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% Document MUST be built with XeLaTeX!
\documentclass[11pt,a4paper]{article}
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\usepackage{cancel} % Зачертаване
\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
\usepackage{indentfirst} % Индентация на параграф след секция
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\setlist[enumerate]{label=\arabic*.}
\renewcommand{\thesection}{Решение}
\renewcommand{\thesubsection}{(\roman{subsection})}
\newcommand{\im}{\operatorname{im}}
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\setlist[enumerate]{label=\Roman*.}
\usepackage[pdfauthor={Камен Димитров Младенов, КН Летен семестър, 2023},
pdftitle={Домашна работа 1 по А2, Зад 1},
pdfsubject={},
pdfkeywords={},
colorlinks=true,
urlcolor=blue]{hyperref}
\title{\vspace{-1.5cm}Домашно №1, Задача 1 по Алгебра 2 на Компютърни Науки}
\author{Камен Димитров Младенов, Група 6}
\date{7 Април 2023}
\begin{document}
\maketitle
\subsection{Подточка}
Ще докажем, че $G$ е подгрупа на $GL_2(\mathbb{R})$, съгласно
\begin{align*}
\forall A = \left( \begin{array}{cc}
a & c \\
0 & b
\end{array} \right) \in G, \text{ знаем че } a, b \in \mathbb{R}^* \implies \det(A) = a.b - c.0 = ab \neq 0 \text{ и } G \subseteq GL_2(\mathbb{R})
\end{align*}
Тоест ще докажем, че $\forall A,B \in G \Rightarrow A.B \in G, B^{-1} \in G$. Нека
\begin{align*}
A = \left( \begin{array}{cc}
a_1 & c_1 \\
0 & b_1
\end{array} \right) \qquad
B = \left( \begin{array}{cc}
a_2 & c_2 \\
0 & b_2
\end{array} \right)
\end{align*}
\begin{align*}
A.B =
\left( \begin{array}{cc}
a_1 & c_1 \\
0 & b_1
\end{array} \right)
\left( \begin{array}{cc}
a_2 & c_2 \\
0 & b_2
\end{array} \right) =
\left( \begin{array}{cc}
a_1 a_2 & a_1 c_2 + c_1 b_2 \\
0 & b_1 b_2
\end{array} \right) = C
\end{align*}
\begin{align*}
a_1 a_2, b_1 b_2 \in \mathbb{R}^* \implies C \in G \text{ и } \det(C) \neq 0
\end{align*}
\begin{align*}
B^{-1} = \left( \begin{array}{cc}
a_2 & c_2 \\
0 & b_2
\end{array} \right)^{-1} = \frac{1}{\det(B)}
\left( \begin{array}{cc}
B_{11} & B_{12} \\
B_{21} & B_{22}
\end{array} \right)^\top = \frac{1}{\det(B)}
\left( \begin{array}{cc}
b_2 & 0 \\
c_2 & a_2
\end{array} \right)^\top = \frac{1}{\det(B)}
\left( \begin{array}{cc}
b_2 & c_2 \\
0 & a_2
\end{array} \right)
\end{align*}
\begin{align*}
a_2, b_2 \in \mathbb{R}^* \text{ и } \det(B) \neq 0 \implies B^{-1} \in G \text{ и } \det(B^{-1}) \neq 0
\end{align*}
\subsection{Подточка}
Ще докажем, че изображението $\varphi$ е хомоморфизъм на групи, като
\begin{align*}
\varphi : G \rightarrow \mathbb{R}^*, \quad \forall
\left( \begin{array}{cc}
a & c \\
0 & b
\end{array} \right) \in G: \;
\varphi
\left( \begin{array}{cc}
a & c \\
0 & b
\end{array} \right) = a^2b
\end{align*}
\begin{align*}
\varphi \left(
\left( \begin{array}{cc}
a_1 & c_1 \\
0 & b_1
\end{array} \right)
\left( \begin{array}{cc}
a_2 & c_2 \\
0 & b_2
\end{array} \right)
\right) & =
\varphi \left(
\left( \begin{array}{cc}
a_1 a_2 & a_1 c_2 + c_1 b_2 \\
0 & b_1 b_2
\end{array} \right)
\right) = (a_1a_2)^2(b_1b_2) = (a_1^2b_1)(a_2^2b_2) \\
& = \varphi \left(
\left( \begin{array}{cc}
a_1 & c_1 \\
0 & b_1
\end{array} \right) \right)
\varphi \left(
\left( \begin{array}{cc}
a_2 & c_2 \\
0 & b_2
\end{array} \right) \right) \implies \varphi \text{ е хомоморфизъм}
\end{align*}
\begin{align*}
\forall a \in \mathbb{R}^*, \forall c \in \mathbb{R}: \; \varphi \left(
\left( \begin{array}{cc}
a & c \\
0 & a^{-1}
\end{array} \right) \right) = a^2a^{-1} = a \implies \varphi \text{ е върху}
\end{align*}
Забелязваме, че
\begin{align*}
\forall a \in \mathbb{R}^* \forall c \in \mathbb{R}: \; \varphi \left(
\left( \begin{array}{cc}
a & c \\
0 & a^{-2}
\end{array} \right) \right) = a^2a^{-2} = 1
\implies \ker(\varphi) = \left\{ \left( \left. \begin{array}{cc}
a & c \\
0 & a^{-2}
\end{array} \right) \right| a \in \mathbb{R}^*, c \in \mathbb{R} \right\} = H
\end{align*}
\begin{align*}
\forall a, b \in \mathbb{R}^*, \forall c \in \mathbb{R}: \; \varphi
\left( \begin{array}{cc}
a & c \\
0 & b
\end{array} \right) = a^2b \in \mathbb{R}^* \implies \im(\varphi) = \mathbb{R}^*
\end{align*}
От теоремата за хомоморфизмите, следва че $G / H = G / \ker(\varphi) \simeq \im (\varphi) = \mathbb{R}^*$
\end{document}
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