theorem exercise_3_22 (D : set ℂ) (hD : D = ball 0 1) (f : ℂ → ℂ)
    (hf : differentiable_on ℂ f D) (hfc : continuous_on f (closure D)) :
    ¬ ∀ z ∈ (sphere (0 : ℂ) 1), f z = 1 / z :=

Show that there is no holomorphic function $f$ in the unit disc $D$ that extends continuously to $\partial D$ such that $f(z) = 1/z$ for $z \in \partial D$.

\begin{proof}

    Consider $g(r)=\int_{|z|=r} f(z) d z$. Cauchy theorem implies that $g(r)=0$ for all $r<1$. Now since $\left.f\right|_{\partial D}=1 / z$ we have $\lim _{r \rightarrow 1} \int_{|z|=r} f(z) d z=\int_{|z|=1} \frac{1}{z} d z=\frac{2}{\pi i} \neq 0$. Contradiction.

\end{proof}


import .common

open real topological_space filter polynomial
open_locale topology big_operators complex_conjugate filter ennreal

\begin{proof}

    It will suffice to show that for any $z_1, z_2, z_3, z_4 \in \mathbb{C}$ of modulus 1 such that $|3-z_1-z_2-z_3-z_4| = |z_1z_2z_3z_4|$, at least one of $z_1, z_2, z_3$ is equal to 1.

To this end, let $z_1=e^{\alpha i}, z_2=e^{\beta i}, z_3=e^{\gamma i}$ and

\[

f(\alpha, \beta, \gamma)=|3-z_1-z_2-z_3|^2-|1-z_1z_2z_3|^2.

\]

 A routine calculation shows that

\begin{align*}

f(\alpha, \beta, \gamma)&=

10 - 6\cos(\alpha) - 6\cos(\beta) - 6\cos(\gamma) \\

&\quad + 2\cos(\alpha + \beta + \gamma) + 2\cos(\alpha - \beta) \\

&\quad + 2\cos(\beta - \gamma) + 2\cos(\gamma - \alpha).

\end{align*}

Since the function $f$ is continuously differentiable, and periodic in each variable, $f$ has a maximum and a minimum and it attains these values only at points where $\nabla f=(0,0,0)$.  A routine calculation now shows that

\begin{align*}

\frac{\partial f}{\partial \alpha} + \frac{\partial f}{\partial \beta} + \frac{\partial f}{\partial \gamma} &=

6(\sin(\alpha) +\sin(\beta)+\sin(\gamma)-  \sin(\alpha + \beta + \gamma)) \\

&=

24\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\beta+\gamma}{2}\right)

\sin\left(\frac{\gamma+\alpha}{2}\right).

\end{align*}

Hence every critical point of $f$ must satisfy one of $z_1z_2=1$, $z_2z_3=1$, or $z_3z_1=1$. By symmetry, let us assume that $z_1z_2=1$. Then

\[

f = |3-2\mathrm{Re}(z_1)-z_3|^2-|1-z_3|^2;

\]

since $3-2\mathrm{Re}(z_1)\ge 1$, $f$ is nonnegative and can be zero only if the real part of $z_1$, and hence also $z_1$ itself, is equal to $1$.

\end{proof}


theorem exercise_2018_b2 (n : ℕ) (hn : n > 0) (f : ℕ → ℂ → ℂ)
  (hf : ∀ n : ℕ, f n = λ z, (∑ (i : fin n), (n-i)* z^(i : ℕ))) :
  ¬ (∃ z : ℂ, ‖z‖ ≤ 1 ∧ f n z = 0) :=

Let $n$ be a positive integer, and let $f_{n}(z)=n+(n-1) z+$ $(n-2) z^{2}+\cdots+z^{n-1}$. Prove that $f_{n}$ has no roots in the closed unit disk $\{z \in \mathbb{C}:|z| \leq 1\}$.

\begin{proof}

    Note first that $f_n(1) > 0$, so $1$ is not a root of $f_n$.

Next, note that

\[

(z-1)f_n(z) = z^n + \cdots + z - n;

\]

however, for $\left| z \right| \leq 1$, we have

$\left| z^n + \cdots + z \right| \leq n$ by the triangle inequality;

equality can only occur if $z,\dots,z^n$ have norm 1 and the same argument, which only happens for $z=1$.

Thus there can be no root of $f_n$ with $|z| \leq 1$.

\end{proof}


Suppose that $f(x)=\sum_{i=0}^{\infty} c_{i} x^{i}$ is a power series for which each coefficient $c_{i}$ is 0 or 1 . Show that if $f(2 / 3)=3 / 2$, then $f(1 / 2)$ must be irrational.

\begin{proof}

    Suppose by way of contradiction that $f(1/2)$ is rational. Then $\sum_{i=0}^{\infty} c_i 2^{-i}$ is the binary expansion of a rational number, and hence must be eventually periodic; that is, there exist some integers $m,n$ such that

$c_i = c_{m+i}$ for all $i \geq n$. We may then write

\[

f(x) = \sum_{i=0}^{n-1} c_i x^i + \frac{x^n}{1-x^m} \sum_{i=0}^{m-1} c_{n+i} x^i.

\]

Evaluating at $x = 2/3$, we may equate $f(2/3) = 3/2$ with

\[

\frac{1}{3^{n-1}} \sum_{i=0}^{n-1} c_i 2^i 3^{n-i-1} + \frac{2^n 3^m}{3^{n+m-1}(3^m-2^m)} \sum_{i=0}^{m-1} c_{n+i} 2^i 3^{m-1-i};

\]

since all terms on the right-hand side have odd denominator, the same must be true of the sum, a contradiction.

\end{proof}


Prove that for each positive integer $n$, the number $10^{10^{10^n}}+10^{10^n}+10^n-1$ is not prime.

\begin{proof}

    Put

\[

N = 10^{10^{10^n}} + 10^{10^n} + 10^n - 1.

\]

Write $n = 2^m k$ with $m$ a nonnegative integer and $k$ a positive odd integer.

For any nonnegative integer $j$,

\[

10^{2^m j} \equiv (-1)^j \pmod{10^{2^m} + 1}.

\]

Since $10^n \geq n \geq 2^m \geq m+1$, $10^n$ is divisible by $2^n$ and hence by $2^{m+1}$,

and similarly $10^{10^n}$ is divisible by $2^{10^n}$ and hence by $2^{m+1}$. It follows that

\[

N \equiv 1 + 1 + (-1) + (-1) \equiv 0 \pmod{10^{2^m} + 1}.

\]

Since $N \geq 10^{10^n} > 10^n + 1 \geq 10^{2^m} + 1$, it follows that $N$ is composite.

\end{proof}


Prove that there exist infinitely many integers $n$ such that $n, n+1, n+2$ are each the sum of the squares of two integers.

\begin{proof}

    It is well-known that the equation $x^2-2y^2=1$ has infinitely

many solutions (the so-called ``Pell'' equation).  Thus setting

$n=2y^2$ (so that $n=y^2+y^2$, $n+1=x^2+0^2$, $n+2=x^2+1^2$)

yields infinitely many $n$ with the desired property.

\end{proof}


theorem exercise_1998_a3 (f : ℝ → ℝ) (hf : cont_diff ℝ 3 f) :
  ∃ a : ℝ, (f a) * (deriv f a) * (iterated_deriv 2 f a) * (iterated_deriv 3 f a) ≥ 0 :=

Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that


theorem exercise_2_12a (f : ℕ → ℕ) (p : ℕ → ℝ) (a : ℝ)
  (hf : injective f) (hp : tendsto p at_top (𝓝 a)) :
  tendsto (λ n, p (f n)) at_top (𝓝 a) :=

import .common

open set real filter function ring_hom topological_space

open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory

Let $(p_n)$ be a sequence and $f:\mathbb{N}\to\mathbb{N}$. The sequence $(q_k)_{k\in\mathbb{N}}$ with $q_k=p_{f(k)}$ is called a rearrangement of $(p_n)$. Show that if $f$ is an injection, the limit of a sequence is unaffected by rearrangement.

\begin{proof}

    Let $\varepsilon>0$. Since $p_n \rightarrow L$, we have that, for all $n$ except $n \leq N$, $d\left(p_n, L\right)<\epsilon$. Let $S=\{n \mid f(n) \leq N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall n>n_0 f(n)>N$ which implies that $p_{f(n)} \rightarrow L$, as required.

\end{proof}


theorem exercise_2_29 (M : Type*) [metric_space M]
  (O C : set (set M))
  (hO : O = {s | is_open s})
  (hC : C = {s | is_closed s}) :
  ∃ f : O → C, bijective f :=

Let $\mathcal{T}$ be the collection of open subsets of a metric space $\mathrm{M}$, and $\mathcal{K}$ the collection of closed subsets. Show that there is a bijection from $\mathcal{T}$ onto $\mathcal{K}$.

\begin{proof}

   The bijection given by $x\mapsto X^C$ suffices.

\end{proof}


theorem exercise_2_41 (m : ℕ) {X : Type*} [normed_space ℝ ((fin m) → ℝ)] :
  is_compact (metric.closed_ball 0 1) :=

Let $\|\cdot\|$ be any norm on $\mathbb{R}^{m}$ and let $B=\left\{x \in \mathbb{R}^{m}:\|x\| \leq 1\right\}$. Prove that $B$ is compact.

\begin{proof}

    Let us call $\|\cdot\|_E$ the Euclidean norm in $\mathbb{R}^m$. We start by claiming that there exist constants $C_1, C_2>0$ such that

$$

C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E, \forall x \in \mathbb{R}^m .

$$

Assuming (1) to be true, let us finish the problem. First let us show that $B$ is bounded w.r.t. $d_E$, which is how we call the Euclidean distance in $\mathbb{R}^m$. Indeed, given $x \in B,\|x\|_E \leq \frac{1}{C_1}\|x\| \leq \frac{1}{C_1}$. Hence $B \subset\left\{x \in \mathbb{R}^m: d_E(x, 0)<\frac{1}{C_1}+1\right\}$, which means $B$ is bounded w.r.t $d_E$.

Now let us show that $B$ is closed w.r.t. $d_E$. Let $x_n \rightarrow x$ w.r.t. $d_E$, where $x_n \in B$. Notice that this implies that $x_n \rightarrow x$ w.r.t. $d(x, y)=\|x-y\|$, the distance coming from $\|\cdot\|$, since by (1) we have

$$

d\left(x_n, x\right)=\left\|x_n-x\right\| \leq C_2\left\|x_n-x\right\|_E \rightarrow 0 .

$$

Also, notice that

$$

\|x\| \leq\left\|x_n-x\right\|+\left\|x_n\right\| \leq\left\|x_n-x\right\|+1,

$$

hence passing to the limit we obtain that $\|x\| \leq 1$, therefore $x \in B$ and so $B$ is closed w.r.t. $d_E$. Since $B$ is closed and bounded w.r.t. $d_E$, it must be compact. Now we claim that the identity function, $i d:\left(\mathbb{R}^m, d_E\right) \rightarrow\left(\mathbb{R}^m, d\right)$ where $\left(\mathbb{R}^m, d_E\right)$ means we are using the distance $d_E$ in $\mathbb{R}^m$ and $\left(\mathbb{R}^m, d\right)$ means we are using the distance $d$ in $\mathbb{R}^m$, is a homeomorphism. This follows by (1), since $i d$ is always a bijection, and it is continuous and its inverse is continuous by (1) (if $x_n \rightarrow x$ w.r.t. $d_E$, then $x_n \rightarrow x$ w.r.t. $d$ and vice-versa, by (1)). By a result we saw in class, since $B$ is compact in $\left(\mathbb{R}^m, d_E\right)$ and $i d$ is a homeomorphism, then $i d(B)=B$ is compact w.r.t. $d$.

We are left with proving (1). Notice that it suffices to prove that $C_1 \leq\|x\| \leq$ $C_2, \forall x \in \mathbb{R}^m$ with $\|x\|_E=1$. Indeed, if this is true, given $x \in \mathbb{R}^m$, either $\|x\|_E=0$ (which implies $x=0$ and (1) holds in this case), or $x /\|x\|_E=y$ is such that $\|y\|_E=1$, so $C_1 \leq\|y\| \leq C_2$, which implies $C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E$.

We want to show now that $\|\cdot\|$ is continuous w.r.t. $d_E$, that is, given $\varepsilon>0$ and $x \in \mathbb{R}^m$, there exists $\delta>0$ such that if $d_E(x, y)<\delta$, then $\|\mid x\|-\|y\| \|<\varepsilon$.

By the triangle inequality, $\|x\|-\|y\| \leq\|x-y\|$, and $\|y\|-\|x\| \leq\|x-y\|$, therefore

$$

|\|x||-\| y|\|\leq\| x-y \| .

$$

Writing now $x=\sum_{i=1}^m a_i e_i, y=\sum_{i=1}^m b_i e_i$, where $e_i=(0, \ldots, 1,0, \ldots, 0)$ (with 1 in the i-th component), we obtain by the triangle inequality,

$$

\begin{aligned}

\|x-y\| & =\left\|\sum_{i=1}^m\left(a_i-b_i\right) e_i\right\| \leq \sum_{i=1}^m\left|a_i-b_i\left\|\left|\left\|e_i\right\| \leq \max _{i=1, \ldots, m}\left\|e_i\right\| \sum_{i=1}^m\right| a_i-b_i \mid\right.\right. \\

& =\max _{i=1, \ldots, m}\left\|e_i\right\| d_{s u m}(x, y) \leq \max _{i=1, \ldots, m}\left\|e_i\right\| m d_{\max }(x, y) \\

& \leq \max _{i=1, \ldots, m}\left\|e_i\right\| m d_E(x, y) .

\end{aligned}

$$

Let $\delta=\frac{\varepsilon}{m \max _{i=1, \ldots, m}\left\|e_i\right\|}$. Then if $d_E(x, y)<\delta,\|x\|-\|y\|||<\varepsilon$.

Since $\|\cdot\|$ is continuous w.r.t. $d_E$ and $K=\left\{x \in \mathbb{R}^m:\|x\|_E=1\right\}$ is compact w.r.t. $d_E$, then the function $\|\cdot\|$ achieves a maximum and a minimum value on $K$. Call $C_1=\min _{x \in K}\|x\|, C_2=\max _{x \in K}\|x\|$. Then

$$

C_1 \leq\|x\| \leq C_2, \forall x \in \mathbb{R}^m \text { such that }\|x\|_E=1,

$$

which is what we needed.

\end{proof}


theorem exercise_2_57 {X : Type*} [topological_space X]
  : ∃ (S : set X), is_connected S ∧ ¬ is_connected (interior S) :=

Show that if $S$ is connected, it is not true in general that its interior is connected.

\begin{proof}

    Consider $X=\mathbb{R}^2$ and

$$

A=([-2,0] \times[-2,0]) \cup([0,2] \times[0,2])

$$

which is connected, while $\operatorname{int}(A)$ is not connected.

To see this consider the continuous function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x, y)=x+y$. Let $U=f^{-1}(0,+\infty)$ which is open in $\mathbb{R}^2$ and so $U \cap \operatorname{int}(A)$ is open in $\operatorname{int}(A)$. Also, since $(0,0) \notin \operatorname{int}(A)$, so for all $(x, y) \in \operatorname{int}(A), f(x, y) \neq 0$ and $U \cap \operatorname{int}(A)=f^{-1}[0,+\infty) \cap \operatorname{int}(A)$ is closed in $\operatorname{int}(A)$. Furthermore, $(1,1)=f^{-1}(2) \in U \cap \operatorname{int}(A)$ shows that $U \cap \operatorname{int}(A) \neq \emptyset$ while $(-1,-1) \in \operatorname{int}(A)$ and $(-1,-1) \notin U$ shows that $U \cap \operatorname{int}(A) \neq \operatorname{int}(A)$.

\end{proof}


theorem exercise_2_126 {E : set ℝ}
  (hE : ¬ set.countable E) : ∃ (p : ℝ), cluster_pt p (𝓟 E) :=

Suppose that $E$ is an uncountable subset of $\mathbb{R}$. Prove that there exists a point $p \in \mathbb{R}$ at which $E$ condenses.

\begin{proof}

    I think this is the proof by contrapositive that you were getting at.

Suppose that $E$ has no limit points at all. Pick an arbitrary point $x \in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$ :

$$

B_\delta(x) \cap E=\{x\}

$$

Call this "point 1 ". For the next point, take the closest element to $x$ and on its left; that is, choose the point

$$

\max [E \cap(-\infty, x)]

$$

if it exists (that is important - if not, skip to the next step). Note that by the argument above, this supremum, should it exist, cannot equal $x$ and is therefore a new point in $E$.

Call this "point 2 ". Now take the first point to the right of $x$ for "point 3 ". Take the first point to the left of point 2 for "point 4 ". And so on, ad infinitum.

This gives a countable list of unique points; we must show that it exhausts the entire set $E$. Suppose not. Suppose there is some element $a<x$ which is never included in the list (picking $a$ on the negative side of $x$ is arbitrary, and the same argument would work for the second case). Then the element closest and to the right of $a$ in $E$ (which exists, by the no-limit-points argument at the beginning) is also not in the list; if it was, $a$ would have been in one of the next two spots. And same with that point (call it $a_1$ ); there is a closest $a_2>a_1 \in E$ such that $a_2$ is not in the list. Repeating, we generate an infinite monotone-increasing sequence $\left\{a_i\right\}$ of elements in $E$ and not in the list, which is clearly bounded above by $x$. By the Monotone

Convergence Theorem this sequence has a limit. But that means the sequence $\left\{a_i\right\} \subset E$ converges to a limit, and hence $E$ has a limit point, contradicting the assumption. Therefore our list exhausts $E$, and we have enumerated all its elements.

\end{proof}


theorem exercise_3_4 (n : ℕ) :
  tendsto (λ n, (sqrt (n + 1) - sqrt n)) at_top (𝓝 0) :=

Prove that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$.

\begin{proof}

    $$

\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2 \sqrt{n}}

$$

\end{proof}


theorem exercise_3_63b (p : ℝ) (f : ℕ → ℝ) (hp : p ≤ 1)
  (h : f = λ k, (1 : ℝ) / (k * (log k) ^ p)) :
  ¬ ∃ l, tendsto f at_top (𝓝 l) :=

Prove that $\sum 1/k(\log(k))^p$ diverges when $p \leq 1$.

\begin{proof}

    Using the integral test, for a set $a$, we see

$$

\lim _{b \rightarrow \infty} \int_a^b \frac{1}{x \log (x)^c} d x=\lim _{b \rightarrow \infty}\left(\frac{\log (b)^{1-c}}{1-c}-\frac{\log (a)^{1-c}}{1-c}\right)

$$

which goes to infinity if $c \leq 1$ and converges if $c>1$. Thus,

$$

\sum_{n=2}^{\infty} \frac{1}{n \log (n)^c}

$$

converges if and only if $c>1$.

\end{proof}


theorem exercise_2_1_21 (G : Type*) [group G] [fintype G]
  (hG : card G = 5) :
  comm_group G :=

import .common

open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory

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}


theorem exercise_2_1_27 {G : Type*} [group G]
  [fintype G] : ∃ (m : ℕ), ∀ (a : G), a ^ m = 1 :=

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}


theorem exercise_2_2_5 {G : Type*} [group G]
  (h : ∀ (a b : G), (a * b) ^ 3 = a ^ 3 * b ^ 3 ∧ (a * b) ^ 5 = a ^ 5 * b ^ 5) :
  comm_group G :=

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}


theorem exercise_2_3_17 {G : Type*} [has_mul G] [group G] (a x : G) :
  set.centralizer {x⁻¹*a*x} =
  (λ g : G, x⁻¹*g*x) '' (set.centralizer {a}) :=

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}


theorem exercise_2_4_36 {a n : ℕ} (h : a > 1) :
  n ∣ (a ^ n - 1).totient :=

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}


theorem exercise_2_5_30 {G : Type*} [group G] [fintype G]
  {p m : ℕ} (hp : nat.prime p) (hp1 : ¬ p ∣ m) (hG : card G = p*m)
  {H : subgroup G} [fintype H] [H.normal] (hH : card H = p):
  characteristic H :=

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}


theorem exercise_2_5_37 (G : Type*) [group G] [fintype G]
  (hG : card G = 6) (hG' : is_empty (comm_group G)) :
  G ≃* equiv.perm (fin 3) :=

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}


theorem exercise_2_5_44 {G : Type*} [group G] [fintype G] {p : ℕ}
  (hp : nat.prime p) (hG : card G = p^2) :
  ∃ (N : subgroup G) (fin : fintype N), @card N fin = p ∧ N.normal :=

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}


theorem exercise_2_6_15 {G : Type*} [comm_group G] {m n : ℕ}
  (hm : ∃ (g : G), order_of g = m)
  (hn : ∃ (g : G), order_of g = n)
  (hmn : m.coprime n) :
  ∃ (g : G), order_of g = m * n :=

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}


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 : is_empty(comm_group G)) (hH1 : is_empty (comm_group H)) :
  G ≃* H :=

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}


theorem exercise_2_9_2 {G H : Type*} [fintype G] [fintype H] [group G]
  [group H] (hG : is_cyclic G) (hH : is_cyclic H) :
  is_cyclic (G × H) ↔ (card G).coprime (card H) :=

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}


theorem exercise_2_11_6 {G : Type*} [group G] {p : ℕ} (hp : nat.prime p)
  {P : sylow p G} (hP : P.normal) :
  ∀ (Q : sylow p G), P = Q :=

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}


theorem exercise_2_11_22 {p : ℕ} {n : ℕ} {G : Type*} [fintype G]
  [group G] (hp : nat.prime p) (hG : card G = p ^ n) {K : subgroup G}
  [fintype K] (hK : card K = p ^ (n-1)) :
  K.normal :=

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}


theorem exercise_4_1_19 : infinite {x : quaternion ℝ | x^2 = -1} :=

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}


theorem exercise_4_2_5 {R : Type*} [ring R]
  (h : ∀ x : R, x ^ 3 = x) : comm_ring R :=

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}


theorem exercise_4_2_9 {p : ℕ} (hp : nat.prime p) (hp1 : odd p) :
  ∃ (a b : ℤ), (a / b : ℚ) = ∑ i in finset.range p, 1 / (i + 1) → ↑p ∣ a :=

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}


theorem exercise_4_3_25 (I : ideal (matrix (fin 2) (fin 2) ℝ)) :
  I = ⊥ ∨ I = ⊤ :=

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}


theorem exercise_4_5_16 {p n: ℕ} (hp : nat.prime p)
  {q : polynomial (zmod p)} (hq : irreducible q) (hn : q.degree = n) :
  ∃ is_fin : fintype $ polynomial (zmod p) ⧸ ideal.span ({q} : set (polynomial $ zmod p)),
  @card (polynomial (zmod p) ⧸ ideal.span {q}) is_fin = p ^ n ∧
  is_field (polynomial $ zmod p):=

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}


theorem exercise_4_5_25 {p : ℕ} (hp : nat.prime p) :
  irreducible (∑ i : finset.range p, X ^ p : polynomial ℚ) :=

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}


theorem exercise_4_6_3 :
  infinite {a : ℤ | irreducible (X^7 + 15*X^2 - 30*X + a : polynomial ℚ)} :=

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}


theorem exercise_5_2_20 {F V ι: Type*} [infinite F] [field F]
  [add_comm_group V] [module F V] {u : ι → submodule F V}
  (hu : ∀ i : ι, u i ≠ ⊤) :
  (⋃ i : ι, (u i : set V)) ≠ ⊤ :=

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}


theorem exercise_5_3_10 : is_algebraic ℚ (cos (real.pi / 180)) :=

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}


theorem exercise_5_5_2 : irreducible (X^3 - 3*X - 1 : polynomial ℚ) :=

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}


theorem exercise_2_2_9 {G : Type*} [group G] {a b : G}
  (h : a * b = b * a) :
  ∀ x y : closure {x | x = a ∨ x = b}, x*y = y*x :=

import .common

open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory

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}


theorem exercise_2_4_19 {G : Type*} [group G] {x : G}
  (hx : order_of x = 2) (hx1 : ∀ y, order_of y = 2 → y = x) :
  x ∈ center G :=

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}


theorem exercise_2_11_3 {G : Type*} [group G] [fintype G]
  (hG : even (card G)) : ∃ x : G, order_of x = 2 :=

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}


theorem exercise_3_5_6 {K V : Type*} [field K] [add_comm_group V]
  [module K V] {S : set V} (hS : set.countable S)
  (hS1 : span K S = ⊤) {ι : Type*} (R : ι → V)
  (hR : linear_independent K R) : countable ι :=

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}


theorem exercise_6_1_14 (G : Type*) [group G]
  (hG : is_cyclic $ G ⧸ (center G)) :
  center G = ⊤  :=

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}


theorem exercise_6_4_3 {G : Type*} [group G] [fintype G] {p q : ℕ}
  (hp : prime p) (hq : prime q) (hG : card G = p^2 *q) :
  is_simple_group G → false :=

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}


theorem exercise_6_8_1 {G : Type*} [group G]
  (a b : G) : closure ({a, b} : set G) = closure {b*a*b^2, b*a*b^3} :=

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}


theorem exercise_10_2_4 :
  span ({2} : set $ polynomial ℤ) ⊓ (span {X}) =
  span ({2 * X} : set $ polynomial ℤ) :=

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}


theorem exercise_10_4_6 {R : Type*} [comm_ring R]
  [no_zero_divisors R] {I J : ideal R} (x : I ⊓ J) :
  is_nilpotent ((ideal.quotient.mk (I*J)) x) :=

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}


theorem exercise_10_7_10 {R : Type*} [ring R]
  (M : ideal R) (hM : ∀ (x : R), x ∉ M → is_unit x) :
  is_maximal M ∧ ∀ (N : ideal R), is_maximal N → N = M :=

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}


theorem exercise_11_4_1b {F : Type*} [field F] [fintype F] (hF : card F = 2) :
  irreducible (12 + 6 * X + X ^ 3 : polynomial F) :=

Prove that $x^3 + 6x + 12$ is irreducible in $\mathbb{Q}$.

\begin{proof}

    Apply Eisenstein's criterion with $p=3$.

\end{proof}


theorem exercise_11_4_6b {F : Type*} [field F] [fintype F] (hF : card F = 31) :
  irreducible (X ^ 3 - 9 : polynomial F) :=

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}


theorem exercise_11_4_8 {p : ℕ} (hp : prime p) (n : ℕ) :
  irreducible (X ^ n - p : polynomial ℚ) :=

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}


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}


theorem exercise_1_1_3 (n : ℤ) :
  ∀ (a b c : ℤ), (a+b)+c ≡ a+(b+c) [ZMOD n] :=

import .common

open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix

open_locale pointwise
open_locale big_operators
noncomputable theory

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}


theorem exercise_1_1_5 (n : ℕ) (hn : 1 < n) :
  is_empty (group (zmod n)) :=

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}


theorem exercise_1_1_16 {G : Type*} [group G]
  (x : G) (hx : x ^ 2 = 1) :
  order_of x = 1 ∨ order_of x = 2 :=

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}


theorem exercise_1_1_18 {G : Type*} [group G]
  (x y : G) : x * y = y * x ↔ y⁻¹ * x * y = x ↔ x⁻¹ * y⁻¹ * x * y = 1 :=

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}


theorem exercise_1_1_22a {G : Type*} [group G] (x g : G) :
  order_of x = order_of (g⁻¹ * x * g) :=

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}


theorem exercise_1_1_25 {G : Type*} [group G]
  (h : ∀ x : G, x ^ 2 = 1) : ∀ a b : G, a*b = b*a :=

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}


theorem exercise_1_1_34 {G : Type*} [group G] {x : G}
  (hx_inf : order_of x = 0) (n m : ℤ) :
  x ^ n ≠ x ^ m :=

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}


theorem exercise_1_6_4 :
  is_empty (multiplicative ℝ ≃* multiplicative ℂ) :=

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}


theorem exercise_1_6_17 {G : Type*} [group G] (f : G → G)
  (hf : f = λ g, g⁻¹) :
  ∀ x y : G, f x * f y = f (x*y) ↔ ∀ x y : G, x*y = y*x :=

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}


theorem exercise_2_1_5 {G : Type*} [group G] [fintype G]
  (hG : card G > 2) (H : subgroup G) [fintype H] :
  card H ≠ card G - 1 :=

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}


theorem exercise_2_4_4 {G : Type*} [group G] (H : subgroup G) :
  subgroup.closure ((H : set G) \ {1}) = ⊤ :=

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}


theorem exercise_2_4_16b {n : ℕ} {hn : n ≠ 0}
  {R : subgroup (dihedral_group n)}
  (hR : R = subgroup.closure {dihedral_group.r 1}) :
  R ≠ ⊤ ∧
  ∀ K : subgroup (dihedral_group n), R ≤ K → K = R ∨ K = ⊤ :=

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}


theorem exercise_3_1_3a {A : Type*} [comm_group A] (B : subgroup A) :
  ∀ a b : A ⧸ B, a*b = b*a :=

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}


theorem exercise_3_1_22b {G : Type*} [group G] (I : Type*)
  (H : I → subgroup G) (hH : ∀ i : I, subgroup.normal (H i)) :
  subgroup.normal (⨅ (i : I), H i):=

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}


theorem exercise_3_2_11 {G : Type*} [group G] {H K : subgroup G}
  (hHK : H ≤ K) :
  H.index = K.index * H.relindex K :=

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}


theorem exercise_3_2_21a (H : add_subgroup ℚ) (hH : H ≠ ⊤) : H.index = 0 :=

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}


theorem exercise_3_4_1 (G : Type*) [comm_group G] [is_simple_group G] :
    is_cyclic G ∧ ∃ G_fin : fintype G, nat.prime (@card G G_fin) :=

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}


theorem exercise_3_4_5a {G : Type*} [group G]
  (H : subgroup G) [is_solvable G] : is_solvable H :=

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}


theorem exercise_3_4_11 {G : Type*} [group G] [is_solvable G]
  {H : subgroup G} (hH : H ≠ ⊥) [H.normal] :
  ∃ A ≤ H, A.normal ∧ ∀ a b : A, a*b = b*a :=

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}


theorem exercise_4_3_26 {α : Type*} [fintype α] (ha : fintype.card α > 1)
  (h_tran : ∀ a b: α, ∃ σ : equiv.perm α, σ a = b) :
  ∃ σ : equiv.perm α, ∀ a : α, σ a ≠ a :=

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}


theorem exercise_4_2_14 {G : Type*} [fintype G] [group G]
  (hG : ¬ (card G).prime) (hG1 : ∀ k ∣ card G,
  ∃ (H : subgroup G) (fH : fintype H), @card H fH = k) :
  ¬ is_simple_group G :=

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}


theorem exercise_4_4_6a {G : Type*} [group G] (H : subgroup G)
  [subgroup.characteristic H] : subgroup.normal H  :=

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}


theorem exercise_4_4_7 {G : Type*} [group G] {H : subgroup G} [fintype H]
  (hH : ∀ (K : subgroup G) (fK : fintype K), card H = @card K fK → H = K) :
  H.characteristic :=

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}


theorem exercise_4_5_1a {p : ℕ} {G : Type*} [group G]
  {P : subgroup G} (hP : is_p_group p P) (H : subgroup G)
  (hH : P ≤ H) : is_p_group p H :=

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}


theorem exercise_4_5_14 {G : Type*} [group G] [fintype G]
  (hG : card G = 312) :
  ∃ (p : ℕ) (P : sylow p G), P.normal :=

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}


theorem exercise_4_5_16 {p q r : ℕ} {G : Type*} [group G]
  [fintype G]  (hpqr : p < q ∧ q < r)
  (hpqr1 : p.prime ∧ q.prime ∧ r.prime)(hG : card G = p*q*r) :
  nonempty (sylow p G) ∨ nonempty(sylow q G) ∨ nonempty(sylow r G) :=

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}


theorem exercise_4_5_18 {G : Type*} [fintype G] [group G]
  (hG : card G = 200) :
  ∃ N : sylow 5 G, N.normal :=

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}


theorem exercise_4_5_20 {G : Type*} [fintype G] [group G]
  (hG : card G = 1365) : ¬ is_simple_group G :=

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}


theorem exercise_4_5_22 {G : Type*} [fintype G] [group G]
  (hG : card G = 132) : ¬ is_simple_group G :=

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}


theorem exercise_4_5_28 {G : Type*} [group G] [fintype G]
  (hG : card G = 105) (P : sylow 3 G) [hP : P.normal] :
  comm_group G :=

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}


theorem exercise_5_4_2 {G : Type*} [group G] (H : subgroup G) :
  H.normal ↔ ⁅(⊤ : subgroup G), H⁆ ≤ H :=

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}


theorem exercise_7_1_11 {R : Type*} [ring R] [is_domain R]
  {x : R} (hx : x^2 = 1) : x = 1 ∨ x = -1 :=

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}


theorem exercise_7_1_15 {R : Type*} [ring R] (hR : ∀ a : R, a^2 = a) :
  comm_ring R :=

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}


theorem exercise_7_2_12 {R G : Type*} [ring R] [group G] [fintype G] :
  ∑ g : G, monoid_algebra.of R G g ∈ center (monoid_algebra R G) :=

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}


theorem exercise_7_3_37 {R : Type*} {p m : ℕ} (hp : p.prime)
  (N : ideal $ zmod $ p^m) :
  is_nilpotent N ↔  is_nilpotent (ideal.span ({p} : set $ zmod $ p^m)) :=

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}


theorem exercise_8_1_12 {N : ℕ} (hN : N > 0) {M M': ℤ} {d : ℕ}
  (hMN : M.gcd N = 1) (hMd : d.gcd N.totient = 1)
  (hM' : M' ≡ M^d [ZMOD N]) :
  ∃ d' : ℕ, d' * d ≡ 1 [ZMOD N.totient] ∧
  M ≡ M'^d' [ZMOD N] :=

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}


theorem exercise_8_3_4 {R : Type*} {n : ℤ} {r s : ℚ}
  (h : r^2 + s^2 = n) :
  ∃ a b : ℤ, a^2 + b^2 = n :=

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}


theorem exercise_8_3_6a {R : Type*} [ring R]
  (hR : R = (gaussian_int ⧸ ideal.span ({⟨0, 1⟩} : set gaussian_int))) :
  is_field R ∧ ∃ finR : fintype R, @card R finR = 2 :=

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}


theorem exercise_9_1_6 : ¬ is_principal
  (ideal.span ({X 0, X 1} : set (mv_polynomial (fin 2) ℚ))) :=

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}


theorem exercise_9_3_2 {f g : polynomial ℚ} (i j : ℕ)
  (hfg : ∀ n : ℕ, ∃ a : ℤ, (f*g).coeff = a) :
  ∃ a : ℤ, f.coeff i * g.coeff j = a :=

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}


theorem exercise_9_4_2b : irreducible
  (X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : polynomial ℤ) :=

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}


theorem exercise_9_4_2d {p : ℕ} (hp : p.prime ∧ p > 2)
  {f : polynomial ℤ} (hf : f = (X + 2)^p):
  irreducible (∑ n in (f.support \ {0}), (f.coeff n) * X ^ (n-1) :
  polynomial ℤ) :=

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}


theorem exercise_9_4_11 :
  irreducible ((X 0)^2 + (X 1)^2 - 1 : mv_polynomial (fin 2) ℚ) :=

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}