v\rangle=0$ if and only if $\|u\| \leq\|u+a v\|$ for all $a \in \mathbf{F}$. | \begin{proof}
First off, let us suppose that $(u, v)=0$.
Now, let $a \in \mathbb{F}$. Next, $u, a v$ are orthogonal.
The Pythagorean theorem thus implies that
$$
\begin{aligned}
\|u+a v\|^2 & =\|u\|^2+\|a v\|^2 \\
& \geq\|u\|^2
\end{aligned}
$$
So, by taking the square roots, this will now give us $\|u\| ... | [
"Axler"
] | validation | Axler_exercise_6_2 | 32490aaf8b79dbe0 | F : Type u_2
V : Type u_1
inst✝⁴ : NormedAddCommGroup V
inst✝³ : NormedField F
inst✝² : RCLike F
inst✝¹ : Module F V
inst✝ : InnerProductSpace F V
u v : V
⊢ ⟪u, v⟫_F = 0 ↔ ∀ (a : F), ‖u‖ ≤ ‖u + a • v‖ |
Axler|exercise_6_7 | theorem exercise_6_7 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] (u v : V) :
⟪u, v⟫_ℂ = (‖u + v‖^2 - ‖u - v‖^2 + I*‖u + I•v‖^2 - I*‖u-I•v‖^2) / 4 := by
| import Mathlib
open InnerProductSpace ContinuousLinearMap Complex
open scoped BigOperators
| Prove that if $V$ is a complex inner-product space, then $\langle u, v\rangle=\frac{\|u+v\|^{2}-\|u-v\|^{2}+\|u+i v\|^{2} i-\|u-i v\|^{2} i}{4}$ for all $u, v \in V$. | \begin{proof}
Let $V$ be an inner-product space and $u, v\in V$. Then
$$
\begin{aligned}
\|u+v\|^2 & =\langle u+v, v+v\rangle \\
& =\|u\|^2+\langle u, v\rangle+\langle v, u\rangle+\|v\|^2 \\
-\|u-v\|^2 & =-\langle u-v, u-v\rangle \\
& =-\|u\|^2+\langle u, v\rangle+\langle v, u\rangle-\|v\|^2 \\
i\|u+i v\|^2 & ... | [
"Axler"
] | validation | Axler_exercise_6_7 | a3936f09601fcba4 | V : Type u_1
inst✝¹ : NormedAddCommGroup V
inst✝ : InnerProductSpace ℂ V
u v : V
⊢ ⟪u, v⟫_ℂ = (↑‖u + v‖ ^ 2 - ↑‖u - v‖ ^ 2 + I * ↑‖u + I • v‖ ^ 2 - I * ↑‖u - I • v‖ ^ 2) / 4 |
Axler|exercise_6_16 | theorem exercise_6_16 {K V : Type*} [RCLike K] [NormedAddCommGroup V] [InnerProductSpace K V]
{U : Submodule K V} :
U.orthogonal = ⊥ ↔ U = ⊤ := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Suppose $U$ is a subspace of $V$. Prove that $U^{\perp}=\{0\}$ if and only if $U=V$ | \begin{proof}
$V=U \bigoplus U^{\perp}$, therefore $U^\perp = \{0\}$ iff $U=V$.
\end{proof} | [
"Axler"
] | validation | Axler_exercise_6_16 | 754b1fb4ef44cbda | K : Type u_1
V : Type u_2
inst✝² : RCLike K
inst✝¹ : NormedAddCommGroup V
inst✝ : InnerProductSpace K V
U : Submodule K V
⊢ Uᗮ = ⊥ ↔ U = ⊤ |
Axler|exercise_7_6 | theorem exercise_7_6 {V : Type*} [NormedAddCommGroup V] [RCLike F] [InnerProductSpace F V]
[FiniteDimensional F V] (T : End F V)
(hT : T * adjoint T = adjoint T * T) :
range T = range (adjoint T) := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Prove that if $T \in \mathcal{L}(V)$ is normal, then $\operatorname{range} T=\operatorname{range} T^{*}.$ | \begin{proof}
Let $T \in \mathcal{L}(V)$ to be a normal operator.
Suppose $u \in \operatorname{null} T$. Then, by $7.20$,
$$
0=\|T u\|=\left\|T^* u\right\|,
$$
which implies that $u \in \operatorname{null} T^*$.
Hence
$$
\operatorname{null} T=\operatorname{null} T^*
$$
because $\left(T^*\right)^*=T$ and ... | [
"Axler"
] | validation | Axler_exercise_7_6 | a75639ff90a8ecc2 | F : Type u_2
V : Type u_1
inst✝³ : NormedAddCommGroup V
inst✝² : RCLike F
inst✝¹ : InnerProductSpace F V
inst✝ : FiniteDimensional F V
T : End F V
hT : T * adjoint T = adjoint T * T
⊢ range T = (adjoint T).range |
Axler|exercise_7_10 | theorem exercise_7_10 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V]
[FiniteDimensional ℂ V] (T : End ℂ V)
(hT : T * adjoint T = adjoint T * T) (hT1 : T^9 = T^8) :
IsSelfAdjoint T ∧ T^2 = T := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Suppose $V$ is a complex inner-product space and $T \in \mathcal{L}(V)$ is a normal operator such that $T^{9}=T^{8}$. Prove that $T$ is self-adjoint and $T^{2}=T$. | \begin{proof}
Based on the complex spectral theorem, there is an orthonormal basis of $\left(e_1, \ldots, e_n\right)$ of $V$ consisting of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues. Therefore,
$$
T e_1=\lambda_j e_j
$$
for $j=1 \ldots n$.
Next, by applying... | [
"Axler"
] | validation | Axler_exercise_7_10 | 2f6f8e46ed1aaf19 | V : Type u_1
inst✝² : NormedAddCommGroup V
inst✝¹ : InnerProductSpace ℂ V
inst✝ : FiniteDimensional ℂ V
T : End ℂ V
hT : T * adjoint T = adjoint T * T
hT1 : T ^ 9 = T ^ 8
⊢ IsSelfAdjoint T ∧ T ^ 2 = T |
Axler|exercise_7_14 | theorem exercise_7_14 {𝕜 V : Type*} [RCLike 𝕜] [NormedAddCommGroup V]
[InnerProductSpace 𝕜 V] [FiniteDimensional 𝕜 V]
{T : Module.End 𝕜 V} (hT : IsSelfAdjoint T)
{l : 𝕜} {ε : ℝ} (he : ε > 0) : (∃ v : V, ‖v‖= 1 ∧ ‖T v - l • v‖ < ε) →
(∃ l' : T.Eigenvalues, ‖l - l'‖ < ε) := by
| import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| Suppose $T \in \mathcal{L}(V)$ is self-adjoint, $\lambda \in \mathbf{F}$, and $\epsilon>0$. Prove that if there exists $v \in V$ such that $\|v\|=1$ and $\|T v-\lambda v\|<\epsilon,$ then $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda-\lambda^{\prime}\right|<\epsilon$. | \begin{proof}
Let $T \in \mathcal{L}(V)$ be a self-adjoint, and let $\lambda \in \mathbf{F}$ and $\epsilon>0$.
By the Spectral Theorem, there is $e_1, \ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvectors of $T$ and let $\lambda_1, \ldots, \lambda_n$ denote their corresponding eigenvalues.
Choose a... | [
"Axler"
] | validation | Axler_exercise_7_14 | d27096e9d90ac787 | 𝕜 : Type u_1
V : Type u_2
inst✝³ : RCLike 𝕜
inst✝² : NormedAddCommGroup V
inst✝¹ : InnerProductSpace 𝕜 V
inst✝ : FiniteDimensional 𝕜 V
T : End 𝕜 V
hT : IsSelfAdjoint T
l : 𝕜
ε : ℝ
he : ε > 0
⊢ (∃ v, ‖v‖ = 1 ∧ ‖T v - l • v‖ < ε) → ∃ l', ‖l - ↑T 1 l'‖ < ε |
Dummit-Foote|exercise_1_1_3 | theorem exercise_1_1_3 (n : ℕ) :
∀ (x y z : ZMod n), (x + y) + z = x + (y + z) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that the addition of residue classes $\mathbb{Z}/n\mathbb{Z}$ is associative. | \begin{proof}
We have
$$
\begin{aligned}
(\bar{a}+\bar{b})+\bar{c} &=\overline{a+b}+\bar{c} \\
&=\overline{(a+b)+c} \\
&=\overline{a+(b+c)} \\
&=\bar{a}+\overline{b+c} \\
&=\bar{a}+(\bar{b}+\bar{c})
\end{aligned}
$$
since integer addition is associative.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_1_3 | bef847fa15eb5f34 | n : ℕ
⊢ ∀ (x y z : ZMod n), x + y + z = x + (y + z) |
Dummit-Foote|exercise_1_1_5 | theorem exercise_1_1_5 (n : ℕ) (hn : 1 < n) :
IsEmpty (Group (ZMod n)) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that for all $n>1$ that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication of residue classes. | \begin{proof}
Note that since $n>1, \overline{1} \neq \overline{0}$. Now suppose $\mathbb{Z} /(n)$ contains a multiplicative identity element $\bar{e}$. Then in particular,
$$
\bar{e} \cdot \overline{1}=\overline{1}
$$
so that $\bar{e}=\overline{1}$. Note, however, that
$$
\overline{0} \cdot \bar{k}=\overlin... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_1_5 | 847530888a7ba5a4 | n : ℕ
hn : 1 < n
⊢ IsEmpty (Group (ZMod n)) |
Dummit-Foote|exercise_1_1_16 | theorem exercise_1_1_16 {G : Type*} [Group G]
(x : G) : x ^ 2 = 1 ↔ (orderOf x = 1 ∨ orderOf x = 2) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $x$ be an element of $G$. Prove that $x^2=1$ if and only if $|x|$ is either $1$ or $2$. | \begin{proof}
$(\Rightarrow)$ Suppose $x^2=1$. Then we have $0<|x| \leq 2$, i.e., $|x|$ is either 1 or 2 .
( $\Leftarrow$ ) If $|x|=1$, then we have $x=1$ so that $x^2=1$. If $|x|=2$ then $x^2=1$ by definition. So if $|x|$ is 1 or 2 , we have $x^2=1$.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_1_16 | 62e0364957022593 | G : Type u_1
inst✝ : Group G
x : G
⊢ x ^ 2 = 1 ↔ orderOf x = 1 ∨ orderOf x = 2 |
Dummit-Foote|exercise_1_1_18 | theorem exercise_1_1_18 {G : Type*} [Group G]
(x y : G) : (x * y = y * x ↔ y⁻¹ * x * y = x) ∧ (y⁻¹ * x * y = x ↔ x⁻¹ * y⁻¹ * x * y = 1) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $x$ and $y$ be elements of $G$. Prove that $xy=yx$ if and only if $y^{-1}xy=x$ if and only if $x^{-1}y^{-1}xy=1$. | \begin{proof}
If $x y=y x$, then $y^{-1} x y=y^{-1} y x=1 x=x$. Multiplying by $x^{-1}$ then gives $x^{-1} y^{-1} x y=1$.
On the other hand, if $x^{-1} y^{-1} x y=1$, then we may multiply on the left by $x$ to get $y^{-1} x y=x$. Then multiplying on the left by $y$ gives $x y=y x$ as desired.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_1_18 | a9ced0e0773eca0c | G : Type u_1
inst✝ : Group G
x y : G
⊢ (x * y = y * x ↔ y⁻¹ * x * y = x) ∧ (y⁻¹ * x * y = x ↔ x⁻¹ * y⁻¹ * x * y = 1) |
Dummit-Foote|exercise_1_1_22a | theorem exercise_1_1_22a {G : Type*} [Group G] (x g : G) :
orderOf x = orderOf (g⁻¹ * x * g) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| If $x$ and $g$ are elements of the group $G$, prove that $|x|=\left|g^{-1} x g\right|$. | \begin{proof}
First we prove a technical lemma:
{\bf Lemma.} For all $a, b \in G$ and $n \in \mathbb{Z},\left(b^{-1} a b\right)^n=b^{-1} a^n b$.
The statement is clear for $n=0$. We prove the case $n>0$ by induction; the base case $n=1$ is clear. Now suppose $\left(b^{-1} a b\right)^n=b^{-1} a^n b$ for som... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_1_22a | 77609d41f4fd2ea7 | G : Type u_1
inst✝ : Group G
x g : G
⊢ orderOf x = orderOf (g⁻¹ * x * g) |
Dummit-Foote|exercise_1_1_25 | theorem exercise_1_1_25 {G : Type*} [Group G]
(h : ∀ x : G, x ^ 2 = 1) : ∀ a b : G, a*b = b*a := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $x^{2}=1$ for all $x \in G$ then $G$ is abelian. | \begin{proof}
Solution: Note that since $x^2=1$ for all $x \in G$, we have $x^{-1}=x$. Now let $a, b \in G$. We have
$$
a b=(a b)^{-1}=b^{-1} a^{-1}=b a .
$$
Thus $G$ is abelian.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_1_25 | 44c199a9bcd76c7d | G : Type u_1
inst✝ : Group G
h : ∀ (x : G), x ^ 2 = 1
⊢ ∀ (a b : G), a * b = b * a |
Dummit-Foote|exercise_1_1_34 | theorem exercise_1_1_34 {G : Type*} [Group G] {x : G}
(hx_inf : orderOf x = 0) (n m : ℤ) (hnm : n ≠ m) :
x ^ n ≠ x ^ m := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| If $x$ is an element of infinite order in $G$, prove that the elements $x^{n}, n \in \mathbb{Z}$ are all distinct. | \begin{proof}
Solution: Suppose to the contrary that $x^a=x^b$ for some $0 \leq a<b \leq n-1$. Then we have $x^{b-a}=1$, with $1 \leq b-a<n$. However, recall that $n$ is by definition the least integer $k$ such that $x^k=1$, so we have a contradiction. Thus all the $x^i$, $0 \leq i \leq n-1$, are distinct. In part... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_1_34 | 8a3fe8a97da9c789 | G : Type u_1
inst✝ : Group G
x : G
hx_inf : orderOf x = 0
n m : ℤ
hnm : n ≠ m
⊢ x ^ n ≠ x ^ m |
Dummit-Foote|exercise_1_6_4 | theorem exercise_1_6_4 :
IsEmpty (Multiplicative ℝ ≃* Multiplicative ℂ) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that the multiplicative groups $\mathbb{R}-\{0\}$ and $\mathbb{C}-\{0\}$ are not isomorphic. | \begin{proof}
Isomorphic groups necessarily have the same number of elements of order $n$ for all finite $n$.
Now let $x \in \mathbb{R}^{\times}$. If $x=1$ then $|x|=1$, and if $x=-1$ then $|x|=2$. If (with bars denoting absolute value) $|x|<1$, then we have
$$
1>|x|>\left|x^2\right|>\cdots,
$$
and in parti... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_6_4 | 00b11808c7db8a12 | ⊢ IsEmpty (Multiplicative ℝ ≃* Multiplicative ℂ) |
Dummit-Foote|exercise_1_6_17 | theorem exercise_1_6_17 {G : Type*} [Group G] (f : G → G)
(hf : f = λ g => g⁻¹) :
(∀ x y : G, f x * f y = f (x*y)) ↔ ∀ x y : G, x*y = y*x := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $G$ be any group. Prove that the map from $G$ to itself defined by $g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian. | \begin{proof}
$(\Rightarrow)$ Suppose $G$ is abelian. Then
$$
\varphi(a b)=(a b)^{-1}=b^{-1} a^{-1}=a^{-1} b^{-1}=\varphi(a) \varphi(b),
$$
so that $\varphi$ is a homomorphism.
$(\Leftarrow)$ Suppose $\varphi$ is a homomorphism, and let $a, b \in G$. Then
$$
a b=\left(b^{-1} a^{-1}\right)^{-1}=\varphi\left(... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_1_6_17 | 03422b27df313d2a | G : Type u_1
inst✝ : Group G
f : G → G
hf : f = fun g => g⁻¹
⊢ (∀ (x y : G), f x * f y = f (x * y)) ↔ ∀ (x y : G), x * y = y * x |
Dummit-Foote|exercise_2_1_5 | theorem exercise_2_1_5 {G : Type*} [Group G] [Fintype G]
(hG : card G > 2) (H : Subgroup G) [Fintype H] :
card H ≠ card G - 1 := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that $G$ cannot have a subgroup $H$ with $|H|=n-1$, where $n=|G|>2$. | \begin{proof}
Solution: Under these conditions, there exists a nonidentity element $x \in H$ and an element $y \notin H$. Consider the product $x y$. If $x y \in H$, then since $x^{-1} \in H$ and $H$ is a subgroup, $y \in H$, a contradiction. If $x y \notin H$, then we have $x y=y$. Thus $x=1$, a contradiction. Th... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_2_1_5 | 2798be37c7f586c9 | G : Type u_1
inst✝² : Group G
inst✝¹ : Fintype G
hG : card G > 2
H : Subgroup G
inst✝ : Fintype ↥H
⊢ card ↥H ≠ card G - 1 |
Dummit-Foote|exercise_2_4_4 | theorem exercise_2_4_4 {G : Type*} [Group G] (H : Subgroup G) :
closure ((H : Set G) \ {1}) = H := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $H$ is a subgroup of $G$ then $H$ is generated by the set $H-\{1\}$. | \begin{proof}
If $H=\{1\}$ then $H-\{1\}$ is the empty set which indeed generates the trivial subgroup $H$. So suppose $|H|>1$ and pick a nonidentity element $h \in H$. Since $1=h h^{-1} \in\langle H-\{1\}\rangle$ (Proposition 9), we see that $H \leq\langle H-\{1\}\rangle$. By minimality of $\langle H-\{1\}\rangle... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_2_4_4 | 0e2034920f39b70c | G : Type u_1
inst✝ : Group G
H : Subgroup G
⊢ Subgroup.closure (↑H \ {1}) = H |
Dummit-Foote|exercise_2_4_16b | theorem exercise_2_4_16b {n : ℕ} {hn : n ≠ 0}
{R : Subgroup (DihedralGroup n)}
(hR : R = Subgroup.closure {DihedralGroup.r 1}) :
R ≠ ⊤ ∧
∀ K : Subgroup (DihedralGroup n), R ≤ K → K = R ∨ K = ⊤ := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Show that the subgroup of all rotations in a dihedral group is a maximal subgroup. | \begin{proof}
Fix a positive integer $n>1$ and let $H \leq D_{2 n}$ consist of the rotations of $D_{2 n}$. That is, $H=\langle r\rangle$. Now, this subgroup is proper since it does not contain $s$. If $H$ is not maximal, then by the previous proof we know there is a maximal subset $K$ containing $H$. Then $K$ must... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_2_4_16b | b581ae0a4466f5c2 | n : ℕ
hn : n ≠ 0
R : Subgroup (DihedralGroup n)
hR : R = Subgroup.closure {DihedralGroup.r 1}
⊢ R ≠ ⊤ ∧ ∀ (K : Subgroup (DihedralGroup n)), R ≤ K → K = R ∨ K = ⊤ |
Dummit-Foote|exercise_3_1_3a | theorem exercise_3_1_3a {A : Type*} [CommGroup A] (B : Subgroup A) :
∀ a b : A ⧸ B, a*b = b*a := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $A$ be an abelian group and let $B$ be a subgroup of $A$. Prove that $A / B$ is abelian. | \begin{proof}
Lemma: Let $G$ be a group. If $|G|=2$, then $G \cong Z_2$.
Proof: Since $G=\{e a\}$ has an identity element, say $e$, we know that $e e=e, e a=a$, and $a e=a$. If $a^2=a$, we have $a=e$, a contradiction. Thus $a^2=e$. We can easily see that $G \cong Z_2$.
If $A$ is abelian, every subgroup of $A$ ... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_3_1_3a | 9a5749899d64a8b3 | A : Type u_1
inst✝ : CommGroup A
B : Subgroup A
⊢ ∀ (a b : A ⧸ B), a * b = b * a |
Dummit-Foote|exercise_3_1_22b | theorem exercise_3_1_22b {G : Type*} [Group G] (I : Type*) [Nonempty I]
(H : I → Subgroup G) (hH : ∀ i : I, Normal (H i)) :
Normal (⨅ (i : I), H i):= by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that the intersection of an arbitrary nonempty collection of normal subgroups of a group is a normal subgroup (do not assume the collection is countable). | \begin{proof}
Let $\left\{H_i \mid i \in I\right\}$ be an arbitrary collection of normal subgroups of $G$ and consider the intersection
$$
\bigcap_{i \in I} H_i
$$
Take an element $a$ in the intersection and an arbitrary element $g \in G$. Then $g a g^{-1} \in H_i$ because $H_i$ is normal for any $i \in H$
By the... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_3_1_22b | c3b5f09a159d0efc | G : Type u_1
inst✝¹ : Group G
I : Type u_2
inst✝ : Nonempty I
H : I → Subgroup G
hH : ∀ (i : I), (H i).Normal
⊢ (⨅ i, H i).Normal |
Dummit-Foote|exercise_3_2_11 | theorem exercise_3_2_11 {G : Type*} [Group G] {H K : Subgroup G}
(hHK : H ≤ K) :
H.index = K.index * H.relindex K := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $H \leq K \leq G$. Prove that $|G: H|=|G: K| \cdot|K: H|$ (do not assume $G$ is finite). | \begin{proof}
Proof. Let $G$ be a group and let $I$ be a nonempty set of indices, not necessarily countable. Consider the collection of subgroups $\left\{N_\alpha \mid \alpha \in I\right\}$, where $N_\alpha \unlhd G$ for each $\alpha \in I$. Let
$$
N=\bigcap_{\alpha \in I} N_\alpha .
$$
We know $N$ is a subgro... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_3_2_11 | 5e753a02eaa816ca | G : Type u_1
inst✝ : Group G
H K : Subgroup G
hHK : H ≤ K
⊢ H.index = K.index * H.relindex K |
Dummit-Foote|exercise_3_2_21a | theorem exercise_3_2_21a (H : AddSubgroup ℚ) (hH : H ≠ ⊤) : H.index = 0 := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that $\mathbb{Q}$ has no proper subgroups of finite index. | \begin{proof}
Solution: We begin with a lemma.
Lemma: If $D$ is a divisible abelian group, then no proper subgroup of $D$ has finite index.
Proof: We saw previously that no finite group is divisible and that every proper quotient $D / A$ of a divisible group is divisible; thus no proper quotient of a divisible g... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_3_2_21a | 56e2041a26c25d3d | H : AddSubgroup ℚ
hH : H ≠ ⊤
⊢ H.index = 0 |
Dummit-Foote|exercise_3_4_1 | theorem exercise_3_4_1 (G : Type*) [CommGroup G] [IsSimpleGroup G] :
IsCyclic G ∧ ∃ G_fin : Fintype G, Nat.Prime (@card G G_fin) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $G$ is an abelian simple group then $G \cong Z_{p}$ for some prime $p$ (do not assume $G$ is a finite group). | \begin{proof}
Solution: Let $G$ be an abelian simple group.
Suppose $G$ is infinite. If $x \in G$ is a nonidentity element of finite order, then $\langle x\rangle<G$ is a nontrivial normal subgroup, hence $G$ is not simple. If $x \in G$ is an element of infinite order, then $\left\langle x^2\right\rangle$ is a no... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_3_4_1 | 55b2f01c42b51d0b | G : Type u_1
inst✝¹ : CommGroup G
inst✝ : IsSimpleGroup G
⊢ IsCyclic G ∧ ∃ G_fin, Nat.Prime (card G) |
Dummit-Foote|exercise_3_4_5a | theorem exercise_3_4_5a {G : Type*} [Group G]
(H : Subgroup G) [IsSolvable G] : IsSolvable H := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that subgroups of a solvable group are solvable. | \begin{proof}
Let $G$ be a solvable group and let $H \leq G$. Since $G$ is solvable, we may find a chain of subgroups
$$
1=G_0 \unlhd G_1 \unlhd G_2 \unlhd \cdots \unlhd G_n=G
$$
so that each quotient $G_{i+1} / G_i$ is abelian. For each $i$, define
$$
H_i=G_i \cap H, \quad 0 \leq i \leq n .
$$
Then $H_i \... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_3_4_5a | 5595003c8270d5bb | G : Type u_1
inst✝¹ : Group G
H : Subgroup G
inst✝ : IsSolvable G
⊢ IsSolvable ↥H |
Dummit-Foote|exercise_3_4_11 | theorem exercise_3_4_11 {G : Type*} [Group G] [IsSolvable G]
{H : Subgroup G} (hH : H ≠ ⊥) [H.Normal] :
∃ A ≤ H, A ≠ ⊥ ∧ A.Normal ∧ ∀ a b : A, a*b = b*a := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $H$ is a nontrivial normal subgroup of the solvable group $G$ then there is a nontrivial subgroup $A$ of $H$ with $A \unlhd G$ and $A$ abelian. | \begin{proof}
Suppose $H$ is a nontrivial normal subgroup of the solvable group $G$.
First, notice that $H$, being a subgroup of a solvable group, is itself solvable. By exercise $8, H$ has a chain of subgroups
$$
1 \leq H_1 \leq \ldots \leq H
$$
such that each $H_i$ is a normal subgroup of $H$ itself and $H_... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_3_4_11 | 427b2b4b506b3513 | G : Type u_1
inst✝² : Group G
inst✝¹ : IsSolvable G
H : Subgroup G
hH : H ≠ ⊥
inst✝ : H.Normal
⊢ ∃ A ≤ H, A ≠ ⊥ ∧ A.Normal ∧ ∀ (a b : ↥A), a * b = b * a |
Dummit-Foote|exercise_4_2_14 | theorem exercise_4_2_14 {G : Type*} [Fintype G] [Group G]
(hG : ¬ (card G).Prime) (hG1 : ∀ k : ℕ, k ∣ card G →
∃ (H : Subgroup G) (fH : Fintype H), @card H fH = k) :
¬ IsSimpleGroup G := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of order $k$ for each positive integer $k$ dividing $n$. Prove that $G$ is not simple. | \begin{proof}
Solution: Let $p$ be the smallest prime dividing $n$, and write $n=p m$. Now $G$ has a subgroup $H$ of order $m$, and $H$ has index $p$. Then $H$ is normal in $G$.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_2_14 | 24d8f3cabfeae477 | G : Type u_1
inst✝¹ : Fintype G
inst✝ : Group G
hG : ¬Nat.Prime (card G)
hG1 : ∀ (k : ℕ), k ∣ card G → ∃ H fH, card ↥H = k
⊢ ¬IsSimpleGroup G |
Dummit-Foote|exercise_4_3_26 | theorem exercise_4_3_26 {α : Type*} [Fintype α] (ha : card α > 1)
(h_tran : ∀ a b: α, ∃ σ : Equiv.Perm α, σ a = b) :
∃ σ : Equiv.Perm α, ∀ a : α, σ a ≠ a := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $G$ be a transitive permutation group on the finite set $A$ with $|A|>1$. Show that there is some $\sigma \in G$ such that $\sigma(a) \neq a$ for all $a \in A$. | \begin{proof}
Let $G$ be a transitive permutation group on the finite set $A,|A|>1$. We want to find an element $\sigma$ which doesn't stabilize anything, that is, we want a $\sigma$ such that
$$
\sigma \notin G_a
$$
for all $a \in A$.
Since the group is transitive, there is always a $g \in G$ such that $b=g ... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_3_26 | 918844c0f3f5315f | α : Type u_1
inst✝ : Fintype α
ha : card α > 1
h_tran : ∀ (a b : α), ∃ σ, σ a = b
⊢ ∃ σ, ∀ (a : α), σ a ≠ a |
Dummit-Foote|exercise_4_4_6a | theorem exercise_4_4_6a {G : Type*} [Group G] (H : Subgroup G)
[Characteristic H] : Normal H := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that characteristic subgroups are normal. | \begin{proof}
Let $H$ be a characterestic subgroup of $G$. By definition $\alpha(H) \subset H$ for every $\alpha \in \operatorname{Aut}(G)$. So, $H$ is in particular invariant under the inner automorphism. Let $\phi_g$ denote the conjugation automorphism by $g$. Then $\phi_g(H) \subset H \Longrightarrow$ $g H g^{-... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_4_6a | 8fc2bfac5cab241b | G : Type u_1
inst✝¹ : Group G
H : Subgroup G
inst✝ : H.Characteristic
⊢ H.Normal |
Dummit-Foote|exercise_4_4_7 | theorem exercise_4_4_7 {G : Type*} [Group G] {H : Subgroup G} [Fintype H]
(hH : ∀ (K : Subgroup G) (fK : Fintype K), card H = @card K fK → H = K) :
H.Characteristic := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| If $H$ is the unique subgroup of a given order in a group $G$ prove $H$ is characteristic in $G$. | \begin{proof}
Let $G$ be group and $H$ be the unique subgroup of order $n$. Now, let $\sigma \in \operatorname{Aut}(G)$. Now Clearly $|\sigma(G)|=n$, because $\sigma$ is a one-one onto map. But then as $H$ is the only subgroup of order $n$, and because of the fact that a automorphism maps subgroups to subgroups, w... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_4_7 | 1f7dd6aa54fd2001 | G : Type u_1
inst✝¹ : Group G
H : Subgroup G
inst✝ : Fintype ↥H
hH : ∀ (K : Subgroup G) (fK : Fintype ↥K), card ↥H = card ↥K → H = K
⊢ H.Characteristic |
Dummit-Foote|exercise_4_5_1a | theorem exercise_4_5_1a {p : ℕ} {G : Type*} [Group G]
{P : Sylow p G} (H : Subgroup G) (hH : P ≤ H) :
IsPGroup p (P.subgroupOf H) ∧
∀ (Q : Subgroup H), IsPGroup p Q → (P.subgroupOf H) ≤ Q → Q = (P.subgroupOf H) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $P \in \operatorname{Syl}_{p}(G)$ and $H$ is a subgroup of $G$ containing $P$ then $P \in \operatorname{Syl}_{p}(H)$. | \begin{proof}
If $P \leq H \leq G$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $[G: P]$. Now $[G: P]=[G: H][H: P]$, so that $p$ does not divide $[H: P]$; hence $P$ is a Sylow $p$-subgroup of $H$.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_5_1a | 3f255f3572046b87 | p : ℕ
G : Type u_1
inst✝ : Group G
P : Sylow p G
H : Subgroup G
hH : ↑P ≤ H
⊢ IsPGroup p ↥((↑P).subgroupOf H) ∧ ∀ (Q : Subgroup ↥H), IsPGroup p ↥Q → (↑P).subgroupOf H ≤ Q → Q = (↑P).subgroupOf H |
Dummit-Foote|exercise_4_5_14 | theorem exercise_4_5_14 {G : Type*} [Group G] [Fintype G]
(hG : card G = 312) :
∃ (p : ℕ) (P : Sylow p G), p.Prime ∧ (p ∣ card G) ∧ P.Normal := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that a group of order 312 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order. | \begin{proof}
Since $|G|=351=3^{2}.13$, $G$ has $3-$Sylow subgroup of order $9$, as well as $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup and $n_{3}$ be the number of $3-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|9$. The choi... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_5_14 | 7b7b856f51ca6459 | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
hG : card G = 312
⊢ ∃ p P, Nat.Prime p ∧ p ∣ card G ∧ (↑P).Normal |
Dummit-Foote|exercise_4_5_16 | theorem exercise_4_5_16 {p q r : ℕ} {G : Type*} [Group G]
[Fintype G] (hpqr : p < q ∧ q < r)
(hpqr1 : p.Prime ∧ q.Prime ∧ r.Prime)(hG : card G = p*q*r) :
(∃ (P : Sylow p G), P.Normal) ∨ (∃ (P : Sylow q G), P.Normal) ∨ (∃ (P : Sylow r G), P.Normal) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $|G|=p q r$, where $p, q$ and $r$ are primes with $p<q<r$. Prove that $G$ has a normal Sylow subgroup for either $p, q$ or $r$. | \begin{proof}
Let $|G|=p q r$. We also assume $p<q<r$. We prove that $G$ has a normal Sylow subgroup of $p$, $q$ or $r$. Now, Let $n_p, n_q, n_r$ be the number of Sylow-p subgroup, Sylow-q subgroup, Sylow-r subgroup resp. So, we have $n_r=1+r k$ such that $1+r k \mid p q$. So, in this case as $r$ is greatest $n_r$... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_5_16 | d5f579a47498e03f | p q r : ℕ
G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
hpqr : p < q ∧ q < r
hpqr1 : Nat.Prime p ∧ Nat.Prime q ∧ Nat.Prime r
hG : card G = p * q * r
⊢ (∃ P, (↑P).Normal) ∨ (∃ P, (↑P).Normal) ∨ ∃ P, (↑P).Normal |
Dummit-Foote|exercise_4_5_18 | theorem exercise_4_5_18 {G : Type*} [Fintype G] [Group G]
(hG : card G = 200) :
∃ N : Sylow 5 G, N.Normal := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that a group of order 200 has a normal Sylow 5-subgroup. | \begin{proof}
Let $G$ be a group of order $200=5^2 \cdot 8$. Note that 5 is a prime not dividing 8 . Let $P \in$ $S y l_5(G)$. [We know $P$ exists since $S y l_5(G) \neq \emptyset$ by Sylow's Theorem]
The number of Sylow 5-subgroups of $G$ is of the form $1+k \cdot 5$, i.e., $n_5 \equiv 1(\bmod 5)$ and $n_5$ di... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_5_18 | 6670f7ff62ac9cc4 | G : Type u_1
inst✝¹ : Fintype G
inst✝ : Group G
hG : card G = 200
⊢ ∃ N, (↑N).Normal |
Dummit-Foote|exercise_4_5_20 | theorem exercise_4_5_20 {G : Type*} [Fintype G] [Group G]
(hG : card G = 1365) : ¬ IsSimpleGroup G := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $|G|=1365$ then $G$ is not simple. | \begin{proof}
Since $|G|=1365=3.5.7.13$, $G$ has $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|3.5.7$. The choices for $k$ is $0$. Hence, there is a unique $13-$Sylow subgroup and hence is normal. so... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_5_20 | 8c99d30b09fe7ae7 | G : Type u_1
inst✝¹ : Fintype G
inst✝ : Group G
hG : card G = 1365
⊢ ¬IsSimpleGroup G |
Dummit-Foote|exercise_4_5_22 | theorem exercise_4_5_22 {G : Type*} [Fintype G] [Group G]
(hG : card G = 132) : ¬ IsSimpleGroup G := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $|G|=132$ then $G$ is not simple. | \begin{proof}
Since $|G|=132=2^{2}.3.11$, $G$ has $2-$Sylow subgroup of order $4$, as well as $11-$Sylow subgroup of order $11$, and $3-$Sylow subgroup of order $3$. Now, we count the number of such subgroups. Let $n_{11}$ be the number of $11-$Sylow subgroup and $n_{3}$ be the number of $3-$Sylow subgroup. Now ... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_5_22 | a693f7ee1a6d48df | G : Type u_1
inst✝¹ : Fintype G
inst✝ : Group G
hG : card G = 132
⊢ ¬IsSimpleGroup G |
Dummit-Foote|exercise_4_5_28 | theorem exercise_4_5_28 {G : Type*} [Group G] [Fintype G]
(hG : card G = 105) (P : Sylow 3 G) [hP : P.Normal] :
∀ a b : G, a*b = b*a := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $G$ be a group of order 105. Prove that if a Sylow 3-subgroup of $G$ is normal then $G$ is abelian. | \begin{proof}
Given that $G$ is a group of order $1575=3^2 .5^2 .7$. Now, Let $n_p$ be the number of Sylow-p subgroups. It is given that Sylow-3 subgroup is normal and hence is unique, so $n_3=1$. First we prove that both Sylow-5 subgroup and Sylow 7-subgroup are normal. Let $P$ be the Sylow3 subgroup. Now, Consid... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_4_5_28 | a59f8c87ee563428 | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
hG : card G = 105
P : Sylow 3 G
hP : (↑P).Normal
⊢ ∀ (a b : G), a * b = b * a |
Dummit-Foote|exercise_5_4_2 | theorem exercise_5_4_2 {G : Type*} [Group G] (H : Subgroup G) :
H.Normal ↔ ⁅(⊤ : Subgroup G), H⁆ ≤ H := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that a subgroup $H$ of $G$ is normal if and only if $[G, H] \leq H$. | \begin{proof}
$H \unlhd G$ is equivalent to $g^{-1} h g \in H, \forall g \in G, \forall h \in H$. We claim that holds if and only if $h^{-1} g^{-1} h g \in H, \forall g \in G, \forall h \in H$, i.e., $\left\{h^{-1} g^{-1} h g: h \in H, g \in G\right\} \subseteq H$. That holds by the following argument:
If $g^{-1}... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_5_4_2 | 8201cb713a09479d | G : Type u_1
inst✝ : Group G
H : Subgroup G
⊢ H.Normal ↔ ⁅⊤, H⁆ ≤ H |
Dummit-Foote|exercise_7_1_11 | theorem exercise_7_1_11 {R : Type*} [CommRing R] [IsDomain R]
{x : R} (hx : x^2 = 1) : x = 1 ∨ x = -1 := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $R$ is an integral domain and $x^{2}=1$ for some $x \in R$ then $x=\pm 1$. | \begin{proof}
Solution: If $x^2=1$, then $x^2-1=0$. Evidently, then,
$$
(x-1)(x+1)=0 .
$$
Since $R$ is an integral domain, we must have $x-1=0$ or $x+1=0$; thus $x=1$ or $x=-1$.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_7_1_11 | 4ae67332d5007d37 | R : Type u_1
inst✝¹ : CommRing R
inst✝ : IsDomain R
x : R
hx : x ^ 2 = 1
⊢ x = 1 ∨ x = -1 |
Dummit-Foote|exercise_7_1_15 | theorem exercise_7_1_15 {R : Type*} [Ring R] (hR : ∀ a : R, a^2 = a) :
∀ a b : R, a*b = b*a := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| A ring $R$ is called a Boolean ring if $a^{2}=a$ for all $a \in R$. Prove that every Boolean ring is commutative. | \begin{proof}
Solution: Note first that for all $a \in R$,
$$
-a=(-a)^2=(-1)^2 a^2=a^2=a .
$$
Now if $a, b \in R$, we have
$$
a+b=(a+b)^2=a^2+a b+b a+b^2=a+a b+b a+b .
$$
Thus $a b+b a=0$, and we have $a b=-b a$. But then $a b=b a$. Thus $R$ is commutative.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_7_1_15 | 963997b5d5514617 | R : Type u_1
inst✝ : Ring R
hR : ∀ (a : R), a ^ 2 = a
⊢ ∀ (a b : R), a * b = b * a |
Dummit-Foote|exercise_7_2_12 | theorem exercise_7_2_12 {R G : Type*} [Ring R] [Group G] [Fintype G] :
∑ g : G, MonoidAlgebra.of R G g ∈ center (MonoidAlgebra R G) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $G=\left\{g_{1}, \ldots, g_{n}\right\}$ be a finite group. Prove that the element $N=g_{1}+g_{2}+\ldots+g_{n}$ is in the center of the group ring $R G$. | \begin{proof}
Let $M=\sum_{i=1}^n r_i g_i$ be an element of $R[G]$. Note that for each $g_i \in G$, the action of $g_i$ on $G$ by conjugation permutes the subscripts. Then we have the following.
$$
\begin{aligned}
N M &=\left(\sum_{i=1}^n g_i\right)\left(\sum_{j=1}^n r_j g_j\right) \\
&=\sum_{j=1}^n \sum_{i=1}... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_7_2_12 | d41b15089cfdff35 | R : Type u_1
G : Type u_2
inst✝² : Ring R
inst✝¹ : Group G
inst✝ : Fintype G
⊢ ∑ g, (MonoidAlgebra.of R G) g ∈ Set.center (MonoidAlgebra R G) |
Dummit-Foote|exercise_7_3_37 | theorem exercise_7_3_37 {p m : ℕ} (hp : p.Prime) :
IsNilpotent (span ({↑p} : Set $ ZMod $ p^m) : Ideal $ ZMod $ p^m) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| An ideal $N$ is called nilpotent if $N^{n}$ is the zero ideal for some $n \geq 1$. Prove that the ideal $p \mathbb{Z} / p^{m} \mathbb{Z}$ is a nilpotent ideal in the ring $\mathbb{Z} / p^{m} \mathbb{Z}$. | \begin{proof}
First we prove a lemma.
Lemma: Let $R$ be a ring, and let $I_1, I_2, J \subseteq R$ be ideals such that $J \subseteq I_1, I_2$. Then $\left(I_1 / J\right)\left(I_2 / J\right)=I_1 I_2 / J$.
Proof: ( $\subseteq$ ) Let
$$
\alpha=\sum\left(x_i+J\right)\left(y_i+J\right) \in\left(I_1 / J\right)\left(I... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_7_3_37 | 02ccb6428fd51b8f | p m : ℕ
hp : Nat.Prime p
⊢ IsNilpotent (span {↑p}) |
Dummit-Foote|exercise_8_1_12 | theorem exercise_8_1_12 {N : ℕ} (hN : N > 0) {M M': ℤ} {d : ℕ}
(hMN : M.gcd N = 1) (hMd : d.gcd N.totient = 1)
(hM' : M' ≡ M^d [ZMOD N]) :
∃ d' : ℕ, d' * d ≡ 1 [ZMOD N.totient] ∧
M ≡ M'^d' [ZMOD N] := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Let $N$ be a positive integer. Let $M$ be an integer relatively prime to $N$ and let $d$ be an integer relatively prime to $\varphi(N)$, where $\varphi$ denotes Euler's $\varphi$-function. Prove that if $M_{1} \equiv M^{d} \pmod N$ then $M \equiv M_{1}^{d^{\prime}} \pmod N$ where $d^{\prime}$ is the inverse of $d \bmod... | \begin{proof}
Note that there is some $k \in \mathbb{Z}$ such that $M^{d d^{\prime}} \equiv M^{k \varphi(N)+1} \equiv\left(M^{\varphi(N)}\right)^k \cdot M \bmod N$. By Euler's Theorem we have $M^{\varphi(N)} \equiv 1 \bmod N$, so that $M_1^{d^{\prime}} \equiv M \bmod N$.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_8_1_12 | 1068802e49266014 | N : ℕ
hN : N > 0
M M' : ℤ
d : ℕ
hMN : M.gcd ↑N = 1
hMd : d.gcd N.totient = 1
hM' : M' ≡ M ^ d [ZMOD ↑N]
⊢ ∃ d', ↑d' * ↑d ≡ 1 [ZMOD ↑N.totient] ∧ M ≡ M' ^ d' [ZMOD ↑N] |
Dummit-Foote|exercise_8_3_4 | theorem exercise_8_3_4 {n : ℤ} {r s : ℚ}
(h : r^2 + s^2 = n) :
∃ a b : ℤ, a^2 + b^2 = n := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if an integer is the sum of two rational squares, then it is the sum of two integer squares. | \begin{proof}
Let $n=\frac{a^2}{b^2}+\frac{c^2}{d^2}$, or, equivalently, $n(b d)^2=a^2 d^2+c^2 b^2$. From this, we see that $n(b d)^2$ can be written as a sum of two squared integers. Therefore, if $q \equiv 3(\bmod 4)$ and $q^i$ appears in the prime power factorization of $n, i$ must be even. Let $j \in \mathbb{N... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_8_3_4 | 283061a83d895218 | n : ℤ
r s : ℚ
h : r ^ 2 + s ^ 2 = ↑n
⊢ ∃ a b, a ^ 2 + b ^ 2 = n |
Dummit-Foote|exercise_8_3_6a | theorem exercise_8_3_6a {R : Type} [Ring R]
(hR : R = (GaussianInt ⧸ span ({⟨1, 1⟩} : Set GaussianInt))) :
IsField R ∧ ∃ finR : Fintype R, @card R finR = 2 := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that the quotient ring $\mathbb{Z}[i] /(1+i)$ is a field of order 2. | \begin{proof}
Let $a+b i \in \mathbb{Z}[i]$. If $a \equiv b \bmod 2$, then $a+b$ and $b-a$ are even and $(1+i)\left(\frac{a+b}{2}+\frac{b-a}{2} i\right)=a+b i \in\langle 1+i\rangle$. If $a \not \equiv b \bmod 2$ then $a-1+b i \in\langle 1+i\rangle$. Therefore every element of $\mathbb{Z}[i]$ is in either $\langle ... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_8_3_6a | 57f0c2e05c1ed205 | R : Type
inst✝ : Ring R
hR : R = (GaussianInt ⧸ span {{ re := 1, im := 1 }})
⊢ IsField R ∧ ∃ finR, card R = 2 |
Dummit-Foote|exercise_9_1_6 | theorem exercise_9_1_6 : ¬ Submodule.IsPrincipal
(span ({MvPolynomial.X 0, MvPolynomial.X 1} : Set (MvPolynomial (Fin 2) ℚ))) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that $(x, y)$ is not a principal ideal in $\mathbb{Q}[x, y]$. | \begin{proof}
Suppose, to the contrary, that $(x, y)=p$ for some polynomial $p \in \mathbb{Q}[x, y]$. From $x, y \in$ $(x, y)=(p)$ there are $s, t \in \mathbb{Q}[x, y]$ such that $x=s p$ and $y=t p$.
Then:
$$
\begin{aligned}
& 0=\operatorname{deg}_y(x)=\operatorname{deg}_y(s)+\operatorname{deg}_y(p) \text { so... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_9_1_6 | 24cda8e3dc952444 | ⊢ ¬Submodule.IsPrincipal (span {MvPolynomial.X 0, MvPolynomial.X 1}) |
Dummit-Foote|exercise_9_3_2 | theorem exercise_9_3_2 {f g : Polynomial ℚ} (i j : ℕ)
(hfg : ∀ n : ℕ, ∃ a : ℤ, (f*g).coeff = a) :
∃ a : ℤ, f.coeff i * g.coeff j = a := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that if $f(x)$ and $g(x)$ are polynomials with rational coefficients whose product $f(x) g(x)$ has integer coefficients, then the product of any coefficient of $g(x)$ with any coefficient of $f(x)$ is an integer. | \begin{proof}
Let $f(x), g(x) \in \mathbb{Q}[x]$ be such that $f(x) g(x) \in \mathbb{Z}[x]$.
By Gauss' Lemma there exists $r, s \in \mathbb{Q}$ such that $r f(x), s g(x) \in \mathbb{Z}[x]$, and $(r f(x))(s g(x))=r s f(x) g(x)=f(x) g(x)$. From this last relation we can conclude that $s=r^{-1}$.
Therefore for an... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_9_3_2 | 939c67ff3c41e640 | f g : ℚ[X]
i j : ℕ
hfg : ∀ (n : ℕ), ∃ a, (f * g).coeff = ↑a
⊢ ∃ a, f.coeff i * g.coeff j = ↑a |
Dummit-Foote|exercise_9_4_2b | theorem exercise_9_4_2b : Irreducible
(X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : Polynomial ℤ) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that $x^6+30x^5-15x^3 + 6x-120$ is irreducible in $\mathbb{Z}[x]$. | \begin{proof}
$$
x^6+30 x^5-15 x^3+6 x-120
$$
The coefficients of the low order.: $30,-15,0,6,-120$
They are divisible by the prime 3 , but $3^2=9$ doesn 't divide $-120$. So this polynomial is irreducible over $\mathbb{Z}$.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_9_4_2b | 897da198689f2500 | ⊢ Irreducible (X ^ 6 + 30 * X ^ 5 - 15 * X ^ 3 + 6 * X - 120) |
Dummit-Foote|exercise_9_4_2d | theorem exercise_9_4_2d {p : ℕ} (hp : p.Prime ∧ p > 2)
{f : Polynomial ℤ} (hf : f = (X + 2)^p):
Irreducible (∑ n ∈ (f.support \ {0}), (f.coeff n : Polynomial ℤ) * X ^ (n-1) :
Polynomial ℤ) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that $\frac{(x+2)^p-2^p}{x}$, where $p$ is an odd prime, is irreducible in $\mathbb{Z}[x]$. | \begin{proof}
$\frac{(x+2)^p-2^p}{x} \quad \quad p$ is on add pprime $Z[x]$
$$
\frac{(x+2)^p-2^p}{x} \quad \text { as a polynomial we expand }(x+2)^p
$$
$2^p$ cancels with $-2^p$, every remaining term has $x$ as $a$ factor
$$
\begin{aligned}
& x^{p-1}+2\left(\begin{array}{l}
p \\
1
\end{array}\right) x^{p-2}... | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_9_4_2d | 7a1fde582b493e23 | p : ℕ
hp : Nat.Prime p ∧ p > 2
f : ℤ[X]
hf : f = (X + 2) ^ p
⊢ Irreducible (∑ n ∈ f.support \ {0}, ↑(f.coeff n) * X ^ (n - 1)) |
Dummit-Foote|exercise_9_4_11 | theorem exercise_9_4_11 :
Irreducible ((MvPolynomial.X 0)^2 + (MvPolynomial.X 1)^2 - 1 : MvPolynomial (Fin 2) ℚ) := by
| import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| Prove that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. | \begin{proof}
$$
p(x)=x^2+y^2-1 \in Q[y][x] \cong Q[y, x]
$$
We have that $y+1 \in Q[y]$ is prime and $Q[y]$ is an UFD, since $p(x)=x^2+y^2-1=x^2+$ $(y+1)(y-1)$ by the Eisenstein criterion $x^2+y^2-1$ is irreducibile in $Q[x, y]$.
\end{proof} | [
"Dummit-Foote"
] | validation | Dummit-Foote_exercise_9_4_11 | 0daa0b37413562ca | ⊢ Irreducible (MvPolynomial.X 0 ^ 2 + MvPolynomial.X 1 ^ 2 - 1) |
Herstein|exercise_2_1_21 | theorem exercise_2_1_21 (G : Type*) [Group G] [Fintype G]
(hG : card G = 5) :
∀ a b : G, a*b = b*a := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Show that a group of order 5 must be abelian. | \begin{proof}
Suppose $G$ is a group of order 5 which is not abelian. Then there exist two non-identity elements $a, b \in G$ such that $a * b \neq$ $b * a$. Further we see that $G$ must equal $\{e, a, b, a * b, b * a\}$. To see why $a * b$ must be distinct from all the others, not that if $a *$ $b=e$, then $a$ an... | [
"Herstein"
] | validation | Herstein_exercise_2_1_21 | e5307cf9afb05274 | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
hG : card G = 5
⊢ ∀ (a b : G), a * b = b * a |
Herstein|exercise_2_1_27 | theorem exercise_2_1_27 {G : Type*} [Group G]
[Fintype G] : ∃ (m : ℕ), m > 0 ∧ ∀ (a : G), a ^ m = 1 := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $G$ is a finite group, prove that there is an integer $m > 0$ such that $a^m = e$ for all $a \in G$. | \begin{proof}
Let $n_1, n_2, \ldots, n_k$ be the orders of all $k$ elements of $G=$ $\left\{a_1, a_2, \ldots, a_k\right\}$. Let $m=\operatorname{lcm}\left(n_1, n_2, \ldots, n_k\right)$. Then, for any $i=$ $1, \ldots, k$, there exists an integer $c$ such that $m=n_i c$. Thus
$$
a_i^m=a_i^{n_i c}=\left(a_i^{n_i}\r... | [
"Herstein"
] | validation | Herstein_exercise_2_1_27 | 175170ea1cda9c36 | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
⊢ ∃ m > 0, ∀ (a : G), a ^ m = 1 |
Herstein|exercise_2_2_5 | theorem exercise_2_2_5 {G : Type*} [Group G]
(h : ∀ (a b : G), (a * b) ^ 3 = a ^ 3 * b ^ 3 ∧ (a * b) ^ 5 = a ^ 5 * b ^ 5) :
∀ a b : G, a*b = b*a := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Let $G$ be a group in which $(a b)^{3}=a^{3} b^{3}$ and $(a b)^{5}=a^{5} b^{5}$ for all $a, b \in G$. Show that $G$ is abelian. | \begin{proof}
We have
$$
\begin{aligned}
& (a b)^3=a^3 b^3, \text { for all } a, b \in G \\
\Longrightarrow & (a b)(a b)(a b)=a\left(a^2 b^2\right) b \\
\Longrightarrow & a(b a)(b a) b=a\left(a^2 b^2\right) b \\
\Longrightarrow & (b a)^2=a^2 b^2, \text { by cancellation law. }
\end{aligned}
$$
Again,
$$
\b... | [
"Herstein"
] | validation | Herstein_exercise_2_2_5 | e0298aec60eb43fe | G : Type u_1
inst✝ : Group G
h : ∀ (a b : G), (a * b) ^ 3 = a ^ 3 * b ^ 3 ∧ (a * b) ^ 5 = a ^ 5 * b ^ 5
⊢ ∀ (a b : G), a * b = b * a |
Herstein|exercise_2_3_17 | theorem exercise_2_3_17 {G : Type*} [Mul G] [Group G] (a x : G) :
centralizer {x⁻¹*a*x} =
(λ g : G => x⁻¹*g*x) '' (centralizer {a}) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $G$ is a group and $a, x \in G$, prove that $C\left(x^{-1} a x\right)=x^{-1} C(a) x$ | \begin{proof}
Note that
$$
C(a):=\{x \in G \mid x a=a x\} .
$$
Let us assume $p \in C\left(x^{-1} a x\right)$. Then,
$$
\begin{aligned}
& p\left(x^{-1} a x\right)=\left(x^{-1} a x\right) p \\
\Longrightarrow & \left(p x^{-1} a\right) x=x^{-1}(a x p) \\
\Longrightarrow & x\left(p x^{-1} a\right)=(a x p) x^... | [
"Herstein"
] | validation | Herstein_exercise_2_3_17 | 0f97dc154a4e33ea | G : Type u_1
inst✝¹ : Mul G
inst✝ : Group G
a x : G
⊢ {x⁻¹ * a * x}.centralizer = (fun g => x⁻¹ * g * x) '' {a}.centralizer |
Herstein|exercise_2_4_36 | theorem exercise_2_4_36 {a n : ℕ} (h : a > 1) :
n ∣ (a ^ n - 1).totient := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $a > 1$ is an integer, show that $n \mid \varphi(a^n - 1)$, where $\phi$ is the Euler $\varphi$-function. | \begin{proof}
Proof: We have $a>1$. First we propose to prove that
$$
\operatorname{Gcd}\left(a, a^n-1\right)=1 .
$$
If possible, let us assume that
$\operatorname{Gcd}\left(a, a^n-1\right)=d$, where $d>1$.
Then
$d$ divides $a$ as well as $a^n-1$.
Now,
$d$ divides $a \Longrightarrow d$ divides $a^n$.
Thi... | [
"Herstein"
] | validation | Herstein_exercise_2_4_36 | 85d45b45a2423098 | a n : ℕ
h : a > 1
⊢ n ∣ (a ^ n - 1).totient |
Herstein|exercise_2_5_30 | theorem exercise_2_5_30 {G : Type*} [Group G] [Fintype G]
{p m : ℕ} (hp : Nat.Prime p) (hp1 : ¬ p ∣ m) (hG : card G = p*m)
{H : Subgroup G} [Fintype H] [H.Normal] (hH : card H = p):
Subgroup.Characteristic H := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Suppose that $|G| = pm$, where $p \nmid m$ and $p$ is a prime. If $H$ is a normal subgroup of order $p$ in $G$, prove that $H$ is characteristic. | \begin{proof}
Let $G$ be a group of order $p m$, such that $p \nmid m$. Now, Given that $H$ is a normal subgroup of order $p$. Now we want to prove that $H$ is a characterestic subgroup, that is $\phi(H)=H$ for any automorphism $\phi$ of $G$. Now consider $\phi(H)$. Clearly $|\phi(H)|=p$. Suppose $\phi(H) \neq H$,... | [
"Herstein"
] | validation | Herstein_exercise_2_5_30 | 18c2cf06d6f7af2c | G : Type u_1
inst✝³ : Group G
inst✝² : Fintype G
p m : ℕ
hp : Nat.Prime p
hp1 : ¬p ∣ m
hG : card G = p * m
H : Subgroup G
inst✝¹ : Fintype ↥H
inst✝ : H.Normal
hH : card ↥H = p
⊢ H.Characteristic |
Herstein|exercise_2_5_37 | theorem exercise_2_5_37 (G : Type*) [Group G] [Fintype G]
(hG : card G = 6) (hG' : IsEmpty (CommGroup G)) :
Nonempty (G ≃* Equiv.Perm (Fin 3)) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $G$ is a nonabelian group of order 6, prove that $G \simeq S_3$. | \begin{proof}
Suppose $G$ is a non-abelian group of order 6 . We need to prove that $G \cong S_3$. Since $G$ is non-abelian, we conclude that there is no element of order 6. Now all the nonidentity element has order either 2 or 3 . All elements cannot be order 3 .This is because except the identity elements there ... | [
"Herstein"
] | validation | Herstein_exercise_2_5_37 | c208c2fb232fd987 | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
hG : card G = 6
hG' : IsEmpty (CommGroup G)
⊢ Nonempty (G ≃* Equiv.Perm (Fin 3)) |
Herstein|exercise_2_5_44 | theorem exercise_2_5_44 {G : Type*} [Group G] [Fintype G] {p : ℕ}
(hp : Nat.Prime p) (hG : card G = p^2) :
∃ (N : Subgroup G) (Fin : Fintype N), @card N Fin = p ∧ N.Normal := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Prove that a group of order $p^2$, $p$ a prime, has a normal subgroup of order $p$. | \begin{proof}
We use the result from problem 40 which is as follows: Suppose $G$ is a group, $H$ is a subgroup and $|G|=n$ and $n \nmid\left(i_G(H)\right) !$. Then there exists a normal subgroup $K \neq \{ e \}$ and $K \subseteq H$.
So, we have now a group $G$ of order $p^2$. Suppose that the group is cyclic, t... | [
"Herstein"
] | validation | Herstein_exercise_2_5_44 | d21b18d453a7f537 | G : Type u_1
inst✝¹ : Group G
inst✝ : Fintype G
p : ℕ
hp : Nat.Prime p
hG : card G = p ^ 2
⊢ ∃ N Fin, card ↥N = p ∧ N.Normal |
Herstein|exercise_2_6_15 | theorem exercise_2_6_15 {G : Type*} [CommGroup G] {m n : ℕ}
(hm : ∃ (g : G), orderOf g = m)
(hn : ∃ (g : G), orderOf g = n)
(hmn : m.Coprime n) :
∃ (g : G), orderOf g = m * n := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $G$ is an abelian group and if $G$ has an element of order $m$ and one of order $n$, where $m$ and $n$ are relatively prime, prove that $G$ has an element of order $mn$. | \begin{proof}
Let $G$ be an abelian group, and let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively, where $m$ and $n$ are relatively prime. We will show that the product $ab$ has order $mn$ in $G$, which will prove that $G$ has an element of order $mn$.
To show that $ab$ has order $mn$, let $k$ be... | [
"Herstein"
] | validation | Herstein_exercise_2_6_15 | 3eea46d2b4bc4fd1 | G : Type u_1
inst✝ : CommGroup G
m n : ℕ
hm : ∃ g, orderOf g = m
hn : ∃ g, orderOf g = n
hmn : m.Coprime n
⊢ ∃ g, orderOf g = m * n |
Herstein|exercise_2_8_12 | theorem exercise_2_8_12 {G H : Type*} [Fintype G] [Fintype H]
[Group G] [Group H] (hG : card G = 21) (hH : card H = 21)
(hG1 : IsEmpty (CommGroup G)) (hH1 : IsEmpty (CommGroup H)) :
Nonempty (G ≃* H) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Prove that any two nonabelian groups of order 21 are isomorphic. | \begin{proof}
By Cauchy's theorem we have that if $G$ is a group of order 21 then it has an element $a$ of order 3 and an element $b$ of order 7. By exercise 2.5.41 we have that the subgroup generated by $b$ is normal, so there is some $i=0,1,2,3,4,5,6$ such that $a b a^{-1}=b^i$. We know $i \neq$ 0 since that imp... | [
"Herstein"
] | validation | Herstein_exercise_2_8_12 | 66b57ec42231099d | G : Type u_1
H : Type u_2
inst✝³ : Fintype G
inst✝² : Fintype H
inst✝¹ : Group G
inst✝ : Group H
hG : card G = 21
hH : card H = 21
hG1 : IsEmpty (CommGroup G)
hH1 : IsEmpty (CommGroup H)
⊢ Nonempty (G ≃* H) |
Herstein|exercise_2_9_2 | theorem exercise_2_9_2 {G H : Type*} [Fintype G] [Fintype H] [Group G]
[Group H] (hG : IsCyclic G) (hH : IsCyclic H) :
IsCyclic (G × H) ↔ (card G).Coprime (card H) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $G_1$ and $G_2$ are cyclic groups of orders $m$ and $n$, respectively, prove that $G_1 \times G_2$ is cyclic if and only if $m$ and $n$ are relatively prime. | \begin{proof}
The order of $G \times H$ is $n$. $m$. Thus, $G \times H$ is cyclic iff it has an element with order n. $m$. Suppose $\operatorname{gcd}(n . m)=1$. This implies that $g^m$ has order $n$, and analogously $h^n$ has order $m$. That is, $g \times h$ has order $n$. $m$, and therefore $G \times H$ is cycli... | [
"Herstein"
] | validation | Herstein_exercise_2_9_2 | d6094afd0dec1d48 | G : Type u_1
H : Type u_2
inst✝³ : Fintype G
inst✝² : Fintype H
inst✝¹ : Group G
inst✝ : Group H
hG : IsCyclic G
hH : IsCyclic H
⊢ IsCyclic (G × H) ↔ (card G).Coprime (card H) |
Herstein|exercise_2_11_6 | theorem exercise_2_11_6 {G : Type*} [Group G] {p : ℕ} (hp : Nat.Prime p)
{P : Sylow p G} (hP : P.Normal) :
∀ (Q : Sylow p G), P = Q := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $P$ is a $p$-Sylow subgroup of $G$ and $P \triangleleft G$, prove that $P$ is the only $p$-Sylow subgroup of $G$. | \begin{proof}
let $G$ be a group and $P$ a sylow-p subgroup. Given $P$ is normal. By sylow second theorem the sylow-p subgroups are conjugate. Let $K$ be any other sylow-p subgroup. Then there exists $g \in G$ such that $K=g P g^{-1}$. But since $P$ is normal $K=g P g^{-1}=P$. Hence the sylow-p subgroup is unique.... | [
"Herstein"
] | validation | Herstein_exercise_2_11_6 | 21f020ad9b24e0eb | G : Type u_1
inst✝ : Group G
p : ℕ
hp : Nat.Prime p
P : Sylow p G
hP : (↑P).Normal
⊢ ∀ (Q : Sylow p G), P = Q |
Herstein|exercise_2_11_22 | theorem exercise_2_11_22 {p : ℕ} {n : ℕ} {G : Type*} [Fintype G]
[Group G] (hp : Nat.Prime p) (hG : card G = p ^ n) {K : Subgroup G}
[Fintype K] (hK : card K = p ^ (n-1)) :
K.Normal := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Show that any subgroup of order $p^{n-1}$ in a group $G$ of order $p^n$ is normal in $G$. | \begin{proof}
Proof: First we prove the following lemma.
\textbf{Lemma:} If $G$ is a finite $p$-group with $|G|>1$, then $Z(G)$, the center of $G$, has more than one element; that is, if $|G|=p^k$ with $k\geq 1$, then $|Z(G)|>1$.
\textit{Proof of the lemma:} Consider the class equation
$$
|G|=|Z(G)|+\sum_{a \n... | [
"Herstein"
] | validation | Herstein_exercise_2_11_22 | 7cb27bc272dfa642 | p n : ℕ
G : Type u_1
inst✝² : Fintype G
inst✝¹ : Group G
hp : Nat.Prime p
hG : card G = p ^ n
K : Subgroup G
inst✝ : Fintype ↥K
hK : card ↥K = p ^ (n - 1)
⊢ K.Normal |
Herstein|exercise_4_1_19 | theorem exercise_4_1_19 : Infinite {x : Quaternion ℝ | x^2 = -1} := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Show that there is an infinite number of solutions to $x^2 = -1$ in the quaternions. | \begin{proof}
Let $x=a i+b j+c k$ then
$$
x^2=(a i+b j+c k)(a i+b j+c k)=-a^2-b^2-c^2=-1
$$
This gives $a^2+b^2+c^2=1$ which has infinitely many solutions for $-1<a, b, c<1$.
\end{proof} | [
"Herstein"
] | validation | Herstein_exercise_4_1_19 | 466d1798e2e8b7ac | ⊢ Infinite ↑{x | x ^ 2 = -1} |
Herstein|exercise_4_2_5 | theorem exercise_4_2_5 {R : Type*} [Ring R]
(h : ∀ x : R, x ^ 3 = x) : Nonempty (CommRing R) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Let $R$ be a ring in which $x^3 = x$ for every $x \in R$. Prove that $R$ is commutative. | \begin{proof}
To begin with
$$
2 x=(2 x)^3=8 x^3=8 x .
$$
Therefore $6 x=0 \quad \forall x$.
Also
$$
(x+y)=(x+y)^3=x^3+x^2 y+x y x+y x^2+x y^2+y x y+y^2 x+y^3
$$
and
$$
(x-y)=(x-y)^3=x^3-x^2 y-x y x-y x^2+x y^2+y x y+y^2 x-y^3
$$
Subtracting we get
$$
2\left(x^2 y+x y x+y x^2\right)=0
$$
Multiply the ... | [
"Herstein"
] | validation | Herstein_exercise_4_2_5 | e433e29d27f8949c | R : Type u_1
inst✝ : Ring R
h : ∀ (x : R), x ^ 3 = x
⊢ Nonempty (CommRing R) |
Herstein|exercise_4_2_9 | theorem exercise_4_2_9 {p : ℕ} (hp : Nat.Prime p) (hp1 : Odd p) :
∀ (a b : ℤ), (a / b : ℚ) = ∑ i ∈ Finset.range (p-1), (1 / (i + 1) : ℚ) → ↑p ∣ a := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Let $p$ be an odd prime and let $1 + \frac{1}{2} + ... + \frac{1}{p - 1} = \frac{a}{b}$, where $a, b$ are integers. Show that $p \mid a$. | \begin{proof}
First we prove for prime $p=3$ and then for all prime $p>3$.
Let us take $p=3$. Then the sum
$$
\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{(p-1)}
$$
becomes
$$
1+\frac{1}{3-1}=1+\frac{1}{2}=\frac{3}{2} .
$$
Therefore in this case $\quad \frac{a}{b}=\frac{3}{2} \quad$ implies $3 \mid a$, i.e. $p... | [
"Herstein"
] | validation | Herstein_exercise_4_2_9 | bb511fde41f3e5e9 | p : ℕ
hp : Nat.Prime p
hp1 : Odd p
⊢ ∀ (a b : ℤ), ↑a / ↑b = ∑ i ∈ Finset.range (p - 1), 1 / (↑i + 1) → ↑p ∣ a |
Herstein|exercise_4_3_25 | theorem exercise_4_3_25 (I : Ideal (Matrix (Fin 2) (Fin 2) ℝ)) :
I = ⊥ ∨ I = ⊤ := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Let $R$ be the ring of $2 \times 2$ matrices over the real numbers; suppose that $I$ is an ideal of $R$. Show that $I = (0)$ or $I = R$. | \begin{proof}
Suppose that $I$ is a nontrivial ideal of $R$, and let
$$
A=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
$$
where not all of $a, b, c d$ are zero. Suppose, without loss of generality -- our steps would be completely analogous, modulo some different placement of 1 s in our matrices... | [
"Herstein"
] | validation | Herstein_exercise_4_3_25 | 28cf69ea8b3ab47a | I : Ideal (Matrix (Fin 2) (Fin 2) ℝ)
⊢ I = ⊥ ∨ I = ⊤ |
Herstein|exercise_4_5_16 | theorem exercise_4_5_16 {p n: ℕ} (hp : Nat.Prime p)
{q : Polynomial (ZMod p)} (hq : Irreducible q) (hn : q.degree = n) :
(∃ is_fin : Fintype $ Polynomial (ZMod p) ⧸ span ({q}),
@card (Polynomial (ZMod p) ⧸ span {q}) is_fin = p ^ n) ∧
IsField (Polynomial (ZMod p) ⧸ span {q}) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Let $F = \mathbb{Z}_p$ be the field of integers $\mod p$, where $p$ is a prime, and let $q(x) \in F[x]$ be irreducible of degree $n$. Show that $F[x]/(q(x))$ is a field having at exactly $p^n$ elements. | \begin{proof}
In the previous problem we have shown that any for any $p(x) \in F[x]$, we have that
$$
p(x)+(q(x))=a_{n-1} x^{n-1}+\cdots+a_1 x+a_0+(q(x))
$$
for some $a_{n-1}, \ldots, a_0 \in F$, and that there are $p^n$ choices for these numbers, so that $F[x] /(q(x)) \leq p^n$. In order to show that equality... | [
"Herstein"
] | validation | Herstein_exercise_4_5_16 | de618f3413374855 | p n : ℕ
hp : Nat.Prime p
q : (ZMod p)[X]
hq : Irreducible q
hn : q.degree = ↑n
⊢ (∃ is_fin, card ((ZMod p)[X] ⧸ span {q}) = p ^ n) ∧ IsField ((ZMod p)[X] ⧸ span {q}) |
Herstein|exercise_4_5_25 | theorem exercise_4_5_25 {p : ℕ} (hp : Nat.Prime p) :
Irreducible (∑ i ∈ Finset.range p, X ^ i : Polynomial ℚ) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| If $p$ is a prime, show that $q(x) = 1 + x + x^2 + \cdots x^{p - 1}$ is irreducible in $Q[x]$. | \begin{proof}
Lemma: Let $F$ be a field and $f(x) \in F[x]$. If $c \in F$ and $f(x+c)$ is irreducible in $F[x]$, then $f(x)$ is irreducible in $F[x]$.
Proof of the Lemma: Suppose that $f(x)$ is reducible, i.e., there exist non-constant $g(x), h(x) \in F[x]$ so that
$$
f(x)=g(x) h(x) .
$$
In particular, then w... | [
"Herstein"
] | validation | Herstein_exercise_4_5_25 | e49adad7eae97b42 | p : ℕ
hp : Nat.Prime p
⊢ Irreducible (∑ i ∈ Finset.range p, X ^ i) |
Herstein|exercise_4_6_3 | theorem exercise_4_6_3 :
Infinite {a : ℤ | Irreducible (X^7 + 15*X^2 - 30*X + (a : Polynomial ℚ) : Polynomial ℚ)} := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Show that there is an infinite number of integers a such that $f(x) = x^7 + 15x^2 - 30x + a$ is irreducible in $Q[x]$. | \begin{proof}
Via Eisenstein's criterion and observation that 5 divides 15 and $-30$, it is sufficient to find infinitely many $a$ such that 5 divides $a$, but $5^2=25$ doesn't divide $a$. For example $5 \cdot 2^k$ for $k=0,1, \ldots$ is one such infinite sequence.
\end{proof} | [
"Herstein"
] | validation | Herstein_exercise_4_6_3 | c10d1131317f55b6 | ⊢ Infinite ↑{a | Irreducible (X ^ 7 + 15 * X ^ 2 - 30 * X + ↑a)} |
Herstein|exercise_5_2_20 | theorem exercise_5_2_20 {F V ι: Type*} [Infinite F] [Field F]
[AddCommGroup V] [Module F V] [Finite ι] {u : ι → Submodule F V}
(hu : ∀ i : ι, u i ≠ ⊤) :
(⋃ i : ι, (u i : Set V)) ≠ ⊤ := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Let $V$ be a vector space over an infinite field $F$. Show that $V$ cannot be the set-theoretic union of a finite number of proper subspaces of $V$. | \begin{proof}
Assume that $V$ can be written as the set-theoretic union of $n$ proper subspaces $U_1, U_2, \ldots, U_n$. Without loss of generality, we may assume that no $U_i$ is contained in the union of other subspaces.
Let $u \in U_i$ but $u \notin \bigcup_{j \neq i} U_j$ and $v \notin U_i$. Then, we have $... | [
"Herstein"
] | validation | Herstein_exercise_5_2_20 | 116849479186ba32 | F : Type u_1
V : Type u_2
ι : Type u_3
inst✝⁴ : Infinite F
inst✝³ : Field F
inst✝² : AddCommGroup V
inst✝¹ : Module F V
inst✝ : Finite ι
u : ι → Submodule F V
hu : ∀ (i : ι), u i ≠ ⊤
⊢ ⋃ i, ↑(u i) ≠ ⊤ |
Herstein|exercise_5_3_10 | theorem exercise_5_3_10 : IsAlgebraic ℚ (cos (Real.pi / 180)) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Prove that $\cos 1^{\circ}$ is algebraic over $\mathbb{Q}$. | \begin{proof}
Since $\left(\cos \left(1^{\circ}\right)+i \sin \left(1^{\circ}\right)\right)^{360}=1$, the number $\cos \left(1^{\circ}\right)+i \sin \left(1^{\circ}\right)$ is algebraic. And the real part and the imaginary part of an algebraic number are always algebraic numbers.
\end{proof} | [
"Herstein"
] | validation | Herstein_exercise_5_3_10 | e9c672858f834554 | ⊢ IsAlgebraic ℚ (cos (π / 180)) |
Herstein|exercise_5_5_2 | theorem exercise_5_5_2 : Irreducible (X^3 - 3*X - 1 : Polynomial ℚ) := by
| import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| Prove that $x^3 - 3x - 1$ is irreducible over $\mathbb{Q}$. | \begin{proof}
Let $p(x)=x^3-3 x-1$. Then
$$
p(x+1)=(x+1)^3-3(x+1)-1=x^3+3 x^2-3
$$
We have $3|3,3| 0$ but $3 \nmid 1$ and $3^2 \nmid 3$. Thus the polynomial is irreducible over $\mathbb{Q}$ by 3 -Eisenstein criterion.
\end{proof} | [
"Herstein"
] | validation | Herstein_exercise_5_5_2 | c5f1e7eda6bcd0f0 | ⊢ Irreducible (X ^ 3 - 3 * X - 1) |
Ireland-Rosen|exercise_1_27 | theorem exercise_1_27 {n : ℕ} (hn : Odd n) : 8 ∣ (n^2 - 1) := by
| import Mathlib
open Real
open scoped BigOperators
| For all odd $n$ show that $8 \mid n^{2}-1$. | \begin{proof}
We have $n^2-1=(n+1)(n-1)$. Since $n$ is odd, both $n+1, n-1$ are even, and moreso, one of these must be divisible by 4 , as one of the two consecutive odd numbers is divisible by 4 . Thus, their product is divisible by 8 . Similarly, if 3 does not divide $n$, it must divide one of $n-1, n+1$, otherw... | [
"Ireland-Rosen"
] | validation | Ireland-Rosen_exercise_1_27 | 8980ccd5cba0e7af | n : ℕ
hn : Odd n
⊢ 8 ∣ n ^ 2 - 1 |
Ireland-Rosen|exercise_1_31 | theorem exercise_1_31 : (⟨1, 1⟩ : GaussianInt) ^ 2 ∣ 2 := by
| import Mathlib
open Real
open scoped BigOperators
| Show that 2 is divisible by $(1+i)^{2}$ in $\mathbb{Z}[i]$. | \begin{proof}
We have $(1+i)^2=1+2 i-1=2 i$, so $2=-i(1+i)^2$.
\end{proof} | [
"Ireland-Rosen"
] | validation | Ireland-Rosen_exercise_1_31 | 48b73c1b2b789ebb | ⊢ { re := 1, im := 1 } ^ 2 ∣ 2 |
Ireland-Rosen|exercise_2_21 | theorem exercise_2_21 {l : ℕ → ℝ}
(hl : ∀ p n : ℕ, p.Prime → l (p^n) = log p )
(hl1 : ∀ m : ℕ, ¬ IsPrimePow m → l m = 0) :
l = λ n => ∑ d : Nat.divisors n, ArithmeticFunction.moebius (n/d) * log d := by
| import Mathlib
open Real
open scoped BigOperators
| Define $\wedge(n)=\log p$ if $n$ is a power of $p$ and zero otherwise. Prove that $\sum_{A \mid n} \mu(n / d) \log d$ $=\wedge(n)$. | \begin{proof}
$$
\left\{
\begin{array}{cccl}
\land(n)& = & \log p & \mathrm{if}\ n =p^\alpha,\ \alpha \in \mathbb{N}^* \\
& = & 0 & \mathrm{otherwise }.
\end{array}
\right.
$$
Let $n = p_1^{\alpha_1}\cdots p_t^{\alpha_t}$ the decomposition of $n$ in prime factors. As $\land(d) = 0$ for all divi... | [
"Ireland-Rosen"
] | validation | Ireland-Rosen_exercise_2_21 | 643dc4e3e6143699 | l : ℕ → ℝ
hl : ∀ (p n : ℕ), Nat.Prime p → l (p ^ n) = log ↑p
hl1 : ∀ (m : ℕ), ¬IsPrimePow m → l m = 0
⊢ l = fun n => ∑ d, ↑(ArithmeticFunction.moebius (n / ↑d)) * log ↑↑d |
Ireland-Rosen|exercise_3_1 | theorem exercise_3_1 : Infinite {p : Nat.Primes // p ≡ -1 [ZMOD 6]} := by
| import Mathlib
open Real
open scoped BigOperators
| Show that there are infinitely many primes congruent to $-1$ modulo 6 . | \begin{proof}
Let $n$ any integer such that $n\geq 3$, and $N = n! -1 = 2 \times 3 \times\cdots\times n - 1 >1$.
Then $N \equiv -1 \pmod 6$. As $6k +2, 6k +3, 6k +4$ are composite for all integers $k$, every prime factor of $N$ is congruent to $1$ or $-1$ modulo $6$. If every prime factor of $N$ was congrue... | [
"Ireland-Rosen"
] | validation | Ireland-Rosen_exercise_3_1 | da99e60cc1d425a6 | ⊢ Infinite { p // ↑↑p ≡ -1 [ZMOD 6] } |
Ireland-Rosen|exercise_3_5 | theorem exercise_3_5 : ¬ ∃ x y : ℤ, 7*x^3 + 2 = y^3 := by
| import Mathlib
open Real
open scoped BigOperators
| Show that the equation $7 x^{3}+2=y^{3}$ has no solution in integers. | \begin{proof}
If $7x^2 + 2 = y^3,\ x,y \in \mathbb{Z}$, then $y^3 \equiv 2 \pmod 7$ (so $y \not \equiv 0 \pmod 7$)
From Fermat's Little Theorem, $y^6 \equiv 1 \pmod 7$, so $2^2 \equiv y^6 \equiv 1 \pmod 7$, which implies $7 \mid 2^2-1 = 3$ : this is a contradiction. Thus the equation $7x^2 + 2 = y^3$ has no sol... | [
"Ireland-Rosen"
] | validation | Ireland-Rosen_exercise_3_5 | 8d4c056beacecefc | ⊢ ¬∃ x y, 7 * x ^ 3 + 2 = y ^ 3 |
Ireland-Rosen|exercise_3_14 | theorem exercise_3_14 {p q n : ℕ} (hp0 : p.Prime ∧ p > 2)
(hq0 : q.Prime ∧ q > 2) (hpq0 : p ≠ q) (hpq1 : p - 1 ∣ q - 1)
(hn : n.gcd (p*q) = 1) :
n^(q-1) ≡ 1 [MOD p*q] := by
| import Mathlib
open Real
open scoped BigOperators
| Let $p$ and $q$ be distinct odd primes such that $p-1$ divides $q-1$. If $(n, p q)=1$, show that $n^{q-1} \equiv 1(p q)$. | \begin{proof}
As $n \wedge pq = 1, n\wedge p=1, n \wedge q = 1$, so from Fermat's Little Theorem
$$n^{q-1} \equiv 1 \pmod q,\qquad n^{p-1} \equiv 1 \pmod p.$$
$p-1 \mid q-1$, so there exists $k \in \mathbb{Z}$ such that $q-1 = k(p-1)$.
Thus
$$n^{q-1} = (n^{p-1})^k \equiv 1 \pmod p.$$
$p \mid n^{q-1} - 1, q \m... | [
"Ireland-Rosen"
] | validation | Ireland-Rosen_exercise_3_14 | 9f752d472011e536 | p q n : ℕ
hp0 : Nat.Prime p ∧ p > 2
hq0 : Nat.Prime q ∧ q > 2
hpq0 : p ≠ q
hpq1 : p - 1 ∣ q - 1
hn : n.gcd (p * q) = 1
⊢ n ^ (q - 1) ≡ 1 [MOD p * q] |
End of preview. Expand in Data Studio
ProofNet#-SATP v4.27 — Lean 4 (371 problems)
ProofNet# (corrected reference
formalizations of ProofNet) normalized to the ChristianZ97/putnambench-satp-v4.27
schema for the SATP-DSP-Eval pipeline.
Provenance
- Source:
PAug/ProofNetSharp@a8da405fbd1e348a87445c2e562c747b7e26dc8f(MIT), 371 rows = 185 valid + 186 test. Introduced in Reliable Evaluation and Benchmarks for Statement Autoformalization (Poiroux, Weiss, Kunčak, Bosselut; EMNLP 2025 main; arXiv:2406.07222) — corrects 118/371 (31.8%) faulty reference formalizations found in circulating Lean 4 ports of ProofNet. - Original benchmark: ProofNet (Azerbayev, Piotrowski, Schoelkopf, Ayers, Radev, Avigad; arXiv:2302.12433).
Compile gate
Every row was compiled standalone with an appended sorry
(header + formal_statement + " trace_state\n sorry") under the pinned
SATP-DSP-Eval environment — toolchain leanprover/lean4:v4.27.0, LeanSATP@b9b30c322f55 (mathlib vendored at deps/mathlib4) —
and goal_state is the pretty-printed goal Lean reports at the start of by
(trace_state). 371/371 pass after the compatibility patches below; a row
failing the gate aborts the port.
v4.27 compatibility patches (25 rows)
Upstream is well-typed for Lean 4.7.0–4.16.0-rc2; on the pin above, 25
rows fail to elaborate (pre-4.19 big-operator binder sugar, and Mathlib renames:
QuotientMap → IsQuotientMap, TopologicalGroup → IsTopologicalGroup,
complex abs → ‖·‖, ambiguous sqrt → Real.sqrt, explicit-𝕜 inner,
removed IsNormalSubgroup → ∃ N : Subgroup G, N.Normal ∧ ↑N = C). Every
patched row and its exact rewrite:
Dummit-Foote|exercise_9_4_2d: big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Herstein|exercise_4_2_9: big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Herstein|exercise_4_5_25: big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Munkres|exercise_22_2a:QuotientMap→IsQuotientMapMunkres|exercise_22_2b:QuotientMap→IsQuotientMapMunkres|exercise_23_11:QuotientMap→IsQuotientMapMunkres|exercise_25_9:[TopologicalGroup G]→[IsTopologicalGroup G];IsNormalSubgroup C→∃ N : Subgroup G, N.Normal ∧ ↑N = CRudin|exercise_1_11a:abs w = 1→‖w‖ = 1Rudin|exercise_1_12:abs (∑ i in range n, f i) ≤ ∑ i in range n, abs (f i)→‖∑ i in range n, f i‖ ≤ ∑ i in range n, ‖f i‖; big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Rudin|exercise_1_13:|(abs x) - (abs y)| ≤ abs (x - y)→|‖x‖ - ‖y‖| ≤ ‖x - y‖Rudin|exercise_1_14:abs z = 1→‖z‖ = 1;(abs (1 + z)) ^ 2 + (abs (1 - z)) ^ 2 = 4→‖1 + z‖ ^ 2 + ‖1 - z‖ ^ 2 = 4Rudin|exercise_1_18a:(inner x y)→(inner ℝ x y)Rudin|exercise_3_13: big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Rudin|exercise_3_2a:sqrt→Real.sqrtRudin|exercise_3_3:sqrt→Real.sqrtRudin|exercise_3_6a:sqrt→Real.sqrt; big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Rudin|exercise_3_7:sqrt→Real.sqrt; big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Rudin|exercise_3_8: big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Rudin|exercise_5_4: big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Shakarchi|exercise_1_13c:abs (f z) = c→‖f z‖ = cShakarchi|exercise_1_19a:abs z = 1→‖z‖ = 1; big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Shakarchi|exercise_1_19b:abs z = 1→‖z‖ = 1; big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Shakarchi|exercise_1_19c:abs z = 1→‖z‖ = 1; big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Shakarchi|exercise_2_13: big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)Shakarchi|exercise_5_1:(1 - abs (zeros i))→(1 - ‖zeros i‖); big-operator binder∑ x in s→∑ x ∈ s(pre-4.19 sugar)
All other rows are PAug/ProofNetSharp verbatim.
The one real reformulation: Munkres|exercise_25_9
Every patch above is a pure rename or notation swap except this row. Its
upstream conclusion uses IsNormalSubgroup C — Mathlib's unbundled
Set-predicate API, removed from Mathlib long before our pin — so the
conclusion is restated through the bundled Subgroup API:
- upstream:
… (C : Set G) (h : C = connectedComponent 1) : IsNormalSubgroup C - this repo:
… : ∃ N : Subgroup G, N.Normal ∧ ↑N = C
Both formalize "the connected component of the identity is a normal subgroup";
the restatement asserts the existence of a normal Subgroup whose carrier is
exactly C. This is the only row where the formal shape (not just names or
notation) differs from upstream.
Differences from upstream (PAug/ProofNetSharp)
| Field | Upstream | This repo |
|---|---|---|
lean4_src_header |
separate column | header; 13 rows' auxiliary defs moved here (split at the last theorem decl, mirroring minif2f-satp-v4.27's vendored-header pattern) |
lean4_formalization |
ends := |
formal_statement, ends := by |
id (Artin|exercise_2_3_2) |
— | problem_name verbatim; name = | → _ (bare Lean decl names collide across textbooks within a split; decl names inside formal_statement are upstream-verbatim) |
nl_statement / nl_proof |
— | informal_statement / informal_solution |
| — | — | tags = [textbook], split valid → validation, uuid = sha256(canonical(formal_statement))[:16] (putnam-satp-v4.27 recipe), goal_state freshly lake-computed |
Schema
Columns match ChristianZ97/putnambench-satp-v4.27:
problem_name, formal_statement, header, informal_statement, informal_solution, tags, split, name, uuid, goal_state.
Citation
@inproceedings{poiroux-etal-2025-reliable,
title = "Reliable Evaluation and Benchmarks for Statement Autoformalization",
author = "Poiroux, Auguste and
Weiss, Gail and
Kun{\v{c}}ak, Viktor and
Bosselut, Antoine",
editor = "Christodoulopoulos, Christos and
Chakraborty, Tanmoy and
Rose, Carolyn and
Peng, Violet",
booktitle = "Proceedings of the 2025 Conference on Empirical Methods in Natural Language Processing",
month = nov,
year = "2025",
address = "Suzhou, China",
publisher = "Association for Computational Linguistics",
url = "https://aclanthology.org/2025.emnlp-main.907/",
doi = "10.18653/v1/2025.emnlp-main.907",
pages = "17947--17969",
ISBN = "979-8-89176-332-6",
abstract = "Evaluating statement autoformalization, translating natural language mathematics into formal languages like Lean 4, remains a significant challenge, with few metrics, datasets, and standards to robustly measure progress. In this work, we present a comprehensive approach combining improved metrics, robust benchmarks, and systematic evaluation, to fill this gap. First, we introduce BEq+, an automated metric that correlates strongly with human judgment, along with ProofNetVerif, a new dataset for assessing the quality of evaluation metrics, containing 3,752 annotated examples. Second, we develop two new autoformalization benchmarks: ProofNet{\#}, a corrected version of ProofNet, and RLM25, with 619 new pairs of research-level mathematics from six formalization projects. Through systematic experimentation across these benchmarks, we find that current techniques can achieve up to 45.1{\%} accuracy on undergraduate mathematics but struggle with research-level content without proper context. Our work establishes a reliable foundation for evaluating and advancing autoformalization systems."
}
@article{azerbayev2023proofnet,
title={ProofNet: Autoformalizing and Formally Proving Undergraduate-Level
Mathematics},
author={Azerbayev, Zhangir and Piotrowski, Bartosz and Schoelkopf, Hailey and
Ayers, Edward W. and Radev, Dragomir and Avigad, Jeremy},
journal={arXiv preprint arXiv:2302.12433},
year={2023}
}
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