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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) :
β x y : closure {x | x = a β¨ 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 : β y, orderOf y = 2 β y = x) :
x β 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)) : β 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 = β€) {ΞΉ : Type*} (R : ΞΉ β V)
(hR : LinearIndependent K R) : Countable ΞΉ := |
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 β§Έ (center G)) :
center G = β€ := |
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 : β}
(hp : Prime p) (hq : Prime q) (hG : card G = p^2 *q) :
IsSimpleGroup G β 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 β€) β (span {X}) =
span ({2 * X} : Set $ Polynomial β€) := |
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 : β(I β 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 : β (x : R), x β M β IsUnit x)
(hProper : β x : R, x β M) :
IsMaximal M β§ β (N : Ideal R), IsMaximal N β 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 β) := |
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 : β) (hp : Prime p) (n : β) (hn : n > 0) :
Irreducible (X ^ n - (p : Polynomial β) : Polynomial β) := |
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 : β} {hp : Nat.Prime p} (h : β r : β, p = 2 ^ r + 1) :
β (k : β), 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 : β U : Set (β Γ β),
(U β β
) β§
(β (u v : β Γ β), u β U β§ v β U β u + v β U) β§
(β (u : β Γ β), u β U β -u β U) β§
(β U' : Submodule β (β Γ β), U β βU') := |
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] {ΞΉ : Type*} (u : ΞΉ β Submodule F V) :
β U : Submodule F V, (β (i : ΞΉ), (u i).carrier) = βU := |
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 ββ[F] V) (hT : finrank F V = 1) :
β c : F, β v : V, T v = c β’ 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 β) :
p.degree = @card (rootSet p β) (rootSetFintype p β) β
Disjoint
(@card (rootSet (derivative p) β) (rootSetFintype (derivative p) β))
(@card (rootSet p β) (rootSetFintype p β)) := |
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 ββ[F] V) (hST : S β T = T β S) (c : F):
Submodule.map S (ker (T - c β’ LinearMap.id)) = ker (T - c β’ 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 : β v : V, β c : F, v β eigenspace S c) :
β c : F, S = c β’ 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 : β v : V, (β c : F, v β eigenspace S c) β (β c : F, v β 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) :
βͺu, vβ«_F = 0 β β (a : F), βuβ β€ βu + a β’ vβ := |
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 β V] (u v : V) :
βͺu, vβ«_β = (βu + vβ^2 - βu - vβ^2 + I*βu + Iβ’vβ^2 - I*βu-Iβ’vβ^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 = β₯ β U = β€ := |
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 β V]
[FiniteDimensional β V] (T : End β V)
(hT : T * adjoint T = adjoint T * T) (hT1 : T^9 = T^8) :
IsSelfAdjoint T β§ 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 {π V : Type*} [RCLike π] [NormedAddCommGroup V]
[InnerProductSpace π V] [FiniteDimensional π V]
{T : Module.End π V} (hT : IsSelfAdjoint T)
{l : π} {Ξ΅ : β} (he : Ξ΅ > 0) : (β v : V, βvβ= 1 β§ βT v - l β’ vβ < Ξ΅) β
(β l' : T.Eigenvalues, βl - l'β < Ξ΅) := |
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 : β) :
β (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 : β) (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 β (orderOf x = 1 β¨ 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 β yβ»ΒΉ * x * y = x) β§ (yβ»ΒΉ * x * y = x β xβ»ΒΉ * yβ»ΒΉ * 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 (gβ»ΒΉ * 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 : β x : G, x ^ 2 = 1) : β 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 : β€) (hnm : n β m) :
x ^ n β 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 β β* Multiplicative β) := |
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 β G)
(hf : f = Ξ» g => gβ»ΒΉ) :
(β x y : G, f x * f y = f (x*y)) β β 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 β 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 : β} {hn : n β 0}
{R : Subgroup (DihedralGroup n)}
(hR : R = Subgroup.closure {DihedralGroup.r 1}) :
R β β€ β§
β K : Subgroup (DihedralGroup n), R β€ K β K = R β¨ K = β€ := |
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) :
β a b : A β§Έ 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 β Subgroup G) (hH : β i : I, Normal (H i)) :
Normal (β¨
(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 β€ 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 β) (hH : H β β€) : 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 β§ β 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 β β₯) [H.Normal] :
β A β€ H, A β β₯ β§ A.Normal β§ β 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 : Β¬ (card G).Prime) (hG1 : β k : β, k β£ card G β
β (H : Subgroup G) (fH : Fintype H), @card H fH = k) :
Β¬ 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 {Ξ± : Type*} [Fintype Ξ±] (ha : card Ξ± > 1)
(h_tran : β a b: Ξ±, β Ο : Equiv.Perm Ξ±, Ο a = b) :
β Ο : Equiv.Perm Ξ±, β a : Ξ±, Ο a β 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 : β (K : Subgroup G) (fK : Fintype K), card H = @card K fK β 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 : β} {G : Type*} [Group G]
{P : Sylow p G} (H : Subgroup G) (hH : P β€ H) :
IsPGroup p (P.subgroupOf H) β§
β (Q : Subgroup H), IsPGroup p Q β (P.subgroupOf H) β€ Q β 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) :
β (p : β) (P : Sylow p G), p.Prime β§ (p β£ card G) β§ 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 : β} {G : Type*} [Group G]
[Fintype G] (hpqr : p < q β§ q < r)
(hpqr1 : p.Prime β§ q.Prime β§ r.Prime)(hG : card G = p*q*r) :
(β (P : Sylow p G), P.Normal) β¨ (β (P : Sylow q G), P.Normal) β¨ (β (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) :
β 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) : Β¬ 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) : Β¬ 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] :
β 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 β β
(β€ : Subgroup G), Hβ β€ 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 β¨ 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 : β a : R, a^2 = a) :
β 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] :
β g : G, MonoidAlgebra.of R G g β 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 : β} (hp : p.Prime) :
IsNilpotent (span ({βp} : 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 : β} (hN : N > 0) {M M': β€} {d : β}
(hMN : M.gcd N = 1) (hMd : d.gcd N.totient = 1)
(hM' : M' β‘ M^d [ZMOD N]) :
β d' : β, d' * d β‘ 1 [ZMOD N.totient] β§
M β‘ 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 : β€} {r s : β}
(h : r^2 + s^2 = n) :
β a b : β€, 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 β§Έ span ({β¨1, 1β©} : Set GaussianInt))) :
IsField R β§ β 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 : Β¬ Submodule.IsPrincipal
(span ({MvPolynomial.X 0, MvPolynomial.X 1} : Set (MvPolynomial (Fin 2) β))) := |
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 β} (i j : β)
(hfg : β n : β, β a : β€, (f*g).coeff = a) :
β a : β€, 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 β€) := |
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 : β} (hp : p.Prime β§ p > 2)
{f : Polynomial β€} (hf : f = (X + 2)^p):
Irreducible (β n in (f.support \ {0}), (f.coeff n : Polynomial β€) * X ^ (n-1) :
Polynomial β€) := |
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) β) := |
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) :
β 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] : β (m : β), m > 0 β§ β (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 : β (a b : G), (a * b) ^ 3 = a ^ 3 * b ^ 3 β§ (a * b) ^ 5 = a ^ 5 * b ^ 5) :
β 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 {xβ»ΒΉ*a*x} =
(Ξ» g : G => xβ»ΒΉ*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 : β} (h : a > 1) :
n β£ (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 : β} (hp : Nat.Prime p) (hp1 : Β¬ p β£ 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 β* 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 : β}
(hp : Nat.Prime p) (hG : card G = p^2) :
β (N : Subgroup G) (Fin : Fintype N), @card N Fin = p β§ 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 : β}
(hm : β (g : G), orderOf g = m)
(hn : β (g : G), orderOf g = n)
(hmn : m.Coprime n) :
β (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 β* 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 Γ H) β (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 : β} (hp : Nat.Prime p)
{P : Sylow p G} (hP : P.Normal) :
β (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 : β} {n : β} {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 β | 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 : β 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 : β} (hp : Nat.Prime p) (hp1 : Odd p) :
β (a b : β€), (a / b : β) = β i in Finset.range (p-1), (1 / (i + 1) : β) β βp β£ 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) β)) :
I = β₯ β¨ I = β€ := |
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: β} (hp : Nat.Prime p)
{q : Polynomial (ZMod p)} (hq : Irreducible q) (hn : q.degree = n) :
(β is_fin : Fintype $ Polynomial (ZMod p) β§Έ span ({q}),
@card (Polynomial (ZMod p) β§Έ span {q}) is_fin = p ^ n) β§
IsField (Polynomial (ZMod p) β§Έ 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 : β} (hp : Nat.Prime p) :
Irreducible (β i in Finset.range p, X ^ i : Polynomial β) := |
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 : β€ | Irreducible (X^7 + 15*X^2 - 30*X + (a : Polynomial β) : Polynomial β)} := |
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 ΞΉ: Type*} [Infinite F] [Field F]
[AddCommGroup V] [Module F V] [Finite ΞΉ] {u : ΞΉ β Submodule F V}
(hu : β i : ΞΉ, u i β β€) :
(β i : ΞΉ, (u i : Set V)) β β€ := |
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 β (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 β) := |
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 : β} (hn : Odd n) : 8 β£ (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 : (β¨1, 1β© : GaussianInt) ^ 2 β£ 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 : β β β}
(hl : β p n : β, p.Prime β l (p^n) = log p )
(hl1 : β m : β, Β¬ IsPrimePow m β l m = 0) :
l = Ξ» n => β 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 β‘ -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 : Β¬ β x y : β€, 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 : β} (hp0 : p.Prime β§ p > 2)
(hq0 : q.Prime β§ q > 2) (hpq0 : p β q) (hpq1 : p - 1 β£ q - 1)
(hn : n.gcd (p*q) = 1) :
n^(q-1) β‘ 1 [MOD p*q] := |
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