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Artin|exercise_2_2_9 | 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} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| 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 := |
Artin|exercise_2_4_19 | 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} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| 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 := |
Artin|exercise_2_11_3 | 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 $e=e^{-1} \Longleftrightarrow e^2=e$ ), so since the number of elements of $G$ is even there must be at least one element more, say $e \neq a \in G$, without a pairing, and thus $a=a^{-1} \Longleftrightarrow a^2=e$
\end{proof} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| theorem exercise_2_11_3 {G : Type*} [Group G] [Fintype G]
(hG : Even (card G)) : ∃ x : G, orderOf x = 2 := |
Artin|exercise_3_5_6 | 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 involved in the decomposition of each element of $A$, the set $C$ is countable. But $C$ also clearly generates the vector space $V$. This contradicts the fact that it is a proper subset of the basis $B$ (since $B$ is uncountable).
\end{proof} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| 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 := |
Artin|exercise_6_1_14 | 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 $H \leq G$, we have by definition too that $a H=b H$ if and only if $b^{-1} a \in H$.
In our case, we have that $g Z(G)=x^m Z(G)$, and this happens if and only if $\left(x^m\right)^{-1} g \in Z(G)$.
Then, there's a $z \in Z(G)$ such that $\left(x^m\right)^{-1} g=z$, and so $g=x^m z$.
$g, h \in G$ implies that $g=x^{a_1} z_1$ and $h=x^{a_2} z_2$, so
$$
\begin{aligned}
g h & =\left(x^{a_1} z_1\right)\left(x^{a_2} z_2\right) \\
& =x^{a_1} x^{a_2} z_1 z_2 \\
& =x^{a_1+a_2} z_2 z_1 \\
& =\ldots=\left(x^{a_2} z_2\right)\left(x^{a_1} z_1\right)=h g .
\end{aligned}
$$
Therefore, $G$ is abelian.
\end{proof} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| theorem exercise_6_1_14 (G : Type*) [Group G]
(hG : IsCyclic $ G ⧸ (center G)) :
center G = ⊤ := |
Artin|exercise_6_4_3 | 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 simple.
\end{proof} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| 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 := |
Artin|exercise_6_8_1 | 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} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| 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} := |
Artin|exercise_10_2_4 | 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
$$
(2 x) \subseteq(2) \cap(x)
$$
On the other hand, let $p(x) \in(2) \cap(x)$. Since $p(x) \in(2)$, there exists some polynomial $h(x) \in \mathbb{Z}[x]$ such that
$$
p(x)=2 h(x)
$$
Furthermore, $p(x) \in(x)$, so
$$
p(x)=x h_2(x)
$$
So, $2 h(x)=x h_2(x)$, for some $h_2(x) \in \mathbb{Z}[x]$. This means that $h(0)=0$, so $x$ divides $h(x)$; that is,
$$
h(x)=x q(x)
$$
for some $q(x) \in \mathbb{Z}[x]$, and
$$
p(x)=2 x q(x)
$$
Thus, $p(x) \in(2 x)$, and
$$
\text { (2) } \cap(x) \subseteq(2 x)
$$
Finally,
(2) $\cap(x)=(2 x)$,
as required.
\end{proof} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| theorem exercise_10_2_4 :
span ({2} : Set $ Polynomial ℤ) ⊓ (span {X}) =
span ({2 * X} : Set $ Polynomial ℤ) := |
Artin|exercise_10_4_6 | 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} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| theorem exercise_10_4_6 {R : Type*} [CommRing R]
(I J : Ideal R) (x : ↑(I ⊓ J)) :
IsNilpotent ((Ideal.Quotient.mk (I*J)) x) := |
Artin|exercise_10_7_10 | 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 we cannot have $M^\prime \subsetneq M \subsetneq R$, therefore $M=M^\prime$.
\end{proof} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| 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 := |
Artin|exercise_11_4_1b | Prove that $x^3 + 6x + 12$ is irreducible in $\mathbb{Q}$. | \begin{proof}
Apply Eisenstein's criterion with $p=3$.
\end{proof} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| theorem exercise_11_4_1b :
Irreducible (12 + 6 * X + X ^ 3 : Polynomial ℚ) := |
Artin|exercise_11_4_6b | 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} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| theorem exercise_11_4_6b {F : Type*} [Field F] [Fintype F] (hF : card F = 7) :
Irreducible (X ^ 2 + 1 : Polynomial F) := |
Artin|exercise_11_4_8 | 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} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| theorem exercise_11_4_8 (p : ℕ) (hp : Prime p) (n : ℕ) (hn : n > 0) :
Irreducible (X ^ n - (p : Polynomial ℚ) : Polynomial ℚ) := |
Artin|exercise_13_4_10 | 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} | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators
| 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 := |
Axler|exercise_1_3 | 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} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| theorem exercise_1_3 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] {v : V} : -(-v) = v := |
Axler|exercise_1_6 | 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 fails scalar multiplication. A discrete set like $\mathbb{Z}^2$ does this.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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') := |
Axler|exercise_1_8 | 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 $0$ is in each subspace $V_i$, as they are subspaces and hence contain the additive identity.
Next, we show that the intersection of subspaces is closed under addition. Let $u$ and $v$ be vectors in $V_1 \cap V_2 \cap \ldots \cap V_n$. By definition, $u$ and $v$ belong to each of the subspaces $V_i$. Since each $V_i$ is a subspace and therefore closed under addition, it follows that $u+v$ belongs to each $V_i$. Thus, $u+v$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$.
Finally, we show that the intersection of subspaces is closed under scalar multiplication. Let $a$ be a scalar in $F$ and let $v$ be a vector in $V_1 \cap V_2 \cap \ldots \cap V_n$. Since $v$ belongs to each $V_i$, we have $av$ belongs to each $V_i$ as well, as $V_i$ are subspaces and hence closed under scalar multiplication. Therefore, $av$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$.
Thus, we have shown that $V_1 \cap V_2 \cap \ldots \cap V_n$ is a subspace of $V$.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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 := |
Axler|exercise_3_1 | 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,
$$
\begin{aligned}
T(v) & =T(v \cdot 1) \\
& =v T(1) \ldots .(\text { By the linearity of } T) \\
& =v \lambda
\end{aligned}
$$
Hence, $T v=\lambda v$ for every $v \in V$.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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:= |
Axler|exercise_4_4 | 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 $p^{\prime}$ have no roots in commons, we must now show that $p^{\prime}\left(\lambda_j\right) \neq 0$ for every $j$. So, to do so, just fix $j$. The previous expression for $p$ shows that we can now write $p$ in the form of $p(z)=\left(z-\lambda_j\right) q(z)$, which $q$ is a polynomial such that $q\left(\lambda_j\right) \neq 0$.
When you differentiate both sides of the previous equation, then you would then have $p^{\prime}(z)=(z-$ $\left.\lambda_j\right) q^{\prime}(z)+q(z)$
Therefore: $\left.=p^{\prime}\left(\lambda_j\right)=q \lambda_j\right)$
Equals: $p^{\prime}\left(\lambda_j\right) \neq 0$
Now, to prove the other direction, we would now prove the contrapositive, which means that we will be proving that if $p$ has actually less than $m$ distinct roots, then both $p$ and $p^{\prime}$ have at least one root in common.
Now, for some root of $\lambda$ of $p$, we can write $p$ is in the form of $\left.p(z)=(z-\lambda)^n q(z)\right)$, which is where both $n \geq 2$ and $q$ is a polynomial. When differentiating both sides of the previous equations, we would then have $p^{\prime}(z)=(z-\lambda)^n q^{\prime}(z)+n(z-\lambda)^{n-1} q(z)$.
Therefore, $p^{\prime}(\lambda)=0$, which would make $\lambda$ is a common root of both $p$ and $p^{\prime}$.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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)) := |
Axler|exercise_5_4 | 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{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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) := |
Axler|exercise_5_12 | 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 $T$ is a scalar multiple of the identity, then me must show that $a_v$ is independent of $v$ for $v \in V\{0\}$. We would now want to show that $a_v=a_w$.
First, just make the case of where $(v, w)$ is linearly dependent. Then, there does exist $b \in F$ such that $w=b v$. Now, you would have the following: $a_W w=T w=T(b v)=b T v=b\left(a_v v\right)=a_v w$. This is showing that $a_v=a_w$.
Finally, make the consideration to make $(v, w)$ be linearly independent. Now, we would have the following: $\left.a_{(} v+w\right)(v+w)=T(v+w)=T v+T w=a_v v+a_w w$.
That previous equation implies the following: $\left.\left.\left(a_{(} v+w\right)-a_v\right) v+\left(a_{(} v+w\right)-a_w\right) w=0$. Since $(v, w)$ is linearly independent, this would imply that both $\left.a_{(} v+w\right)=a_v$ and $\left.a_{(} v+w\right)=a_w$. Therefore, $a_v=a_w$.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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 := |
Axler|exercise_5_20 | 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 $v_j$ is also an eigenvector of S, so $S v_j=a_j v_j$ for some $a_j \in F$. For each $j$, we would then have $(S T) v_j=S\left(T v_j\right)=\lambda_j S v_j=a_j \lambda_j v_j$ and $(T S) v_j=T\left(S v_j\right)=a_j T v_j=a_j \lambda_j v_j$. Since both operators, which are $S T$ and $T S$, agree on a basis, then both are equal.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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 := |
Axler|exercise_6_2 | 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 $\|u\| \leq\|u+a v\|$.
Now, to prove the implication in the other direction, we must now let $\|u\| \leq$ $\|u+a v\|$ for all $a \in \mathbb{F}$. Squaring this inequality, we get both:
$$
\begin{gathered}
\|u\|^2 a n d \leq\|u+a v\|^2 \\
=(u+a v, u+a v) \\
=(u, u)+(u, a v)+(a v, u)+(a v, a v) \\
=\|u\|^2+\bar{a}(u, v)+a \overline{(u, v)}+|a|^2\|v\|^2 \\
\|u\|^2+2 \Re \bar{a}(u, v)+|a|^2\|v\|^2
\end{gathered}
$$
for all $a \in \mathbb{F}$.
Therefore,
$$
-2 \Re \bar{a}(u, v) \leq|a|^2\|v\|^2
$$
for all $a \in \mathbb{F}$. In particular, we can let $a$ equal $-t(u, v)$ for $t>0$. Substituting this value for $a$ into the inequality above gives
$$
2 t|(u, v)|^2 \leq t^2|(u, v)|^2\|v\|^2
$$
for all $t>0$.
Step 4
4 of 4
Divide both sides of the inequality above by $t$, getting
$$
2|(u, v)|^2 \leq t \mid(u, v)^2\|v\|^2
$$
for all $t>0$. If $v=0$, then $(u, v)=0$, as desired. If $v \neq 0$, set $t$ equal to $1 /\|v\|^2$ in the inequality above, getting
$$
2|(u, v)|^2 \leq|(u, v)|^2,
$$
which implies that $(u, v)=0$.
\end{proof} | import Mathlib
open InnerProductSpace RCLike ContinuousLinearMap Complex
open scoped BigOperators
| 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‖ := |
Axler|exercise_6_7 | 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
$$
\begin{aligned}
\|u+v\|^2 & =\langle u+v, v+v\rangle \\
& =\|u\|^2+\langle u, v\rangle+\langle v, u\rangle+\|v\|^2 \\
-\|u-v\|^2 & =-\langle u-v, u-v\rangle \\
& =-\|u\|^2+\langle u, v\rangle+\langle v, u\rangle-\|v\|^2 \\
i\|u+i v\|^2 & =i\langle u+i v, u+i v\rangle \\
& =i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle+i\|v\|^2 \\
-i\|u-i v\|^2 & =-i\langle u-i v, u-i v\rangle \\
& =-i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle-i\|v\|^2 .
\end{aligned}
$$
Thus $\left(\|u+v\|^2\right)-\|u-v\|^2+\left(i\|u+i v\|^2\right)-i\|u-i v\|^2=4\langle u, v\rangle.$
\end{proof} | import Mathlib
open InnerProductSpace ContinuousLinearMap Complex
open scoped BigOperators
| 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 := |
Axler|exercise_6_16 | 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} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| theorem exercise_6_16 {K V : Type*} [RCLike K] [NormedAddCommGroup V] [InnerProductSpace K V]
{U : Submodule K V} :
U.orthogonal = ⊥ ↔ U = ⊤ := |
Axler|exercise_7_6 | 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 the same argument can be repeated.
Now we have
$$
\begin{aligned}
\text { range } T & =\left(\text { null } T^*\right)^{\perp} \\
& =(\text { null } T)^{\perp} \\
& =\operatorname{range} T^*,
\end{aligned}
$$
where the first and last equality follow from items (d) and (b) of 7.7.
Hence, range $T=$ range $T^*$.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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) := |
Axler|exercise_7_10 | 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 $T$ repeatedly to both sides of the equation above, we get $T^9 e_j=\left(\lambda_j\right)^9 e_j$ and rei =8ej. Thus $T^8 e_j=\left(\lambda_j\right)^8 e_j$, which implies that $\lambda_j$ equals 0 or 1 . In particular, all the eigenvalues of $T$ are real. This would then imply that $T$ is self-adjoint.
Now, by applying $T$ to both sides of the equation above, we get
$$
\begin{aligned}
T^2 e_j & =\left(\lambda_j\right)^2 e_j \\
& =\lambda_j e_j \\
& =T e_j
\end{aligned}
$$
which is where the second equality holds because $\lambda_j$ equals 0 or 1 . Because $T^2$ and $T$ agree on a basis, they must be equal.
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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 := |
Axler|exercise_7_14 | 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 an eigenvalue $\lambda^{\prime}$ of $T$ such that $\left|\lambda^{\prime}-\lambda\right|^2$ is minimized.
There are $a_1, \ldots, a_n \in \mathbb{F}$ such that
$$
v=a_1 e_1+\cdots+a_n e_n .
$$
Thus, we have
$$
\begin{aligned}
\epsilon^2 & >|| T v-\left.\lambda v\right|^2 \\
& =\left|\left\langle T v-\lambda v, e_1\right\rangle\right|^2+\cdots+\left|\left\langle T v-\lambda v, e_n\right\rangle\right|^2 \\
& =\left|\lambda_1 a_1-\lambda a_1\right|^2+\cdots+\left|\lambda_n a_n-\lambda a_n\right|^2 \\
& =\left|a_1\right|^2\left|\lambda_1-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda_n-\lambda\right|^2 \\
& \geq\left|a_1\right|^2\left|\lambda^{\prime}-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda^{\prime}-\lambda\right|^2 \\
& =\left|\lambda^{\prime}-\lambda\right|^2
\end{aligned}
$$
where the second and fifth lines follow from $6.30$ (the fifth because $\|v\|=1$ ). Now, we taking the square root.
Hence, $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda^{\prime}-\lambda\right|<\epsilon$
\end{proof} | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators
| 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) := |
Dummit-Foote|exercise_1_1_3 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_1_1_3 (n : ℕ) :
∀ (x y z : ZMod n), (x + y) + z = x + (y + z) := |
Dummit-Foote|exercise_1_1_5 | 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
$$
\overline{0} \cdot \bar{k}=\overline{0}
$$
for all k, so that $\overline{0}$ does not have a multiplicative inverse. Hence $\mathbb{Z} /(n)$ is not a group under multiplication.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_1_1_5 (n : ℕ) (hn : 1 < n) :
IsEmpty (Group (ZMod n)) := |
Dummit-Foote|exercise_1_1_16 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_1_1_16 {G : Type*} [Group G]
(x : G) : x ^ 2 = 1 ↔ (orderOf x = 1 ∨ orderOf x = 2) := |
Dummit-Foote|exercise_1_1_18 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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) := |
Dummit-Foote|exercise_1_1_22a | 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 some $n \geq 1$; then
$$
\left(b^{-1} a b\right)^{n+1}=\left(b^{-1} a b\right)\left(b^{-1} a b\right)^n=b^{-1} a b b^{-1} a^n b=b^{-1} a^{n+1} b .
$$
By induction the statement holds for all positive $n$.
Now suppose $n<0$; we have
$$
\left(b^{-1} a b\right)^n=\left(\left(b^{-1} a b\right)^{-n}\right)^{-1}=\left(b^{-1} a^{-n} b\right)^{-1}=b^{-1} a^n b .
$$
Hence, the statement holds for all integers $n$.
Now to the main result. Suppose first that $|x|$ is infinity and that $\left|g^{-1} x g\right|=n$ for some positive integer $n$. Then we have
$$
\left(g^{-1} x g\right)^n=g^{-1} x^n g=1,
$$
and multiplying on the left by $g$ and on the right by $g^{-1}$ gives us that $x^n=1$, a contradiction. Thus if $|x|$ is infinity, so is $\left|g^{-1} x g\right|$. Similarly, if $\left|g^{-1} x g\right|$ is infinite and $|x|=n$, we have
$$
\left(g^{-1} x g\right)^n=g^{-1} x^n g=g^{-1} g=1,
$$
a contradiction. Hence if $\left|g^{-1} x g\right|$ is infinite, so is $|x|$.
Suppose now that $|x|=n$ and $\left|g^{-1} x g\right|=m$ for some positive integers $n$ and $m$. We have
$$
\left(g^{-1} x g\right)^n=g^{-1} x^n g=g^{-1} g=1,
$$
So that $m \leq n$, and
$$
\left(g^{-1} x g\right)^m=g^{-1} x^m g=1,
$$
so that $x^m=1$ and $n \leq m$. Thus $n=m$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_1_1_22a {G : Type*} [Group G] (x g : G) :
orderOf x = orderOf (g⁻¹ * x * g) := |
Dummit-Foote|exercise_1_1_25 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_1_1_25 {G : Type*} [Group G]
(h : ∀ x : G, x ^ 2 = 1) : ∀ a b : G, a*b = b*a := |
Dummit-Foote|exercise_1_1_34 | 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 particular, we have
$$
\left\{x^i \mid 0 \leq i \leq n-1\right\} \subseteq G,
$$
so that $|x|=n \leq|G|$
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_1_6_4 | 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 particular, $1>\left|x^n\right|$ for all $n$. So $x$ has infinite order in $\mathbb{R}^{\times}$.
Similarly, if $|x|>1$ (absolute value) then $x$ has infinite order in $\mathbb{R}^{\times}$. So $\mathbb{R}^{\times}$has 1 element of order 1,1 element of order 2 , and all other elements have infinite order.
In $\mathbb{C}^{\times}$, on the other hand, $i$ has order 4 . Thus $\mathbb{R}^{\times}$and $\mathbb{C}^{\times}$are not isomorphic.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_1_6_4 :
IsEmpty (Multiplicative ℝ ≃* Multiplicative ℂ) := |
Dummit-Foote|exercise_1_6_17 | 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
$$
a b=\left(b^{-1} a^{-1}\right)^{-1}=\varphi\left(b^{-1} a^{-1}\right)=\varphi\left(b^{-1}\right) \varphi\left(a^{-1}\right)=\left(b^{-1}\right)^{-1}\left(a^{-1}\right)^{-1}=b a,
$$
so that $G$ is abelian.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_2_1_5 | 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. Thus no such subgroup exists.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_2_4_4 | 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 $\langle H-\{1\}\rangle$, the reverse inclusion also holds so that $\langle H-\{1\}\rangle=$ $H$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_2_4_4 {G : Type*} [Group G] (H : Subgroup G) :
closure ((H : Set G) \ {1}) = H := |
Dummit-Foote|exercise_2_4_16b | 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 contain a reflection $s r^k$ for $k \in\{0,1, \ldots, n-1\}$. Then since $s r^k \in K$ and $r^{n-k} \in K$, it follows by closure that
$$
s=\left(s r^k\right)\left(r^{n-k}\right) \in K .
$$
But $D_{2 n}=\langle r, s\rangle$, so this shows that $K=D_{2 n}$, which is a contradiction. Therefore $H$ must be maximal.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 = ⊤ := |
Dummit-Foote|exercise_3_1_3a | 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 $A$ is normal; in particular, $B$ is normal, so $A / B$ is a group. Now let $x B, y B \in A / B$. Then
$$
(x B)(y B)=(x y) B=(y x) B=(y B)(x B) .
$$
Hence $A / B$ is abelian.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_3_1_3a {A : Type*} [CommGroup A] (B : Subgroup A) :
∀ a b : A ⧸ B, a*b = b*a := |
Dummit-Foote|exercise_3_1_22b | 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 definition of the intersection, this shows that $g a g^{-1} \in \bigcap_{i \in I} H_i$ and therefore it is a normal subgroup.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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):= |
Dummit-Foote|exercise_3_2_11 | 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 subgroup of $G$.
For any $g \in G$ and any $n \in N$, we must have $n \in N_\alpha$ for each $\alpha$. And since $N_\alpha \unlhd G$, we have $g n g^{-1} \in N_\alpha$ for each $\alpha$. Therefore $g n g^{-1} \in N$, which shows that $g N g^{-1} \subseteq N$ for each $g \in G$. As before, this is enough to complete the proof.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_3_2_11 {G : Type*} [Group G] {H K : Subgroup G}
(hHK : H ≤ K) :
H.index = K.index * H.relindex K := |
Dummit-Foote|exercise_3_2_21a | 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 group is finite. Equivalently, $[D: A]$ is not finite.
Because $\mathbb{Q}$ and $\mathbb{Q} / \mathbb{Z}$ are divisible, the conclusion follows.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_3_2_21a (H : AddSubgroup ℚ) (hH : H ≠ ⊤) : H.index = 0 := |
Dummit-Foote|exercise_3_4_1 | 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 nontrivial normal subgroup, so $G$ is not simple.
Suppose $G$ is finite; say $|G|=n$. If $n$ is composite, say $n=p m$ for some prime $p$ with $m \neq 1$, then by Cauchy's Theorem $G$ contains an element $x$ of order $p$ and $\langle x\rangle$ is a nontrivial normal subgroup. Hence $G$ is not simple. Thus if $G$ is an abelian simple group, then $|G|=p$ is prime. We saw previously that the only such group up to isomorphism is $\mathbb{Z} /(p)$, so that $G \cong \mathbb{Z} /(p)$. Moreover, these groups are indeed simple.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_3_4_1 (G : Type*) [CommGroup G] [IsSimpleGroup G] :
IsCyclic G ∧ ∃ G_fin : Fintype G, Nat.Prime (@card G G_fin) := |
Dummit-Foote|exercise_3_4_5a | 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 $H_i \leq H_{i+1}$ for each $i$. Moreover, if $g \in H_{i+1}$ and $x \in H_i$, then in particular $g \in G_{i+1}$ and $x \in G_i$, so that
$$
g x g^{-1} \in G_i
$$
because $G_i \unlhd G_{i+1}$. But $g$ and $x$ also belong to $H$, so
$$
g x g^{-1} \in H_i,
$$
which shows that $H_i \unlhd H_{i+1}$ for each $i$.
Next, note that
$$
H_i=G_i \cap H=\left(G_i \cap G_{i+1}\right) \cap H=G_i \cap H_{i+1} .
$$
By the Second Isomorphism Theorem, we then have
$$
H_{i+1} / H_i=H_{i+1} /\left(H_{i+1} \cap G_i\right) \cong H_{i+1} G_i / G_i \leq G_{i+1} / G_i .
$$
Since $H_{i+1} / H_i$ is isomorphic to a subgroup of the abelian group $G_{i+1} / G_i$, it follows that $H_{i+1} / H_i$ is also abelian. This completes the proof that $H$ is solvable.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_3_4_5a {G : Type*} [Group G]
(H : Subgroup G) [IsSolvable G] : IsSolvable H := |
Dummit-Foote|exercise_3_4_11 | 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 $H_{i+1} / H_i$ is abelian. But then the first group in the series
$$
H_1 / 1 \cong H
$$
is an abelian subgroup of $H$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_4_2_14 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_4_3_26 | 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 $b=g \cdot a$. Let us see in what relationship the stabilizers of $a$ and $b$ are. We find
$$
\begin{aligned}
G_b & =\{h \in G \mid h \cdot b=b\} \\
& =\{h \in G \mid h g \cdot a=g \cdot a\} \\
& =\left\{h \in G \mid g^{-1} h g \cdot a=a\right\}
\end{aligned}
$$
Putting $h^{\prime}=g^{-1} h g$, we have $h=g h^{\prime} g^{-1}$ and
$$
\begin{aligned}
G_b & =g\left\{h^{\prime} \in H \mid h^{\prime} \cdot a=a\right\} g^{-1} \\
& =g G_a g^{-1}
\end{aligned}
$$
By the above, the stabilizer subgroup of any element is conjugate to some other stabilizer subgroup. Now, the stabilizer cannot be all of $G$ (else $\{a\}$ would be a orbit). Thus it is a proper subgroup of $G$. By the previous exercise, we have
$$
\bigcup_{a \in A} G_a=\bigcup_{g \in G} g G_a g^{-1} \subset G
$$
(the union of conjugates of a proper subgroup can never be all of $G$ ). This shows there is an element $\sigma$ which is not in any stabilizer of any element of $A$. Then $\sigma(a) \neq a$ for all $a \in A$, as we wanted to show.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_4_4_6a | 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 $\phi_g(H) \subset H \Longrightarrow$ $g H g^{-1} \subset H$. So, $H$ is normal.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_4_4_6a {G : Type*} [Group G] (H : Subgroup G)
[Characteristic H] : Normal H := |
Dummit-Foote|exercise_4_4_7 | 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, we have $\sigma(H)=$ $H$ for every $\sigma \in \operatorname{Aut}(G)$. Hence, $H$ is a characterestic subgroup of $G$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_4_5_1a | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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) := |
Dummit-Foote|exercise_4_5_14 | 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 choices for $k$ is $0$. Hence, there is a unique $13-$Sylow subgroup and hence is normal.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_4_5_16 | 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 $n_r$ can be 1 or $p q$. We assume $n_r=p q$. Now we have $n_q=1+q k$ such that $1+q k \mid p r$. Now,as $p<q<r, n_q$ can be 1 or $r$, or $p r$. Assume that $n_q=r$. Now we turn to $n_p$. Again my similar method we can conclude $n_p$ can be $1, q, r$, or $q r$. We assume that $n_p$ is $q$. Now we count the number of elements of order $p, q, r$. Since $n_r=p q$, the number of elements of order $r$ is $p q(r-1)$. Since $n_q=r$, the number of elements of order $q$ is $(q-1) r$. And as $n_p=q$, the number of elements of order $p$ is $(p-1) q$. So, in total we get $p q(r-1)+(q-1) r+(p-$ 1) $q=p q r+q r-r-q=p q r+r(q-1)-r$. But observe that as $q>1, r(q-1)-r>$ 0 . So the number of elements exceeds $p q r$. So, it proves that atleast $n_p$ or $n_q$ or $n_r$ is 1, which ultimately proves the result, because a unique Sylow-p subgroup is always normal.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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) := |
Dummit-Foote|exercise_4_5_18 | 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 $n_5$ divides 8 . The only such number that divides 8 and equals $1 (\bmod 5)$ is 1 so $n_5=1$. Hence $P$ is the unique Sylow 5-subgroup.
Since $P$ is the unique Sylow 5-subgroup, this implies that $P$ is normal in $G$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_4_5_18 {G : Type*} [Fintype G] [Group G]
(hG : card G = 200) :
∃ N : Sylow 5 G, N.Normal := |
Dummit-Foote|exercise_4_5_20 | 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 $G$ is not simple.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_4_5_20 {G : Type*} [Fintype G] [Group G]
(hG : card G = 1365) : ¬ IsSimpleGroup G := |
Dummit-Foote|exercise_4_5_22 | 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 $n_{11}=1+11k$ where $1+11k|12$. The choices for $k$ are $0$ or $1$. If $k=0$, there is only one $11-$Sylow subgroup and hence normal. So, assume now, that there are $12$ $11-$Sylow subgroup(for $k=1$). Now we look at $3-$ Sylow subgroups. $n_{3}=1+3k| 44$. So choice for $k$ are $0$, $1$, and $7$. If $k=0$, there is only one $3-$Sylow subgroup and hence normal. So, assume now, that there are $4$ $2-$Sylow subgroup (for $k=3$). Now we claim that simultaneously, there cannot be $12$ $11-$Sylow subgroup and $4$ $3-$Sylow subgroups provided there is more than one $2-$Sylow subgroups. So, either $2-$Sylow subgroup is normal or if not, then, either $11-$Sylow subgroup is normal being unique, or the $3-$Sylow subgroup is normal(We don't consider the possibility of $22$ $3-$Sylow subgroup because of obvious reason). Now, to prove the claim, we observe that there are $120$ elements of order $11$. Also there are $8$ elements of order $3$. So we already get $120+8+1=129$ distinct elements in the group. Let us count the number of $2-$Sylow subgroups in $G$. $n_{2}=1+2k|33$. The possibilities for $k$ are $0$, $1$, $5$, $16$. Now, assume there is more than one $2-$Sylow subgroups. Let $H_{1}$ and $H_{2}$ be two distinct $2-$Sylow subgroup. Now $|H_{1}|=4$. So we already get $129+3=132$ distinct elements in the group. Now $H_{2}$ being distinct from $H_{1}$, has at least one element which is not in $H_{1}$. This adds one more element in the group, at the least. Now already we have number of elements in the group exceeding the number of element in $G$. This gives a contradiction and proves the claim.
Hence $G$ is not simple.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_4_5_22 {G : Type*} [Fintype G] [Group G]
(hG : card G = 132) : ¬ IsSimpleGroup G := |
Dummit-Foote|exercise_4_5_28 | 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, Consider $G / P$, which has order $5^2 .7$. Now, the number of Sylow $-5$ subgroup of $G / P$ is given by $1+5 k$, where $1+5 k \mid 7$. Clearly $k=0$ is the only choice and hence there is a unique Sylow-5 subgroup of $G / P$, and hence normal. In the same way Sylow-7 subgroup of $G / P$ is also unique and hence normal. Consider now the canonical map $\pi: G \rightarrow G / P$. The inverse image of Sylow-5 subgroup of $G / P$ under $\pi$, call it $H$, is a normal subgroup of $G$, and $|H|=3^2 .5^2$. Similarly, the inverse image of Sylow-7 subgroup of $G / P$ under $\pi$ call it $K$ is also normal in $G$ and $|K|=3^2 .7$. Now, consider $H$. Observe first that the number of Sylow-5 subgroup in $H$ is $1+5 k$ such that $1+5 k \mid 9$. Again $k=0$ and hence $H$ has a unique Sylow-5 subgroup, call it $P_1$. But, it is easy to see that $P_1$ is also a Sylow-5 subgroup of $G$, because $\left|P_1\right|=25$. But now any other Sylow 5 subgroup of $G$ is of the form $g P_1 g^{-1}$ for some $g \in G$. But observe that since $P_1 \subset H$ and $H$ is normal in $G$, so $g P_1 g^1 \subset H$, and $g P_1 g^{-1}$ is also Sylow-5 subgroup of $H$. But, then as Sylow-5 subgroup of $H$ is unique we have $g P_1 g^{-1}=P_1$. This shows that Sylow-5 subgroup of $G$ is unique and hence normal in $G$.
Similarly, one can argue the same for $K$ and deduce that Sylow-7 subgroup of $G$ is unique and hence normal. So, the first part of the problem is done.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_5_4_2 | 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 $g^{-1} h g \in H, \forall g \in G, \forall h \in H$, note that $h^{-1} \in H$, so multiplying them, we also obtain an element of $H$.
On the other hand, if $h^{-1} g^{-1} h g \in H, \forall g \in G, \forall h \in H$, then
$$
h h^{-1} g^{-1} h g=g^{-1} h g \in H, \forall g \in G, \forall h \in H .
$$
Since $\left\{h^{-1} g^{-1} h g: h \in H, g \in G\right\} \subseteq H \Leftrightarrow\left\langle\left\{h^{-1} g^{-1} h g: h \in H, g \in G\right\}\right\rangle \leq H$, we've solved the exercise by definition of $[H, G]$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_5_4_2 {G : Type*} [Group G] (H : Subgroup G) :
H.Normal ↔ ⁅(⊤ : Subgroup G), H⁆ ≤ H := |
Dummit-Foote|exercise_7_1_11 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_7_1_11 {R : Type*} [CommRing R] [IsDomain R]
{x : R} (hx : x^2 = 1) : x = 1 ∨ x = -1 := |
Dummit-Foote|exercise_7_1_15 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_7_1_15 {R : Type*} [Ring R] (hR : ∀ a : R, a^2 = a) :
∀ a b : R, a*b = b*a := |
Dummit-Foote|exercise_7_2_12 | 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.
$$
\begin{aligned}
N M &=\left(\sum_{i=1}^n g_i\right)\left(\sum_{j=1}^n r_j g_j\right) \\
&=\sum_{j=1}^n \sum_{i=1}^n r_j g_i g_j \\
&=\sum_{j=1}^n \sum_{i=1}^n r_j g_j g_j^{-1} g_i g_j \\
&=\sum_{j=1}^n r_j g_j\left(\sum_{i=1}^n g_j^{-1} g_i g_j\right) \\
&=\sum_{j=1}^n r_j g_j\left(\sum_{i=1}^n g_i\right) \\
&=\left(\sum_{j=1}^n r_j g_j\right)\left(\sum_{i=1}^n g_i\right) \\
&=M N .
\end{aligned}
$$
Thus $N \in Z(R[G])$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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) := |
Dummit-Foote|exercise_7_3_37 | 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
$$
\alpha=\sum\left(x_i+J\right)\left(y_i+J\right) \in\left(I_1 / J\right)\left(I_2 / J\right) .
$$
Then
$$
\alpha=\sum\left(x_i y_i+J\right)=\left(\sum x_i y_i\right)+J \in\left(I_1 I_2\right) / J .
$$
Now let $\alpha=\left(\sum x_i y_i\right)+J \in\left(I_1 I_2\right) / J$. Then
$$
\alpha=\sum\left(x_i+J\right)\left(y_i+J\right) \in\left(I_1 / J\right)\left(I_2 / J\right) .
$$
From this lemma and the lemma to Exercise 7.3.36, it follows by an easy induction that
$$
\left(p \mathbb{Z} / p^m \mathbb{Z}\right)^m=(p \mathbb{Z})^m / p^m \mathbb{Z}=p^m \mathbb{Z} / p^m \mathbb{Z} \cong 0 .
$$
Thus $p \mathbb{Z} / p^m \mathbb{Z}$ is nilpotent in $\mathbb{Z} / p^m \mathbb{Z}$.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_7_3_37 {p m : ℕ} (hp : p.Prime) :
IsNilpotent (span ({↑p} : Set $ ZMod $ p^m) : Ideal $ ZMod $ p^m) := |
Dummit-Foote|exercise_8_1_12 | 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 $d \bmod \varphi(N)$: $d d^{\prime} \equiv 1 \pmod {\varphi(N)}$. | \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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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] := |
Dummit-Foote|exercise_8_3_4 | 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 $j \in \mathbb{N} \cup\{0\}$ such that $q^j$ divides $b d$. Then $q^{i-2 j}$ divides $n$. But since $i$ is even, $i-2 j$ is even as well. Consequently, $n$ can be written as a sum of two squared integers.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_8_3_4 {n : ℤ} {r s : ℚ}
(h : r^2 + s^2 = n) :
∃ a b : ℤ, a^2 + b^2 = n := |
Dummit-Foote|exercise_8_3_6a | 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 $\langle 1+i\rangle$ or $1+\langle 1+i\rangle$, so $\mathbb{Z}[i] /\langle 1+i\rangle$ is a finite ring of order 2 , which must be a field.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_9_1_6 | 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:
$$
\begin{aligned}
& 0=\operatorname{deg}_y(x)=\operatorname{deg}_y(s)+\operatorname{deg}_y(p) \text { so } \\
& 0=\operatorname{deg}_y(p) \\
& 0=\operatorname{deg}_x(y)=\operatorname{deg}_x(s)+\operatorname{deg}_x(p) \text { so } \\
& 0=\operatorname{deg}_x(p) \text { so }
\end{aligned}
$$
From : $\quad 0=\operatorname{deg}_y(p)=\operatorname{deg}_x(p)$ we get $\operatorname{deg}(p)=0$ and $p \in \mathbb{Q}$.
But $p \in(p)=(x, y)$ so $p=a x+b y$ for some $a, b \in \mathbb{Q}[x, y]$
$$
\begin{aligned}
\operatorname{deg}(p) & =\operatorname{deg}(a x+b y) \\
& =\min (\operatorname{deg}(a)+\operatorname{deg}(x), \operatorname{deg}(b)+\operatorname{deg}(y)) \\
& =\min (\operatorname{deg}(a)+1, \operatorname{deg}(b)+1) \geqslant 1
\end{aligned}
$$
which contradicts $\operatorname{deg}(p)=0$.
So we conclude that $(x, y)$ is not principal ideal in $\mathbb{Q}[x, y]$
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_9_1_6 : ¬ Submodule.IsPrincipal
(span ({MvPolynomial.X 0, MvPolynomial.X 1} : Set (MvPolynomial (Fin 2) ℚ))) := |
Dummit-Foote|exercise_9_3_2 | 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 any coefficient $f_i$ of $f(x)$ and $g_j$ of $g(x)$ we have that $r f_i, r^{-1} g_j \in$ $\mathbb{Z}$ and by multiplicative closure and commutativity of $\mathbb{Z}$ we have that $r f_i r^{-1} g_j=$ $f_i g_j \in \mathbb{Z}$
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| 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 := |
Dummit-Foote|exercise_9_4_2b | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_9_4_2b : Irreducible
(X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : Polynomial ℤ) := |
Dummit-Foote|exercise_9_4_2d | 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
$$
\begin{aligned}
& x^{p-1}+2\left(\begin{array}{l}
p \\
1
\end{array}\right) x^{p-2}+2^2\left(\begin{array}{l}
p \\
2
\end{array}\right) x^{p-3}+\ldots+2^{p-1}\left(\begin{array}{c}
p \\
p-1
\end{array}\right) \\
& 2^k\left(\begin{array}{l}
p \\
k
\end{array}\right) x^{p-k-1}=2^k \cdot p \cdot(p-1) \ldots(p-k-1), \quad 0<k<p
\end{aligned}
$$
Every lower order coef. has $p$ as a factor but doesnt have $\$ \mathrm{p}^{\wedge} 2 \$$ as a fuction so the polynomial is irreducible by Eisensteins Criterion.
\end{proof} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_9_4_2d {p : ℕ} (hp : p.Prime ∧ p > 2)
{f : Polynomial ℤ} (hf : f = (X + 2)^p):
Irreducible (∑ n in (f.support \ {0}), (f.coeff n : Polynomial ℤ) * X ^ (n-1) :
Polynomial ℤ) := |
Dummit-Foote|exercise_9_4_11 | 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} | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators
| theorem exercise_9_4_11 :
Irreducible ((MvPolynomial.X 0)^2 + (MvPolynomial.X 1)^2 - 1 : MvPolynomial (Fin 2) ℚ) := |
Herstein|exercise_2_1_21 | 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 $a$ and $b$ are inverses and hence $a * b=b * a$.
Contradiction. If $a * b=a$ (or $=b$ ), then $b=e$ (or $a=e$ ) and $e$ commutes with everything. Contradiction. We know by supposition that $a * b \neq b * a$. Hence all the elements $\{e, a, b, a * b, b * a\}$ are distinct.
Now consider $a^2$. It can't equal $a$ as then $a=e$ and it can't equal $a * b$ or $b * a$ as then $b=a$. Hence either $a^2=e$ or $a^2=b$.
Now consider $a * b * a$. It can't equal $a$ as then $b * a=e$ and hence $a * b=b * a$. Similarly it can't equal $b$. It also can't equal $a * b$ or $b * a$ as then $a=e$. Hence $a * b * a=e$.
So then we additionally see that $a^2 \neq e$ because then $a^2=e=$ $a * b * a$ and consequently $a=b * a$ (and hence $b=e$ ). So $a^2=b$. But then $a * b=a * a^2=a^2 * a=b * a$. Contradiction.
Hence starting with the assumption that there exists an order 5 abelian group $G$ leads to a contradiction. Thus there is no such group.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_2_1_21 (G : Type*) [Group G] [Fintype G]
(hG : card G = 5) :
∀ a b : G, a*b = b*a := |
Herstein|exercise_2_1_27 | 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
$$
a_i^m=a_i^{n_i c}=\left(a_i^{n_i}\right)^c=e^c=e
$$
Hence $m$ is a positive integer such that $a^m=e$ for all $a \in G$.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_2_1_27 {G : Type*} [Group G]
[Fintype G] : ∃ (m : ℕ), m > 0 ∧ ∀ (a : G), a ^ m = 1 := |
Herstein|exercise_2_2_5 | 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,
$$
\begin{aligned}
& (a b)^5=a^5 b^5, \text { for all } a, b \in G \\
\Longrightarrow & (a b)(a b)(a b)(a b)(a b)=a\left(a^4 b^4\right) b \\
\Longrightarrow & a(b a)(b a)(b a)(b a) b=a\left(a^4 b^4\right) b \\
\Longrightarrow & (b a)^4=a^4 b^4, \text { by cancellation law. }
\end{aligned}
$$
Now by combining two cases we have
$$
\begin{aligned}
& (b a)^4=a^4 b^4 \\
\Longrightarrow & \left((b a)^2\right)^2=a^2\left(a^2 b^2\right) b^2 \\
\Longrightarrow & \left(a^2 b^2\right)^2=a^2\left(a^2 b^2\right) b^2 \\
\Longrightarrow & \left(a^2 b^2\right)\left(a^2 b^2\right)=a^2\left(a^2 b^2\right) b^2 \\
\Longrightarrow & a^2\left(b^2 a^2\right) b^2=a^2\left(a^2 b^2\right) b^2 \\
\Longrightarrow & b^2 a^2=a^2 b^2, \text { by cancellation law. } \\
\Longrightarrow & b^2 a^2=(b a)^2, \text { since }(b a)^2=a^2 b^2 \\
\Longrightarrow & b(b a) a=(b a)(b a) \\
\Longrightarrow & b(b a) a=b(a b) a \\
\Longrightarrow & b a=a b, \text { by cancellation law. }
\end{aligned}
$$
It follows that, $a b=b a$ for all $a, b \in G$. Hence $G$ is abelian
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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 := |
Herstein|exercise_2_3_17 | 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,
$$
\begin{aligned}
& p\left(x^{-1} a x\right)=\left(x^{-1} a x\right) p \\
\Longrightarrow & \left(p x^{-1} a\right) x=x^{-1}(a x p) \\
\Longrightarrow & x\left(p x^{-1} a\right)=(a x p) x^{-1} \\
\Longrightarrow & \left(x p x^{-1}\right) a=a\left(x p x^{-1}\right) \\
\Longrightarrow & x p x^{-1} \in C(a) .
\end{aligned}
$$
Therefore,
$$
p \in C\left(x^{-1} a x\right) \Longrightarrow x p x^{-1} \in C(a) .
$$
Thus,
$$
C\left(x^{-1} a x\right) \subset x^{-1} C(a) x .
$$
Let us assume
$$
q \in x^{-1} C(a) x .
$$
Then there exists an element $y$ in $C(a)$ such that
$$
q=x^{-1} y x
$$
Now,
$$
y \in C(a) \Longrightarrow y a=a y .
$$
Also,
$$
q\left(x^{-1} a x\right)=\left(x^{-1} y x\right)\left(x^{-1} a x\right)=x^{-1}(y a) x=x^{-1}(y a) x=\left(x^{-1} y x\right)\left(x^{-1} a x\right)=\left(x^{-1} y x\right) q .
$$
Therefore,
$$
q\left(x^{-1} a x\right)=\left(x^{-1} y x\right) q
$$
So,
$$
q \in C\left(x^{-1} a x\right) .
$$
Consequently we have
$$
x^{-1} C(a) x \subset C\left(x^{-1} a x\right) .
$$
It follows from the aforesaid argument
$$
C\left(x^{-1} a x\right)=x^{-1} C(a) x .
$$
This completes the proof.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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}) := |
Herstein|exercise_2_4_36 | 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 $a^n$.
This is an impossibility, since $d$ divides $a^n-1$ by our assumption. Consequently, $d$ divides 1 , which implies $d=1$. Hence we are contradict to the fact that $d>1$. Therefore
$$
\operatorname{Gcd}\left(a, a^n-1\right)=1 .
$$
Then $a \in U_{a^n-1}$, where $U_n$ is a group defined by
$$
U_n:=\left\{\bar{a} \in \mathbb{Z}_n \mid \operatorname{Gcd}(a, n)=1\right\} .
$$
We know that order of an element divides the order of the group. Here order of the group $U_{a^n-1}$ is $\phi\left(a^n-1\right)$ and $a \in U_{a^n-1}$. This follows that $\mathrm{o}(a)$ divides $\phi\left(a^n-1\right)$.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_2_4_36 {a n : ℕ} (h : a > 1) :
n ∣ (a ^ n - 1).totient := |
Herstein|exercise_2_5_30 | 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 $\phi(H) \neq H$, then $H \cap \phi (H)=\{ e\}$. Consider $H \phi(H)$, this is a subgroup of $G$ as $H$ is normal. Also $|H \phi(H)|=p^2$. By lagrange's theorem then $p^2 \mid$ $p m \Longrightarrow p \mid m$ - contradiction. So $\phi(H)=H$, and $H$ is characterestic subgroup of $G$
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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 := |
Herstein|exercise_2_5_37 | 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 are 5 elements, but order 3 elements occur in pair, that is $a, a^2$, both have order 3 , and $a \neq a^2$. So, this is a contradiction, as there are only 5 elements. So, there must be an element of order 2 . All elements of order 2 will imply that $G$ is abelian, hence there is also element of order 3 . Let $a$ be an element of order 2 , and $b$ be an element of order 3 . So we have $e, a, b, b^2$, already 4 elements. Now $a b \neq e, b, b^2$. So $a b$ is another element distinct from the ones already constructed. $a b^2 \neq e, b, a b, b^2, a$. So, we have got another element distinct from the other. So, now $ G=\left\{e, a, b, b^2, a b, a b^2\right\}$. Also, ba must be equal to one of these elements. But $b a \neq e, a, b, b^2$. Also if $b a=a b$, the group will become abelian. so $b a=a b^2$. So what we finally get is $G=\left\langle a, b \mid a^2=e=b^3, b a=a b^2\right\rangle$. Hence $G \cong S_3$.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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)) := |
Herstein|exercise_2_5_44 | 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, then it is abelian and any subgroup of order $p$ is normal. Now let us suppose that $G$ is not cyclic, then there exists an element $a$ of order $p$, and $A=\langle a\rangle$. Now $i_G(A)=p$, so $p^2 \nmid p$ ! , hence by the above result there is a normal subgroup $K$, non-trivial and $K \subseteq A$. But $|A|=p$, a prime order subgroup, hence has no non-trivial subgroup, so $K=A$. so $A$ is normal subgroup.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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 := |
Herstein|exercise_2_6_15 | 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 the order of $ab$ in $G$. We have $a^m = e$, $b^n = e$, and $(ab)^k = e$, where $e$ denotes the identity element of $G$. Since $G$ is abelian, we have
$$(ab)^{mn} = a^{mn}b^{mn} = e \cdot e = e.$$
Thus, $k$ is a divisor of $mn$.
Now, observe that $a^k = b^{-k}$. Since $m$ and $n$ are relatively prime, there exist integers $x$ and $y$ such that $mx + ny = 1$. Taking $kx$ on both sides of the equation, we get $a^{kx} = b^{-kx}$, or equivalently, $(a^k)^x = (b^k)^{-x}$. It follows that $a^{kx} = (a^m)^{xny} = e$, and similarly, $b^{ky} = (b^n)^{mxk} = e$. Therefore, $m$ divides $ky$ and $n$ divides $kx$. Since $m$ and $n$ are relatively prime, it follows that $mn$ divides $k$. Hence, $k = mn$, and $ab$ has order $mn$ in $G$. This completes the proof.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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 := |
Herstein|exercise_2_8_12 | 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 implies $a b=a$ and so that $b=e$, a contradiction, and we know $i \neq 1$ since then $a b=b a$ and this would imply $G$ is abelian, which we are assuming is not the case.
Now, $a$ has order 3 so we must have $b=a^3 b a^{-3}=b^{i^3}$ mod 7 , and so $i$ is restricted by the modular equation $i^3 \equiv 1 \bmod 7$
\begin{center}
\begin{tabular}{|c|c|}
\hline$x$ & $x^3 \bmod 7$ \\
\hline 2 & 1 \\
\hline 3 & 6 \\
\hline 4 & 1 \\
\hline 5 & 6 \\
\hline 6 & 6 \\
\hline
\end{tabular}
\end{center}
Therefore the only options are $i=2$ and $i=4$. Now suppose $G$ is such that $a b a^{-1}=b^2$ and let $G^{\prime}$ be another group of order 21 with an element $c$ of order 3 and an element $d$ of order 7 such that $c d c^{-1}=d^4$. We now prove that $G$ and $G^{\prime}$ are isomorphic. Define
$$
\begin{aligned}
\phi: G & \rightarrow G^{\prime} \\
a & \mapsto c^{-1} \\
b & \mapsto d
\end{aligned}
$$
since $a$ and $c^{-1}$ have the same order and $b$ and $d$ have the same order this is a well defined function. Since
$$
\begin{aligned}
\phi(a) \phi(b) \phi(a)^{-1} & =c^{-1} d c \\
& =\left(c d^{-1} c^{-1}\right)^{-1} \\
& =\left(d^{-4}\right)^{-1} \\
& =d^4 \\
& =\left(d^2\right)^2 \\
& =\phi(b)^2
\end{aligned}
$$
$\phi$ is actually a homomorphism. For any $c^i d^j \in G^{\prime}$ we have $\phi\left(a^{-i} b^j\right)=c^i d^j$ so $\phi$ is onto and $\phi\left(a^i b^j\right)=c^{-i} d^j=e$ only if $i=j=0$, so $\phi$ is 1-to-l. Therefore $G$ and $G^{\prime}$ are isomorphic and so up to isomorphism there is only one nonabelian group of order 21 .
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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) := |
Herstein|exercise_2_9_2 | 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 cyclic.
Suppose now that $\operatorname{gcd}(n . m)>1$. Let $g^k$ be an element of $G$ and $h^j$ be an element of $H$. Since the lowest common multiple of $n$ and $m$ is lower than the product $n . m$, that is, $\operatorname{lcm}(n, m)<n$. $m$, and since $\left(g^k\right)^{l c m(n, m)}=e_G,\left(h^j\right)^{l c m(n, m)}=e_H$, we have $\left(g^k \times h^j\right)^{l c m(n, m)}=e_{G \times H}$. It follows that every element of $G \times H$ has order lower than $n . m$, and therefore $G \times H$ is not cyclic.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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) := |
Herstein|exercise_2_11_6 | 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.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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 := |
Herstein|exercise_2_11_22 | 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
$$
|G|=|Z(G)|+\sum_{a \notin Z(G)}[G: C(a)],
$$
where $C(a)$ denotes the centralizer of $a$ in $G$. If $G=Z(G)$, then the lemma is immediate. Suppose $Z(G)$ is a proper subset of $G$ and consider an element $a\in G$ such that $a\notin Z(G)$. Then $C(a)$ is a proper subgroup of $G$. Since $C(a)$ is a subgroup of a $p$-group, $[G:C(a)]$ is divisible by $p$ for all $a\notin Z(G)$. This implies that $p$ divides $|G|=|Z(G)|+\sum_{a\notin Z(G)} [G:C(a)]$.
Since $p$ also divides $|G|$, it follows that $p$ divides $|Z(G)|$. Hence, $|Z(G)|>1$. $\Box$
This proves our \textbf{lemma}.
We will prove the result by induction on $n$.
If $n=1$, the $G$ is a cyclic group of prime order and hence every subgroup of $G$ is normal in $G$. Thus, the result is true for $n=1$.
Suppose the result is true for all groups of order $p^m$, where $1 \leq m<n$.
Let $H$ be a subgroup of order $p^{n-1}$.
Consider $N(H)=\{g \in H: g H=H g\}$.
If $H \neq N(H)$, then $|N(H)|>p^{n-1}$. Thus, $|N(H)|=p^n$ and $N(H)=G$.
In this case $H$ is normal in $G$.
Let $H=N(H)$. Then $Z(G)$, the center of $G$, is a subset of $H$ and $Z(G) \neq$ $\{e\}$.
By Cauchy's theorem and the above Claim, there exists $a \in Z(G)$ such that $o(a)=p$.
Let $K=\langle a\rangle$, a cyclic group generated by $a$.
Then $K$ is a normal subgroup of $G$ of order $p$. Now, $|H / K|=p^{n-2}$ and $|G / K|=p^{n-1}$.
Thus, by induction hypothesis, $H / K$ is a normal subgroup of $G / K$.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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 := |
Herstein|exercise_4_1_19 | 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} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_4_1_19 : Infinite {x : Quaternion ℝ | x^2 = -1} := |
Herstein|exercise_4_2_5 | 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 last relation by $x$ on the left and right to get
$$
2\left(x y+x^2 y x+x y x^2\right)=0 \quad 2\left(x^2 y x+x y x^2+y x\right)=0 .
$$
Subtracting the last two relations we have
$$
2(x y-y x)=0 .
$$
We then show that $3\left(x+x^2\right)=0 \forall x$. You get this from
$$
x+x^2=\left(x+x^2\right)^3=x^3+3 x^4+3 x^5+x^6=4\left(x+x^2\right) .
$$
In particular
$$
3\left(x+y+(x+y)^2\right)=3\left(x+x^2+y+y^2+x y+y x\right)=0
$$
we end-up with $3(x y+y x)=0$. But since $6 x y=0$, we have $3(x y-y x)=0$. Then subtract $2(x y-y x)=0$ to get $x y-y x=0$.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_4_2_5 {R : Type*} [Ring R]
(h : ∀ x : R, x ^ 3 = x) : Nonempty (CommRing R) := |
Herstein|exercise_4_2_9 | 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. $p \mid a$.
Now for odd prime $p>3$.
Let us consider $f(x)=(x-1)(x-2) \ldots(x-(p-1))$.
Now, by Fermat, we know that the coefficients of $f(x)$ other than the $x^{p-1}$ and $x^0$ are divisible by $p$.
So if,
$$
\begin{array}{r}
f(x)=x^{p-1}+\sum_{i=0}^{p-2} a_i x^i \\
\text { and } p>3 .
\end{array}
$$
Then $p \mid a_2$, and
$$
f(p) \equiv a_1 p+a_0 \quad\left(\bmod p^3\right)
$$
But we see that
$$
f(x)=(-1)^{p-1} f(p-x) \text { for any } x,
$$
so if $p$ is odd,
$$
f(p)=f(0)=a_0,
$$
So it follows that:
$$
0=f(p)-a_0 \equiv a_1 p \quad\left(\bmod p^3\right)
$$
Therefore,
$$
0 \equiv a_1 \quad\left(\bmod p^2\right) .
$$
Hence,
$$
0 \equiv a_1 \quad(\bmod p) .
$$
Now our sum is just $\frac{a_1}{(p-1) !}=\frac{a}{b}$.
It follows that $p$ divides $a$. This completes the proof.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_4_2_9 {p : ℕ} (hp : Nat.Prime p) (hp1 : Odd p) :
∀ (a b : ℤ), (a / b : ℚ) = ∑ i in Finset.range (p-1), (1 / (i + 1) : ℚ) → ↑p ∣ a := |
Herstein|exercise_4_3_25 | 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, if we assumed some other element to be nonzero -- that $a \neq 0$. Then we have that
$$
\left(\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right)\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)=\left(\begin{array}{ll}
a & b \\
0 & 0
\end{array}\right) \in I
$$
and so
$$
\left(\begin{array}{ll}
a & b \\
0 & 0
\end{array}\right)\left(\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right)=\left(\begin{array}{ll}
a & 0 \\
0 & 0
\end{array}\right) \in I
$$
so that
$$
\left(\begin{array}{ll}
x & 0 \\
0 & 0
\end{array}\right) \in I
$$
for any real $x$. Now, also for any real $x$,
$$
\left(\begin{array}{ll}
x & 0 \\
0 & 0
\end{array}\right)\left(\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right)=\left(\begin{array}{ll}
0 & x \\
0 & 0
\end{array}\right) \in I .
$$
Likewise
$$
\left(\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right)\left(\begin{array}{ll}
0 & x \\
0 & 0
\end{array}\right)=\left(\begin{array}{ll}
0 & 0 \\
0 & x
\end{array}\right) \in I
$$
and
$$
\left(\begin{array}{ll}
0 & 0 \\
0 & x
\end{array}\right)\left(\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right)=\left(\begin{array}{ll}
0 & 0 \\
x & 0
\end{array}\right)
$$
Thus, as
$$
\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)=\left(\begin{array}{ll}
a & 0 \\
0 & 0
\end{array}\right)+\left(\begin{array}{ll}
0 & b \\
0 & 0
\end{array}\right)+\left(\begin{array}{ll}
0 & 0 \\
c & 0
\end{array}\right)+\left(\begin{array}{ll}
0 & 0 \\
0 & d
\end{array}\right)
$$
and since all the terms on the right side are in $I$ and $I$ is an additive group, it follows that
$$
\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
$$
for arbitrary $a, b, c, d$ is in $I$, i.e. $I=R$
Note that the intuition for picking these matrices is that, if we denote by $E_{i j}$ the matrix with 1 at position $(i, j)$ and 0 elsewhere, then
$$
E_{i j}\left(\begin{array}{ll}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2}
\end{array}\right) E_{n m}=a_{j, n} E_{i m}
$$
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_4_3_25 (I : Ideal (Matrix (Fin 2) (Fin 2) ℝ)) :
I = ⊥ ∨ I = ⊤ := |
Herstein|exercise_4_5_16 | 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 holds, we have to show that each of these choices induces a different element of $F[x] /(q(x))$; in other words, that each different polynomial of degree $n-1$ or lower belongs to a different coset of $(q(x))$ in $F[x]$.
Suppose now, then, that
$$
a_{n-1} x^{n-1}+\cdots+a_1 x+a_0+(q(x))=b_{n-1} x^{n-1}+\cdots+b_1 x+b_0+(q(x))
$$
which is equivalent with $\left(a_{n-1}-b_{n-1}\right)^{n-1}+\cdots\left(a_1-b_1\right) x+\left(a_0-b_0\right) \in(q(x))$, which is in turn equivalent with there being a $w(x) \in F[x]$ such that
$$
q(x) w(x)=\left(a_{n-1}-b_{n-1}\right)^{n-1}+\cdots\left(a_1-b_1\right) x+\left(a_0-b_0\right) .
$$
Degree of the right hand side is strictly smaller than $n$, while the degree of the left hand side is greater or equal to $n$ except if $w(x)=0$, so that if equality is hold we must have that $w(x)=0$, but then since polynomials are equal iff all of their coefficient are equal we get that $a_{n-1}-b_{n-1}=$ $0, \ldots, a_1-b_1=0, a_0-b_0=0$, i.e.
$$
a_{n-1}=b_{n-1}, \ldots, a_1=b_1, a_0=b_0
$$
which is what we needed to prove.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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}) := |
Herstein|exercise_4_5_25 | 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 we have
$$
f(x+c)=g(x+c) h(x+c) .
$$
Note that $g(x+c)$ and $h(x+c)$ have the same degree at $g(x)$ and $h(x)$ respectively; in particular, they are non-constant polynomials. So our assumption is wrong.
Hence, $f(x)$ is irreducible in $F[x]$. This proves our Lemma.
Now recall the identity
$$
\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\ldots \ldots+x^2+x+1 .
$$
We prove that $f(x+1)$ is $\$$ |textbffirreducible in $\mathbb{Q}[x]$ and then apply the Lemma to conclude that $f(x)$ is irreducible in $\mathbb{Q}[x] .3 \$$ Note that
$$
\begin{aligned}
& f(x+1)=\frac{(x+1)^p-1}{x} \\
& =\frac{x^p+p x^{p-1}+\ldots+p x}{x} \\
& =x^{p-1}+p x^{p-2}+\ldots .+p .
\end{aligned}
$$
Using that the binomial coefficients occurring above are all divisible by $p$, we have that $f(x+1)$ is irreducible $\mathbb{Q}[x]$ by Eisenstein's criterion applied with prime $p$.
Then by the lemma $f(x)$ is irreducible $\mathbb{Q}[x]$. This completes the proof.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_4_5_25 {p : ℕ} (hp : Nat.Prime p) :
Irreducible (∑ i in Finset.range p, X ^ i : Polynomial ℚ) := |
Herstein|exercise_4_6_3 | 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} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_4_6_3 :
Infinite {a : ℤ | Irreducible (X^7 + 15*X^2 - 30*X + (a : Polynomial ℚ) : Polynomial ℚ)} := |
Herstein|exercise_5_2_20 | 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 $(v + Fu) \cap U_i = \varnothing$, and $(v + Fu) \cap U_j$ for $j \neq i$ contains at most one vector, since otherwise $U_j$ would contain $u$.
Therefore, we have $|v + Fu| \leq |F| \leq n-1$. However, since $n$ is a finite natural number, this contradicts the fact that the field $F$ is finite.
Thus, our assumption that $V$ can be written as the set-theoretic union of proper subspaces is wrong, and the claim is proven.
\end{proof} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| 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)) ≠ ⊤ := |
Herstein|exercise_5_3_10 | 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} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_5_3_10 : IsAlgebraic ℚ (cos (Real.pi / 180)) := |
Herstein|exercise_5_5_2 | 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} | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators
| theorem exercise_5_5_2 : Irreducible (X^3 - 3*X - 1 : Polynomial ℚ) := |
Ireland-Rosen|exercise_1_27 | 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 $n-1, n+1$, otherwise it wouldn't divide three consecutive integers, which is impossible. As $n$ is odd, $n+1$ is even, so $(n+1)(n-1)$ is divisible by both 2 and 3 , so it is divisible by 6 .
\end{proof} | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_1_27 {n : ℕ} (hn : Odd n) : 8 ∣ (n^2 - 1) := |
Ireland-Rosen|exercise_1_31 | 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} | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_1_31 : (⟨1, 1⟩ : GaussianInt) ^ 2 ∣ 2 := |
Ireland-Rosen|exercise_2_21 | 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 divisors of $n$, except for $d = p_j^i, i>0, j=1,\ldots t$,
\begin{align*}
\sum_{d \mid n} \land(d)&= \sum_{i=1}^{\alpha_1} \land(p_1^{i}) + \cdots+ \sum_{i=1}^{\alpha_t} \land(p_t^{i})\\
&= \alpha_1 \log p_1+\cdots + \alpha_t \log p_t\\
&= \log n
\end{align*}
By Mobius Inversion Theorem,
$$\land(n) = \sum_{d \mid n} \mu\left (\frac{n}{d}\right ) \log d.$$
\end{proof} | import Mathlib
open Real
open scoped BigOperators
| 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 := |
Ireland-Rosen|exercise_3_1 | 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 congruent to 1, then $N \equiv 1 \pmod 6$ : this is a contradiction because $-1 \not \equiv 1 \pmod 6$. So there exists a prime factor $p$ of $N$ such that $p\equiv -1 \pmod 6$.
If $p\leq n$, then $p \mid n!$, and $p \mid N = n!-1$, so $p \mid 1$. As $p$ is prime, this is a contradiction, so $p>n$.
Conclusion :
for any integer $n$, there exists a prime $p >n$ such that $p \equiv -1 \pmod 6$ : there are infinitely many primes congruent to $-1$ modulo $6$.
\end{proof} | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_3_1 : Infinite {p : Nat.Primes // p ≡ -1 [ZMOD 6]} := |
Ireland-Rosen|exercise_3_5 | 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 solution in integers.
\end{proof} | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_3_5 : ¬ ∃ x y : ℤ, 7*x^3 + 2 = y^3 := |
Ireland-Rosen|exercise_3_14 | 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
$$n^{q-1} = (n^{p-1})^k \equiv 1 \pmod p.$$
$p \mid n^{q-1} - 1, q \mid n^{q-1} - 1$, and $p\wedge q = 1$, so $pq \mid n^{q-1} - 1$ :
$$n^{q-1} \equiv 1 \pmod{pq}.$$
\end{proof} | import Mathlib
open Real
open scoped BigOperators
| 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] := |
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ProofNet#
ProofNet# is a Lean 4 port of the ProofNet benchmark including fixes. A comparison with previous Lean 4 ports can be found at: https://proofnet4-fix.streamlit.app/.
This benchmark is compatible with all Lean versions between v4.7.0 and v4.16.0-rc2.
Original Dataset Summary
ProofNet is a benchmark for autoformalization and formal proving of undergraduate-level mathematics. The ProofNet benchmarks consists of 371 examples, 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.
Tasks
- Statement Autoformalization:
- Input:
nl_statement - Output:
lean4_formalization
- Input:
- Proof Autoformalization:
- Input:
lean4_formalization,nl_proof, - Output: use Lean to check if the generated proof is correct.
- Input:
- Theorem Proving:
- Input:
lean4_formalization - Output: use Lean to check if the generated proof is correct.
- Input:
Data Fields
id: Unique string identifier for the problem.nl_statement: Natural language theorem statement.nl_proof: Mathematical proof in natural language for the theorem statement.lean4_src_header: File header including imports, namespaces, and locales required for the formal statement.lean4_formalization: Formal theorem statement in Lean 4.
Citation
ProofNet# is introduced in Improving Autoformalization using Type Checking.
@misc{poiroux2024improvingautoformalizationusingtype,
title={Improving Autoformalization using Type Checking},
author={Auguste Poiroux and Gail Weiss and Viktor Kunčak and Antoine Bosselut},
year={2024},
eprint={2406.07222},
archivePrefix={arXiv},
primaryClass={cs.CL},
url={https://arxiv.org/abs/2406.07222},
}
Original work where ProofNet has been introduced:
@misc{azerbayev2023proofnet,
title={ProofNet: Autoformalizing and Formally Proving Undergraduate-Level Mathematics},
author={Zhangir Azerbayev and Bartosz Piotrowski and Hailey Schoelkopf and Edward W. Ayers and Dragomir Radev and Jeremy Avigad},
year={2023},
eprint={2302.12433},
archivePrefix={arXiv},
primaryClass={cs.CL}
}
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