index int64 1 367 | name stringlengths 17 29 | textbook stringclasses 10
values | header stringclasses 60
values | helper stringclasses 14
values | informal_stmt stringlengths 42 474 | formal_stmt stringlengths 66 545 | informal_proof stringlengths 40 3.85k | formal_proof stringclasses 1
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1 | Artin_exercise_2_2_9 | Artin | 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. | theorem Artin_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
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
2 | Artin_exercise_2_4_19 | Artin | 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. | theorem Artin_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
sorry | 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 $G$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
3 | Artin_exercise_2_11_3 | Artin | 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 $2 .$ | theorem Artin_exercise_2_11_3 {G : Type*} [Group G] [Fintype G]
(hG : Even (card G)) : ∃ x : G, orderOf x = 2 := by
sorry | 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 $e=e^{-1} \Longleftrightarrow e^2=e$ ), so sin... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
4 | Artin_exercise_3_5_6 | Artin | 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. | theorem Artin_exercise_3_5_6 {K V : Type*} [Field K] [AddCommGroup V]
[Module K V] {S : Set V} (hS : Set.Countable S) (hSinf : S.Infinite)
(hS1 : span K S = ⊤) {ι : Type*} (R : ι → V)
(hR : LinearIndependent K R) : Countable ι := by
sorry | 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 involved in the decomposit... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
5 | Artin_exercise_6_1_14 | Artin | 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 $G=Z$. | theorem Artin_exercise_6_1_14 (G : Type*) [Group G]
(hG : IsCyclic $ G ⧸ (center G)) :
center G = ⊤ := by
sorry | 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 $H \leq G$, we have by de... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
6 | Artin_exercise_6_4_3 | Artin | 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. | theorem Artin_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
sorry | 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 simple. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
7 | Artin_exercise_6_8_1 | Artin | 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 $b a b^2, b a b^3$. | theorem Artin_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
sorry | 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 $\langle a, b\rangle\subset H$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
8 | Artin_exercise_10_2_4 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that in the ring $\mathbb{Z}[x],(2) \cap(x)=(2 x)$. | theorem Artin_exercise_10_2_4 :
span ({2} : Set $ Polynomial ℤ) ⊓ (span {X}) =
span ({2 * X} : Set $ Polynomial ℤ) := by
sorry | 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
$$
(2 x) \subseteq(2) \cap(x)
$$
On the other hand, l... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
9 | Artin_exercise_10_4_6 | Artin | 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. | theorem Artin_exercise_10_4_6 {R : Type*} [CommRing R]
(I J : Ideal R) (x : ↑(I ⊓ J)) :
IsNilpotent ((Ideal.Quotient.mk (I*J)) x) := by
sorry | 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
$$
[x]^2=\left[x^2\right]=[0] \text { in } R / I J
$$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
10 | Artin_exercise_10_7_10 | Artin | 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 $R$. | theorem Artin_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
sorry | 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 we cannot have $... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
11 | Artin_exercise_11_4_1b | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that $x^3 + 6x + 12$ is irreducible in $\mathbb{Q}$. | theorem Artin_exercise_11_4_1b :
Irreducible (12 + 6 * X + X ^ 3 : Polynomial ℚ) := by
sorry | Apply Eisenstein's criterion with $p=3$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
12 | Artin_exercise_11_4_6b | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that $x^2+1$ is irreducible in $\mathbb{F}_7$ | theorem Artin_exercise_11_4_6b {F : Type*} [Field F] [Fintype F] (hF : card F = 7) :
Irreducible (X ^ 2 + 1 : Polynomial F) := by
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
13 | Artin_exercise_11_4_8 | Artin | 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 $\mathbb{Q}[x]$. | theorem Artin_exercise_11_4_8 (p : ℕ) (hp : Prime p) (n : ℕ) (hn : n > 0) :
Irreducible (X ^ n - (p : Polynomial ℚ) : Polynomial ℚ) := by
sorry | Straightforward application of Eisenstein's criterion with $p$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
14 | Artin_exercise_13_4_10 | Artin | 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 $2^{2^k}+1$. | theorem Artin_exercise_13_4_10
{p : ℕ} (hp : Nat.Prime p)
(h : ∃ r : ℕ, 1 ≤ r ∧ p = 2 ^ r + 1) :
∃ k : ℕ, p = 2 ^ (2 ^ k) + 1 := by
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
15 | Axler_exercise_1_3 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | Prove that $-(-v) = v$ for every $v \in V$. | theorem Axler_exercise_1_3 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] {v : V} : -(-v) = v := by
sorry | 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 $-(-v)=v$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
16 | Axler_exercise_1_8 | Axler | 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 $V$. | theorem Axler_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
sorry | 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 $0$ is in each s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
17 | Axler_exercise_3_1 | Axler | 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 $v \in V$. | theorem Axler_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
sorry | 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) \ldots .(\text { By the lineari... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
18 | Axler_exercise_4_4 | Axler | 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. | theorem Axler_exercise_4_4 (p : Polynomial ℂ) (m : ℕ) (hp : p ≠ 0) (hm : m = p.degree) :
@card (p.rootSet ℂ) (rootSetFintype p ℂ) = m ↔
Disjoint (p.rootSet ℂ) ((Polynomial.derivative p).rootSet ℂ) := by
sorry | 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^{\prime}$ have no ro... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
19 | Axler_exercise_5_4 | Axler | 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}$. | theorem Axler_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
sorry | 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$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
20 | Axler_exercise_5_12 | Axler | 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. | theorem Axler_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
sorry | 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$ is a scalar multiple ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
21 | Axler_exercise_5_20 | Axler | 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$. | theorem Axler_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
sorry | 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$ is also an eigenvect... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
22 | Axler_exercise_6_2 | Axler | 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}$. | theorem Axler_exercise_6_2 {V : Type*} [NormedAddCommGroup V] {F : Type*}
[RCLike F] [InnerProductSpace F V]
(u v : V) :
⟪u, v⟫_F = 0 ↔ ∀ (a : F), ‖u‖ ≤ ‖u + a • v‖ := by
sorry | 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}
$$
So, by taking the square roots, this will now give us $\|u\| \leq\|u+a v\|$.
Now, to prov... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
23 | Axler_exercise_6_7 | Axler | import Mathlib
open InnerProductSpace ContinuousLinearMap Complex
open scoped BigOperators | Prove that if $V$ is a complex inner-product space, then $\langle u, v\rangle=\frac{\|u+v\|^{2}-\|u-v\|^{2}+\|u+i v\|^{2} i-\|u-i v\|^{2} i}{4}$ for all $u, v \in V$. | theorem Axler_exercise_6_7 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V]
(u v : V) :
⟪u, v⟫_ℂ =
(‖u + v‖ ^ 2 - ‖u - v‖ ^ 2 +
(‖u - I • v‖ ^ 2 - ‖u + I • v‖ ^ 2) * I) / 4 := by
sorry | Let $V$ be an inner-product space and $u, v\in V$. Then
$$
\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 & =i\langle u+i v, u+i v... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
24 | Axler_exercise_6_16 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators ENNReal ComplexConjugate | Suppose $U$ is a subspace of $V$. Prove that $U^{\perp}=\{0\}$ if and only if $U=V$ | theorem Axler_exercise_6_16 {K V : Type*} [RCLike K] [NormedAddCommGroup V]
[InnerProductSpace K V] [CompleteSpace V] {U : Submodule K V}
(hU : IsClosed (U : Set V)) :
U.orthogonal = ⊥ ↔ U = ⊤ := by
sorry | $V=U \bigoplus U^{\perp}$, therefore $U^\perp = \{0\}$ iff $U=V$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
25 | Axler_exercise_7_6 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators ComplexOrder | Prove that if $T \in \mathcal{L}(V)$ is normal, then $\operatorname{range} T=\operatorname{range} T^{*}.$ | theorem Axler_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
sorry | 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 the same argument can be repe... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
26 | Axler_exercise_7_10 | Axler | 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 $T^{2}=T$. | theorem Axler_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
sorry | 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 $T$ repeatedly to both s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
27 | Axler_exercise_7_14 | Axler | 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 $\left|\lambda-\lambda^{\prime}\right|<\epsilon$. | theorem Axler_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
sorry | 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 an eigenvalue $\lambda... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
28 | Dummit_Foote_exercise_1_1_3 | Dummit Foote | 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. | theorem Dummit_Foote_exercise_1_1_3 (n : ℕ) :
∀ (x y z : ZMod n), (x + y) + z = x + (y + z) := by
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
29 | Dummit_Foote_exercise_1_1_5 | Dummit Foote | 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. | theorem Dummit_Foote_exercise_1_1_5 (n : ℕ) (hn : 1 < n) :
¬ (∀ x : ZMod n, ∃ y : ZMod n, x * y = (1 : ZMod n) ∧ y * x = 1) := by
sorry | 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
$$
\overline{0} \cdot \bar{k}=\overline{0}
$$
for all k, so tha... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
30 | Dummit_Foote_exercise_1_1_16 | Dummit Foote | 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 $2$. | theorem Dummit_Foote_exercise_1_1_16 {G : Type*} [Group G]
(x : G) : x ^ 2 = 1 ↔ (orderOf x = 1 ∨ orderOf x = 2) := by
sorry | $(\Rightarrow)$ 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 $x^2=1$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
31 | Dummit_Foote_exercise_1_1_18 | Dummit Foote | 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 $x^{-1}y^{-1}xy=1$. | theorem Dummit_Foote_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
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
32 | Dummit_Foote_exercise_1_1_22a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | If $x$ and $g$ are elements of the group $G$, prove that $|x|=\left|g^{-1} x g\right|$. | theorem Dummit_Foote_exercise_1_1_22a {G : Type*} [Group G] (x g : G) :
orderOf x = orderOf (g⁻¹ * x * g) := by
sorry | 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 some $n \geq 1$; then
$$
... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
33 | Dummit_Foote_exercise_1_1_25 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $x^{2}=1$ for all $x \in G$ then $G$ is abelian. | theorem Dummit_Foote_exercise_1_1_25 {G : Type*} [Group G]
(h : ∀ x : G, x ^ 2 = 1) : ∀ a b : G, a*b = b*a := by
sorry | Solution: Note that since $x^2=1$ for all $x \in G$, we have $x^{-1}=x$. Now let $a, b \in G$. We have
$$
a b=(a b)^{-1}=b^{-1} a^{-1}=b a .
$$
Thus $G$ is abelian. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
34 | Dummit_Foote_exercise_1_1_34 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | If $x$ is an element of infinite order in $G$, prove that the elements $x^{n}, n \in \mathbb{Z}$ are all distinct. | theorem Dummit_Foote_exercise_1_1_34 {G : Type*} [Group G] {x : G}
(hx_inf : orderOf x = 0) (n m : ℤ) (hnm : n ≠ m) :
x ^ n ≠ x ^ m := by
sorry | Solution: Suppose to the contrary that $x^a=x^b$ for some $0 \leq a<b \leq n-1$. Then we have $x^{b-a}=1$, with $1 \leq b-a<n$. However, recall that $n$ is by definition the least integer $k$ such that $x^k=1$, so we have a contradiction. Thus all the $x^i$, $0 \leq i \leq n-1$, are distinct. In particular, we have
$$
... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
35 | Dummit_Foote_exercise_1_6_4 | Dummit Foote | 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. | theorem Dummit_Foote_exercise_1_6_4 :
IsEmpty (ℝˣ ≃* ℂˣ) := by
sorry | 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 particular, $1>\left|x^n\right... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
36 | Dummit_Foote_exercise_1_6_17 | Dummit Foote | 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. | theorem Dummit_Foote_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
sorry | $(\Rightarrow)$ 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(b^{-1} a^{-1}\right)=\varp... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
37 | Dummit_Foote_exercise_2_1_5 | Dummit Foote | 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$. | theorem Dummit_Foote_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
sorry | 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. Thus no such subgroup... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
38 | Dummit_Foote_exercise_2_4_4 | Dummit Foote | 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\}$. | theorem Dummit_Foote_exercise_2_4_4 {G : Type*} [Group G] (H : Subgroup G) :
Subgroup.closure ((H : Set G) \ {1}) = H := by
sorry | 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$, the reverse incl... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
39 | Dummit_Foote_exercise_2_4_16b | Dummit Foote | 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. | theorem Dummit_Foote_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
sorry | 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 contain a reflecti... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
40 | Dummit_Foote_exercise_3_1_3a | Dummit Foote | 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. | theorem Dummit_Foote_exercise_3_1_3a {A : Type*} [CommGroup A] (B : Subgroup A) :
∀ a b : A ⧸ B, a*b = b*a := by
sorry | 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$ is normal; in particul... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
41 | Dummit_Foote_exercise_3_1_22b | Dummit Foote | 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). | theorem Dummit_Foote_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
sorry | 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
$$
Take an element $a$ in the intersection and an arbitrary element $g \in G$. Then $g a g^{-1} \in H_i$ because $H_i$ is normal for any $i \in H$
By the definition of the i... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
42 | Dummit_Foote_exercise_3_2_11 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | noncomputable def Subgroup.relindex {G : Type*} [Group G] (H K : Subgroup G) : ℕ :=
H.relIndex K | Let $H \leq K \leq G$. Prove that $|G: H|=|G: K| \cdot|K: H|$ (do not assume $G$ is finite). | theorem Dummit_Foote_exercise_3_2_11 {G : Type*} [Group G] {H K : Subgroup G}
(hHK : H ≤ K) :
H.index = K.index * H.relindex K := by
sorry | Proof. Let $G$ be a group and let $I$ be a nonempty set of indices, not necessarily countable. Consider the collection of subgroups $\left\{N_\alpha \mid \alpha \in I\right\}$, where $N_\alpha \unlhd G$ for each $\alpha \in I$. Let
$$
N=\bigcap_{\alpha \in I} N_\alpha .
$$
We know $N$ is a subgroup of $G$.
For any $g ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
43 | Dummit_Foote_exercise_3_2_21a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $\mathbb{Q}$ has no proper subgroups of finite index. | theorem Dummit_Foote_exercise_3_2_21a (H : AddSubgroup ℚ) (hH : H ≠ ⊤) : H.index = 0 := by
sorry | 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 group is finite. Equiv... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
44 | Dummit_Foote_exercise_3_4_1 | Dummit Foote | 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). | theorem Dummit_Foote_exercise_3_4_1 (G : Type*) [CommGroup G] [IsSimpleGroup G] :
IsCyclic G ∧ ∃ G_fin : Fintype G, Nat.Prime (@card G G_fin) := by
sorry | 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 nontrivial normal subg... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
45 | Dummit_Foote_exercise_3_4_5a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that subgroups of a solvable group are solvable. | theorem Dummit_Foote_exercise_3_4_5a {G : Type*} [Group G]
(H : Subgroup G) [IsSolvable G] : IsSolvable H := by
sorry | 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 .
$$
Then $H_i \leq H_{i+1}$ for each $i$. ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
46 | Dummit_Foote_exercise_3_4_11 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $H$ is a nontrivial normal subgroup of the solvable group $G$ then there is a nontrivial subgroup $A$ of $H$ with $A \unlhd G$ and $A$ abelian. | theorem Dummit_Foote_exercise_3_4_11 {G : Type*} [Group G] [IsSolvable G]
{H : Subgroup G} (hH : H ≠ ⊥) [H.Normal] :
∃ A ≤ H, A ≠ ⊥ ∧ A.Normal ∧ ∀ a b : A, a*b = b*a := by
sorry | Suppose $H$ is a nontrivial normal subgroup of the solvable group $G$.
First, notice that $H$, being a subgroup of a solvable group, is itself solvable. By exercise $8, H$ has a chain of subgroups
$$
1 \leq H_1 \leq \ldots \leq H
$$
such that each $H_i$ is a normal subgroup of $H$ itself and $H_{i+1} / H_i$ is abelian.... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
47 | Dummit_Foote_exercise_4_2_14 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of order $k$ for each positive integer $k$ dividing $n$. Prove that $G$ is not simple. | theorem Dummit_Foote_exercise_4_2_14 {G : Type*} [Fintype G] [Group G]
(hG : ¬ (card G).Prime) (hG_ne_one : card G ≠ 1)
(hG1 : ∀ k : ℕ, k ∣ card G →
∃ (H : Subgroup G) (fH : Fintype H), @card H fH = k) :
¬ IsSimpleGroup G := by
sorry | Solution: Let $p$ be the smallest prime dividing $n$, and write $n=p m$. Now $G$ has a subgroup $H$ of order $m$, and $H$ has index $p$. Then $H$ is normal in $G$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
48 | Dummit_Foote_exercise_4_3_26 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $G$ be a transitive permutation group on the finite set $A$ with $|A|>1$. Show that there is some $\sigma \in G$ such that $\sigma(a) \neq a$ for all $a \in A$. | theorem Dummit_Foote_exercise_4_3_26 {α : Type*} [Fintype α] (ha : Fintype.card α > 1)
(G : Subgroup (Equiv.Perm α))
(h_tran : ∀ a b : α, ∃ σ : G, (σ : Equiv.Perm α) a = b) :
∃ σ : G, ∀ a : α, (σ : Equiv.Perm α) a ≠ a := by
sorry | Let $G$ be a transitive permutation group on the finite set $A,|A|>1$. We want to find an element $\sigma$ which doesn't stabilize anything, that is, we want a $\sigma$ such that
$$
\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 \cdot a$. Let us see in ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
49 | Dummit_Foote_exercise_4_4_6a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that characteristic subgroups are normal. | theorem Dummit_Foote_exercise_4_4_6a {G : Type*} [Group G] (H : Subgroup G)
[Characteristic H] : Normal H := by
sorry | Let $H$ be a characteristic 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^{-1} \subset H$. So, ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
50 | Dummit_Foote_exercise_4_4_7 | Dummit Foote | 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$. | theorem Dummit_Foote_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
sorry | 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, we have $\sigma(H)=$... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
51 | Dummit_Foote_exercise_4_5_1a | Dummit Foote | 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)$. | theorem Dummit_Foote_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
sorry | 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$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
52 | Dummit_Foote_exercise_4_5_14 | Dummit Foote | 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. | theorem Dummit_Foote_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
sorry | Let $n_{13}$ be the number of Sylow 13 -subgroup of $G$. Then by Sylow's Theorem, $n_{13} \equiv 1(\bmod 13)$ and $n_{13}$ divides $2^3 \cdot 3=24$. This implies $n_{13}=1$, so that there is only one Sylow 13 -subgroup, which is consequently normal. The last assertion follows from the fact conjugation preserves the ord... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
53 | Dummit_Foote_exercise_4_5_16 | Dummit Foote | 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$. | theorem Dummit_Foote_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
sorry | 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$ can be 1 or $p q$.... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
54 | Dummit_Foote_exercise_4_5_18 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that a group of order 200 has a normal Sylow 5-subgroup. | theorem Dummit_Foote_exercise_4_5_18 {G : Type*} [Fintype G] [Group G]
(hG : card G = 200) :
∃ N : Sylow 5 G, N.Normal := by
sorry | 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$ divides 8 . The only su... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
55 | Dummit_Foote_exercise_4_5_20 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $|G|=1365$ then $G$ is not simple. | theorem Dummit_Foote_exercise_4_5_20 {G : Type*} [Fintype G] [Group G]
(hG : card G = 1365) : ¬ IsSimpleGroup G := by
sorry | 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 $G$ is not simple. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
56 | Dummit_Foote_exercise_4_5_22 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $|G|=132$ then $G$ is not simple. | theorem Dummit_Foote_exercise_4_5_22 {G : Type*} [Fintype G] [Group G]
(hG : card G = 132) : ¬ IsSimpleGroup G := by
sorry | 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 $n_{11}=1+11k$ wher... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
57 | Dummit_Foote_exercise_4_5_28 | Dummit Foote | 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. | theorem Dummit_Foote_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
sorry | 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, Consider $G / P$, which h... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
58 | Dummit_Foote_exercise_5_4_2 | Dummit Foote | 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$. | theorem Dummit_Foote_exercise_5_4_2 {G : Type*} [Group G] (H : Subgroup G) :
H.Normal ↔ ⁅(⊤ : Subgroup G), H⁆ ≤ H := by
sorry | $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} h g \in H, \forall ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
59 | Dummit_Foote_exercise_7_1_11 | Dummit Foote | 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$. | theorem Dummit_Foote_exercise_7_1_11 {R : Type*} [CommRing R] [IsDomain R]
{x : R} (hx : x^2 = 1) : x = 1 ∨ x = -1 := by
sorry | Solution: If $x^2=1$, then $x^2-1=0$. Evidently, then,
$$
(x-1)(x+1)=0 .
$$
Since $R$ is an integral domain, we must have $x-1=0$ or $x+1=0$; thus $x=1$ or $x=-1$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
60 | Dummit_Foote_exercise_7_1_15 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | A ring $R$ is called a Boolean ring if $a^{2}=a$ for all $a \in R$. Prove that every Boolean ring is commutative. | theorem Dummit_Foote_exercise_7_1_15 {R : Type*} [Ring R] (hR : ∀ a : R, a^2 = a) :
∀ a b : R, a*b = b*a := by
sorry | Solution: Note first that for all $a \in R$,
$$
-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 .
$$
Thus $a b+b a=0$, and we have $a b=-b a$. But then $a b=b a$. Thus $R$ is commutative. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
61 | Dummit_Foote_exercise_7_2_12 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $G=\left\{g_{1}, \ldots, g_{n}\right\}$ be a finite group. Prove that the element $N=g_{1}+g_{2}+\ldots+g_{n}$ is in the center of the group ring $R G$. | theorem Dummit_Foote_exercise_7_2_12 {R G : Type*} [Ring R] [Group G] [Fintype G] :
∑ g : G, MonoidAlgebra.of R G g ∈ center (MonoidAlgebra R G) := by
sorry | Let $M=\sum_{i=1}^n r_i g_i$ be an element of $R[G]$. Note that for each $g_i \in G$, the action of $g_i$ on $G$ by conjugation permutes the subscripts. Then we have the following.
$$
\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}^n r_j g_i g_j \\
&=\su... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
62 | Dummit_Foote_exercise_7_3_37 | Dummit Foote | 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 $\mathbb{Z} / p^{m} \mathbb{Z}$. | theorem Dummit_Foote_exercise_7_3_37 {p m : ℕ} (hp : p.Prime) :
IsNilpotent (span ({↑p} : Set $ ZMod $ p^m) : Ideal $ ZMod $ p^m) := by
sorry | 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
$$
\alpha=\sum\left(x_i+J\right)\left(y_i+J\right) \in\left(I_1 / J\right)\left(I_2 / J\right) .
$$
Then... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
63 | Dummit_Foote_exercise_8_1_12 | Dummit Foote | 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... | theorem Dummit_Foote_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
sorry | 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 $M_1^{d^{\prime}} \equiv M \bmod N$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
64 | Dummit_Foote_exercise_8_3_4 | Dummit Foote | 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. | theorem Dummit_Foote_exercise_8_3_4 {n : ℤ} {r s : ℚ}
(h : r^2 + s^2 = n) :
∃ a b : ℤ, a^2 + b^2 = n := by
sorry | 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 $j \in \mathbb{N} \cup\{0\}$ such t... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
65 | Dummit_Foote_exercise_8_3_6a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal Zsqrtd
open scoped BigOperators
set_option synthInstance.maxHeartbeats 40000
set_option maxRecDepth 8000 | Prove that the quotient ring $\mathbb{Z}[i] /(1+i)$ is a field of order 2. | theorem Dummit_Foote_exercise_8_3_6a :
IsField (GaussianInt ⧸ span ({⟨1, 1⟩} : Set GaussianInt)) ∧
∃ finR : Fintype (GaussianInt ⧸ span ({⟨1, 1⟩} : Set GaussianInt)),
@Fintype.card _ finR = 2 := by
sorry | 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 $\langle 1+i\rangle$ or $1+\... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
66 | Dummit_Foote_exercise_9_1_6 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $(x, y)$ is not a principal ideal in $\mathbb{Q}[x, y]$. | theorem Dummit_Foote_exercise_9_1_6 : ¬ Submodule.IsPrincipal
(span ({MvPolynomial.X 0, MvPolynomial.X 1} : Set (MvPolynomial (Fin 2) ℚ))) := by
sorry | 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:
$$
\begin{aligned}
& 0=\operatorname{deg}_y(x)=\operatorname{deg}_y(s)+\operatorname{deg}_y(p) \text { so } \\
& 0=\operatorname... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
67 | Dummit_Foote_exercise_9_3_2 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $f(x)$ and $g(x)$ are polynomials with rational coefficients whose product $f(x) g(x)$ has integer coefficients, then the product of any coefficient of $g(x)$ with any coefficient of $f(x)$ is an integer. | theorem Dummit_Foote_exercise_9_3_2 {f g : Polynomial ℚ} (i j : ℕ)
(hfg : ∀ n : ℕ, ∃ a : ℤ, (f * g).coeff n = a) :
∃ a : ℤ, f.coeff i * g.coeff j = a := by
sorry | Let $f(x), g(x) \in \mathbb{Q}[x]$ be such that $f(x) g(x) \in \mathbb{Z}[x]$.
By Gauss' Lemma there exists $r, s \in \mathbb{Q}$ such that $r f(x), s g(x) \in \mathbb{Z}[x]$, and $(r f(x))(s g(x))=r s f(x) g(x)=f(x) g(x)$. From this last relation we can conclude that $s=r^{-1}$.
Therefore for any coefficient $f_i$ of... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
68 | Dummit_Foote_exercise_9_4_2b | Dummit Foote | import Mathlib
import Mathlib.Tactic.ComputeDegree
import Mathlib.Tactic.IntervalCases
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $x^6+30x^5-15x^3 + 6x-120$ is irreducible in $\mathbb{Z}[x]$. | theorem Dummit_Foote_exercise_9_4_2b : Irreducible
(X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : Polynomial ℤ) := by
sorry | $$
x^6+30 x^5-15 x^3+6 x-120
$$
The coefficients of the low order.: $30,-15,0,6,-120$
They are divisible by the prime 3 , but $3^2=9$ doesn't divide $-120$. So this polynomial is irreducible over $\mathbb{Z}$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
69 | Dummit_Foote_exercise_9_4_2d | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $\frac{(x+2)^p-2^p}{x}$, where $p$ is an odd prime, is irreducible in $\mathbb{Z}[x]$. | theorem Dummit_Foote_exercise_9_4_2d {p : ℕ} (hp : p.Prime ∧ p > 2)
{f : Polynomial ℤ} (hf : f = (X + 2)^p):
Irreducible (∑ n ∈ (f.support \ {0}), (f.coeff n : Polynomial ℤ) * X ^ (n-1) :
Polynomial ℤ) := by
sorry | $\frac{(x+2)^p-2^p}{x} \quad \quad p$ is on add pprime $Z[x]$
$$
\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}+2^2\left(\begin{array}{l... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
70 | Dummit_Foote_exercise_9_4_11 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. | theorem Dummit_Foote_exercise_9_4_11 :
Irreducible ((MvPolynomial.X 0)^2 + (MvPolynomial.X 1)^2 - 1 : MvPolynomial (Fin 2) ℚ) := by
sorry | $$
p(x)=x^2+y^2-1 \in Q[y][x] \cong Q[y, x]
$$
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's Criterion $x^2+y^2-1$ is irreducibile in $Q[x, y]$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
71 | Herstein_exercise_2_1_21 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Show that a group of order 5 must be abelian. | theorem Herstein_exercise_2_1_21 (G : Type*) [Group G] [Fintype G]
(hG : card G = 5) :
∀ a b : G, a*b = b*a := by
sorry | 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 $a$ and $b$ are inverses ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
72 | Herstein_exercise_2_1_27 | Herstein | 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 $a \in G$. | theorem Herstein_exercise_2_1_27 {G : Type*} [Group G]
[Fintype G] : ∃ (m : ℕ), m > 0 ∧ ∀ (a : G), a ^ m = 1 := by
sorry | 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
$$
a_i^m=a_i^{n_i c}=\left(a_i^{n_i}\right)^c=e^c=e
$$
Henc... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
73 | Herstein_exercise_2_2_5 | Herstein | 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. | theorem Herstein_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
sorry | 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,
$$
\begin{aligned}
& (a b)^5=a^5... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
74 | Herstein_exercise_2_3_17 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $G$ is a group and $a, x \in G$, prove that $C\left(x^{-1} a x\right)=x^{-1} C(a) x$ | theorem Herstein_exercise_2_3_17 {G : Type*} [Group G] (a x : G) :
centralizer {x⁻¹ * a * x} = (fun g : G => x⁻¹ * g * x) '' (centralizer {a}) := by
sorry | Note that
$$
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^{-1} \\
\Longrightarrow & \l... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
75 | Herstein_exercise_2_4_36 | Herstein | 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. | theorem Herstein_exercise_2_4_36 {a n : ℕ} (h : a > 1) :
n ∣ (a ^ n - 1).totient := by
sorry | 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 $a^n$.
This is an impossibility, since ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
76 | Herstein_exercise_2_5_30 | Herstein | 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. | theorem Herstein_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
sorry | 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 characteristic subgroup, that is $\phi(H)=H$ for any automorphism $\phi$ of $G$. Now consider $\phi(H)$. Clearly $|\phi(H)|=p$. Suppose $\phi(H) \neq H$, then $H \cap \phi ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
77 | Herstein_exercise_2_5_37 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $G$ is a nonabelian group of order 6, prove that $G \simeq S_3$. | theorem Herstein_exercise_2_5_37 (G : Type*) [Group G] [Fintype G]
(hG : card G = 6) (hG' : ∃ a b : G, a * b ≠ b * a) :
Nonempty (G ≃* Equiv.Perm (Fin 3)) := by
sorry | 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 are 5 elements, but... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
78 | Herstein_exercise_2_5_44 | Herstein | 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 $p$. | theorem Herstein_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
sorry | 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, then it is abelian and... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
79 | Herstein_exercise_2_6_15 | Herstein | 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 $mn$. | theorem Herstein_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
sorry | 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 the order of $ab... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
80 | Herstein_exercise_2_8_12 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Prove that any two nonabelian groups of order 21 are isomorphic. | theorem Herstein_exercise_2_8_12 {G H : Type*} [Fintype G] [Fintype H]
[Group G] [Group H] (hG : card G = 21) (hH : card H = 21)
(hG1 : ∃ a b : G, a * b ≠ b * a) (hH1 : ∃ a b : H, a * b ≠ b * a) :
Nonempty (G ≃* H) := by
sorry | 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 implies $a b=a$ and so... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
81 | Herstein_exercise_2_9_2 | Herstein | 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. | theorem Herstein_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
sorry | 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 cyclic.
Suppose now tha... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
82 | Herstein_exercise_2_11_6 | Herstein | 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 $G$. | theorem Herstein_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
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
83 | Herstein_exercise_2_11_22 | Herstein | 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 $G$. | theorem Herstein_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
sorry | 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
$$
|G|=|Z(G)|+\sum_{a \notin Z(G)}[G: C(a)],
... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
84 | Herstein_exercise_4_1_19 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Show that there is an infinite number of solutions to $x^2 = -1$ in the quaternions. | theorem Herstein_exercise_4_1_19 : Infinite {x : Quaternion ℝ | x^2 = -1} := by
sorry | Let $x=a i+b j+c k$ then
$$
x^2=(a i+b j+c k)(a i+b j+c k)=-a^2-b^2-c^2=-1
$$
This gives $a^2+b^2+c^2=1$ which has infinitely many solutions for $-1<a, b, c<1$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
85 | Herstein_exercise_4_2_5 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Let $R$ be a ring in which $x^3 = x$ for every $x \in R$. Prove that $R$ is commutative. | theorem Herstein_exercise_4_2_5 {R : Type*} [Ring R]
(h : ∀ x : R, x ^ 3 = x) : ∀ x y : R, x * y = y * x := by
sorry | To begin with
$$
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
$$
Multiply the last relation by $x$ on the left... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
86 | Herstein_exercise_4_2_9 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Let $p$ be an odd prime and let $1 + \frac{1}{2} + ... + \frac{1}{p - 1} = \frac{a}{b}$, where $a, b$ are integers. Show that $p \mid a$. | theorem Herstein_exercise_4_2_9 {p : ℕ} (hp : Nat.Prime p) (hp1 : Odd p) :
∀ (a b : ℤ), (a / b : ℚ) = ∑ i ∈ Finset.range (p-1), (1 / (i + 1) : ℚ) → ↑p ∣ a := by
sorry | First we prove for prime $p=3$ and then for all prime $p>3$.
Let us take $p=3$. Then the sum
$$
\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{(p-1)}
$$
becomes
$$
1+\frac{1}{3-1}=1+\frac{1}{2}=\frac{3}{2} .
$$
Therefore in this case $\quad \frac{a}{b}=\frac{3}{2} \quad$ implies $3 \mid a$, i.e. $p \mid a$.
Now for odd prime ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
87 | Herstein_exercise_4_3_25 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Let $R$ be the ring of $2 \times 2$ matrices over the real numbers; suppose that $I$ is an ideal of $R$. Show that $I = (0)$ or $I = R$. | theorem Herstein_exercise_4_3_25
(I : TwoSidedIdeal (Matrix (Fin 2) (Fin 2) ℝ)) :
I = ⊥ ∨ I = ⊤ := by
sorry | Suppose that $I$ is a nontrivial ideal of $R$, and let
$$
A=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
$$
where not all of $a, b, c d$ are zero. Suppose, without loss of generality -- our steps would be completely analogous, modulo some different placement of 1 s in our matrices, if we assumed some other... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
88 | Herstein_exercise_4_5_16 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Let $F = \mathbb{Z}_p$ be the field of integers $\mod p$, where $p$ is a prime, and let $q(x) \in F[x]$ be irreducible of degree $n$. Show that $F[x]/(q(x))$ is a field having at exactly $p^n$ elements. | theorem Herstein_exercise_4_5_16 {p n: ℕ} (hp : Nat.Prime p)
{q : Polynomial (ZMod p)} (hq : Irreducible q) (hn : q.degree = n) :
(∃ is_fin : Fintype $ Polynomial (ZMod p) ⧸ span ({q}),
@card (Polynomial (ZMod p) ⧸ span {q}) is_fin = p ^ n) ∧
IsField (Polynomial (ZMod p) ⧸ span {q}) := by
sorry | In the previous problem we have shown that any for any $p(x) \in F[x]$, we have that
$$
p(x)+(q(x))=a_{n-1} x^{n-1}+\cdots+a_1 x+a_0+(q(x))
$$
for some $a_{n-1}, \ldots, a_0 \in F$, and that there are $p^n$ choices for these numbers, so that $F[x] /(q(x)) \leq p^n$. In order to show that equality holds, we have to show... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
89 | Herstein_exercise_4_5_25 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $p$ is a prime, show that $q(x) = 1 + x + x^2 + \cdots x^{p - 1}$ is irreducible in $Q[x]$. | theorem Herstein_exercise_4_5_25 {p : ℕ} (hp : Nat.Prime p) :
Irreducible (∑ i ∈ Finset.range p, X ^ i : Polynomial ℚ) := by
sorry | Lemma: Let $F$ be a field and $f(x) \in F[x]$. If $c \in F$ and $f(x+c)$ is irreducible in $F[x]$, then $f(x)$ is irreducible in $F[x]$.
Proof of the Lemma: Suppose that $f(x)$ is reducible, i.e., there exist non-constant $g(x), h(x) \in F[x]$ so that
$$
f(x)=g(x) h(x) .
$$
In particular, then we have
$$
f(x+c)=g(x+c) ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
90 | Herstein_exercise_4_6_3 | Herstein | 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 $Q[x]$. | theorem Herstein_exercise_4_6_3 :
Infinite {a : ℤ | Irreducible (X^7 + 15*X^2 - 30*X + (a : Polynomial ℚ) : Polynomial ℚ)} := by
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
91 | Herstein_exercise_5_2_20 | Herstein | 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 $V$. | theorem Herstein_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
sorry | 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 $(v + Fu) \cap U_i = \... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
92 | Herstein_exercise_5_3_10 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Prove that $\cos 1^{\circ}$ is algebraic over $\mathbb{Q}$. | set_option maxRecDepth 2048 in
theorem Herstein_exercise_5_3_10 : IsAlgebraic ℚ (cos (Real.pi / 180)) := by
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
93 | Herstein_exercise_5_5_2 | Herstein | import Mathlib
open Polynomial
open scoped BigOperators | Prove that $x^3 - 3x - 1$ is irreducible over $\mathbb{Q}$. | theorem Herstein_exercise_5_5_2 : Irreducible (X^3 - 3*X - 1 : Polynomial ℚ) := by
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
94 | Ireland_Rosen_exercise_1_27 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | For all odd $n$ show that $8 \mid n^{2}-1$. | theorem Ireland_Rosen_exercise_1_27 {n : ℕ} (hn : Odd n) : 8 ∣ (n^2 - 1) := by
sorry | 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 $n-1, n+1$, otherwise it wouldn't div... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
95 | Ireland_Rosen_exercise_1_31 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Show that 2 is divisible by $(1+i)^{2}$ in $\mathbb{Z}[i]$. | theorem Ireland_Rosen_exercise_1_31 : (⟨1, 1⟩ : GaussianInt) ^ 2 ∣ 2 := by
sorry | We have $(1+i)^2=1+2 i-1=2 i$, so $2=-i(1+i)^2$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
96 | Ireland_Rosen_exercise_2_21 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators
open scoped ArithmeticFunction ArithmeticFunction.Moebius | Define $\wedge(n)=\log p$ if $n$ is a power of $p$ and zero otherwise. Prove that $\sum_{A \mid n} \mu(n / d) \log d$ $=\wedge(n)$. | theorem Ireland_Rosen_exercise_2_21 {l : ℕ → ℝ}
(hl : ∀ p n : ℕ, p.Prime → 0 < n → l (p ^ n) = Real.log p)
(hl1 : ∀ m : ℕ, ¬ IsPrimePow m → l m = 0) :
l = fun n ↦ (∑ d ∈ Nat.divisors n, ((ArithmeticFunction.moebius (n / d) : ℤ) : ℝ) * log d) := by
sorry | $$
\left\{
\begin{array}{cccl}
\land(n)& = & \log p & \mathrm{if}\ n =p^\alpha,\ \alpha \in \mathbb{N}^* \\
& = & 0 & \mathrm{otherwise }.
\end{array}
\right.
$$
Let $n = p_1^{\alpha_1}\cdots p_t^{\alpha_t}$ the decomposition of $n$ in prime factors. As $\land(d) = 0$ for all divisors of $n$, except for $d ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
97 | Ireland_Rosen_exercise_3_1 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Show that there are infinitely many primes congruent to $-1$ modulo 6 . | theorem Ireland_Rosen_exercise_3_1 : Infinite {p : Nat.Primes // p ≡ -1 [ZMOD 6]} := by
sorry | 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 congruent to 1, then $N \equ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
98 | Ireland_Rosen_exercise_3_5 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Show that the equation $7 x^{3}+2=y^{3}$ has no solution in integers. | theorem Ireland_Rosen_exercise_3_5 : ¬ ∃ x y : ℤ, 7*x^3 + 2 = y^3 := by
sorry | 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 solution in integers. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
99 | Ireland_Rosen_exercise_3_14 | Ireland Rosen | 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 $n^{q-1} \equiv 1(p q)$. | theorem Ireland_Rosen_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
sorry | 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
$$n^{q-1} = (n^{p-1})^k \equiv 1 \pmod p.$$
$p \mid n^{q-1} - 1, q \mid n^{q-1} - 1$, and $p\... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
100 | Ireland_Rosen_exercise_4_5 | Ireland Rosen | import Mathlib
open Real Nat
open scoped BigOperators | Consider a prime $p$ of the form $4 t+3$. Show that $a$ is a primitive root modulo $p$ iff $-a$ has order $(p-1) / 2$. | theorem Ireland_Rosen_exercise_4_5 {p t : ℕ} (hp0 : p.Prime) (hp1 : p = 4 * t + 3)
(a : (ZMod p)ˣ) :
IsPrimitiveRoot (a : ZMod p) (p - 1) ↔
IsPrimitiveRoot (- (a : ZMod p)) ((p - 1) / 2) := by
sorry | Let $a$ a primitive root modulo $p$.
As $a^{p-1} \equiv 1(\bmod p), p \mid\left(a^{(p-1) / 2}-1\right)\left(a^{(p-1) / 2}+1\right)$, so $p \mid a^{(p-1) / 2}-1$ or $p \mid$ $a^{(p-1) / 2}+1$. As $a$ is a primitive root modulo $p, a^{(p-1) / 2} \not \equiv 1(\bmod p)$, so
$$
a^{(p-1) / 2} \equiv-1 \quad(\bmod p) .
$$
He... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
End of preview. Expand in Data Studio
ProofNet-Verified
Resource Links:
- Paper: https://openreview.net/forum?id=5c0RSYyIWW, https://openreview.net/forum?id=NjgaeXNit3
- GitHub: https://github.com/marcusm117/ProofNet-Verified
- HuggingFace: https://huggingface.co/datasets/marcusm117/ProofNet-Verified
- Visualization: https://marcusm117.github.io/ProofNet-Verified/
ProofNet-Verified (ProofNet-V) is an audited, corrected, and verified version of the ProofNet dataset that is upgraded to be compatible with Lean v4.28.0. A visualization of all errors, error classifications, and our corrections can be found at https://marcusm117.github.io/ProofNet-Verified.
ProofNet is a benchmark for theorem proving and autoformalization of undergraduate-level mathematics. It originally consists of 371 data points, each consisting of a formal theorem statement in Lean 3, a natural language theorem statement, and a natural language proof. The problems are primarily drawn from popular undergraduate pure mathematics textbooks and cover topics such as real and complex analysis, linear algebra, abstract algebra, and topology. We intend for ProofNet to be a challenging benchmark that will drive progress in autoformalization and automatic theorem proving
ProofNet# is a cleaner version of ProofNet compatible with Lean v4.7.0 to Lean v4.16.0-rc2, correcting 118 errors in previous Lean 4 implementations of ProofNet like ProofNet-lean4.
We gratefully acknowledge the authors of these projects for their effort and making the data publicly available.
Based on ProofNet and ProofNet#, we made the following modifications:
- Filter out 4 non-proof problems that have inconsistent natural language statement and formal statements, resulting in 367 problems in total.
- Axler|exercise_1_6
- Axler|exercise_1_7
- Rudin|exercise_1_14
- Rudin|exercise_1_19
Upgrade the formal statements to be compatible with
Lean v4.28.0, so that more automated proof search tools like Grind, Canonical, and Lean-SMT are available.Audit the benchmark for both errors in the ground-truth formalizations and ambiguity in the informal statements, identifying 204/367 (55.6%) errors in ProofNet and 97/367 (26.4%) errors in ProofNet#.
Confirm each error by proving the negation of incorrect formalizations, providing trivialized proofs of formalization that are provably true but unfaithful to the informal statement, or giving detailed reasoning for why a formalization is unfaithful or an informal statement is ambiguous.
Correct these errors and verify all 367 formal statements by providing formal proofs for each. Additionally, the faithfulness of these corrections is assessed by both human reviewers and an AI agent until they reach a consensus.
Citation
If you find the ProofNet-Verified benchmark useful for your research, please kindly cite:
@inproceedings{
an2026ground,
title={Ground False: Dissecting Errors in Formal Mathematics Benchmarks},
author={One An$^{*}$ and Marcus J. Min$^{*}$ and Xujie Si and Osbert Bastani},
booktitle={3rd AI for Math Workshop: Toward Self-Evolving Scientific Agents},
year={2026},
url={https://openreview.net/forum?id=5c0RSYyIWW}
}
@inproceedings{
min2026divide,
title={Divide and Abstract: Autoformalization via Decomposition and Abstraction Learning},
author={Marcus J. Min and Yeqi Gao and Wilson Sy and Zhaoyu Li and Xujie Si and Osbert Bastani},
booktitle={The Fourteenth International Conference on Learning Representations},
year={2026},
url={https://openreview.net/forum?id=NjgaeXNit3}
}
Benchmark Audit Summary
| ProofNet | ProofNet# | |
|---|---|---|
| Faithful | 163 (44.4%) | 270 (73.6%) |
| Unfaithful | 194 (52.9%) | 88 (24.0%) |
| Ambiguous NL | 9 (2.5%) | 8 (2.2%) |
| Wrong NL | 1 (0.3%) | 1 (0.3%) |
| Total Errors | 204 (55.6%) | 97 (26.4%) |
| Total | 367 | 367 |
Unfaithful Error Type Breakdown
| Unfaithful Error Type | ProofNet | ProofNet# |
|---|---|---|
| Misunderstanding of informal statement | 94 (25.6%) | 27 (7.4%) |
| Missing premise | 30 (8.2%) | 3 (0.8%) |
| Misunderstanding of Mathlib | 26 (7.1%) | 31 (8.4%) |
| Missing implicit premise | 18 (4.9%) | 18 (4.9%) |
| Misunderstanding of Lean | 10 (2.7%) | 8 (2.2%) |
| Adding extra case | 9 (2.5%) | 0 (0.0%) |
| Adding extra conclusion | 7 (1.9%) | 1 (0.3%) |
| Total Unfaithful | 194 (52.9%) | 88 (24.0%) |
Dataset Format
Each entry in proofnet-verified.jsonl contains:
| Field | Description |
|---|---|
index |
Problem number (1-367) |
name |
Identifier, e.g. Artin_exercise_2_2_9 |
textbook |
Source textbook |
header |
Lean imports and open declarations |
helper |
Auxiliary definitions needed by some problems |
informal_stmt |
Natural language theorem statement |
formal_stmt |
Lean 4 formal statement (ends with sorry) |
informal_proof |
Natural language proof |
formal_proof |
Zipped to avoid data contamination |
Example Entry
import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
/-Informal Statement
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.
-/
theorem Artin_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
sorry
Getting Started
Prerequisites
Building
# Clone the repository
git clone https://github.com/marcusm117/ProofNet-Verified
cd ProofNet-Verified
# Fetch dependencies and pre-built Mathlib artifacts
lake update
Compiling All Statements
To type-check all 367 formal statements in parallel:
uv sync
uv run python scripts/verify_compilation.py
This runs lake env lean on each file with 48 parallel workers (configurable via --jobs).
Python Environment
Python scripts (e.g. scripts/verify_compilation.py) are managed with uv:
# Create .venv and install dependencies (joblib, tqdm)
uv sync
# Run a script through the managed environment
uv run python scripts/verify_compilation.py
Using the Benchmark
For Theorem Proving Evaluation
The proofnet-verified/ directory contains 367 Lean files, each with a single theorem statement ending in sorry. Replace sorry with a valid proof to solve the problem. We recommend using Lean Comparator as a rigorous proof checker.
For Autoformalization Evaluation
Each entry in proofnet-verified.jsonl pairs an informal_stmt with a formal_stmt. We recommend using symbolic checkers like BEq+ over LLM-as-a-Judge to compare the ground truth formal_stmt with the generated candidate formalizations.
For Error Analysis
The error_taxonomy/ directory provides detailed per-problem error classifications, useful for studying common failure modes in mathematical formalization (Lean syntax misunderstandings, Mathlib API misuse, missing premises, etc.).
License
This project is licensed under the Apache License 2.0. See the LICENSE file for details.
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