1\documentclass{article}
  2\usepackage[utf8]{inputenc}
  3\usepackage{amsfonts}
  4\usepackage{amsthm}
  5\usepackage{amsmath}
  6\usepackage{enumerate}
  7\usepackage{hyperref}
  8\hypersetup{
  9	colorlinks,
 10	citecolor=black,
 11	filecolor=black,
 12	linkcolor=black,
 13	urlcolor=black
 14}
 15
 16% custom solution environment to set custom numbers
 17\theoremstyle{definition}
 18\newtheorem{innersolution}{Solution}
 19\newenvironment{solution}[1]
 20{\renewcommand\theinnersolution{#1}\innersolution}
 21{\endinnersolution}
 22
 23\title{Seminar exercises}
 24\author{ }
 25\date{February 2022}
 26
 27\begin{document}
 28\maketitle
 29
 30\begin{solution}{1.9}\
 31	\begin{enumerate}[1.]
 32		\item Let $f(a) = u$, then $g(f(a)) = g(u)$, so $g \circ f$ is a function.
 33		\item We can see that composition of functions is associative as follows:\\
 34			we know that $[ f \circ g](x) = f(g(x)), \forall x \in A$,\\
 35			so,
 36			$$(h \circ [g \circ f])(x) = h([g \circ f](x)) = h(g(f(x)))$$
 37			\\
 38			and
 39			$$([h \circ g] \circ f)(x) = [h \circ g](f(x)) = h(g(f(x)))$$
 40			Then, we can see that $$h \circ (g \circ f) = h(g(f(x))) = (h \circ g) \circ f$$
 41	\end{enumerate}
 42\end{solution}
 43
 44\begin{solution}{1.28}\
 45	\emph{(WIP)}\\
 46	It is isomorphic to the cosets of the \emph{nth} roots of unity, which are $\mathbb{G}_n = \{w_k\}^{n-1}_{k=0}$, where $w_k=e^{\frac{2 \pi i}{n}}$.
 47\end{solution}
 48
 49\begin{solution}{2.2}\
 50
 51	To prove that the inverse $x^{-1}$ is unique, assume $x^{-1}$ and $\tilde{x}^{-1}$ are two inverses of $x$.\\
 52	By the definition of the inverse, we know that $x \cdot x^{-1} = e$. And by the definition of the unit element, we know that $x \cdot e = x$.\\
 53	Then, $$x^{-1} \cdot (x \cdot \tilde{x}^{-1}) = x^{-1} \cdot e = x^{-1}$$
 54	and $$(x^{-1} \cdot x) \cdot \tilde{x}^{-1} = e \cdot \tilde{x}^{-1} = \tilde{x}^{-1}$$
 55	By associativity property of groups, we know that
 56	$$x^{-1} \cdot (x \cdot \tilde{x}^{-1}) = (x^{-1} \cdot x) \cdot \tilde{x}^{-1}$$
 57	so, $$x^{-1} \cdot e = e \cdot \tilde{x}^{-1}$$
 58	which is $$x^{-1} = \tilde{x}^{-1}$$
 59	So, for any $x \in G$, the inverse $x^{-1}$ is unique.
 60\end{solution}
 61
 62\begin{solution}{2.5}\
 63
 64	Let $\alpha = (\begin{smallmatrix}1 & 2 & 3\\ 1 & 3 & 2\end{smallmatrix})$, $\beta = (\begin{smallmatrix}1 & 2 & 3\\ 3 & 1 & 2\end{smallmatrix})$, then,
 65	$$
 66	\alpha \cdot \beta = 
 67	(\begin{smallmatrix}1 & 2 & 3\\ 1 & 3 & 2\end{smallmatrix})
 68	\cdot (\begin{smallmatrix}1 & 2 & 3\\ 3 & 1 & 2\end{smallmatrix})
 69	= (\begin{smallmatrix}1 & 2 & 3\\ 3 & 2 & 1\end{smallmatrix})
 70	$$
 71
 72	and
 73	$$
 74	\beta \cdot \alpha = 
 75	(\begin{smallmatrix}1 & 2 & 3\\ 3 & 1 & 2\end{smallmatrix}) \cdot
 76	(\begin{smallmatrix}1 & 2 & 3\\ 1 & 3 & 2\end{smallmatrix})
 77	= (\begin{smallmatrix}1 & 2 & 3\\ 2 & 1 & 3\end{smallmatrix})
 78	$$
 79
 80	So, we can see that
 81	$$
 82	(\begin{smallmatrix}1 & 2 & 3\\ 3 & 2 & 1\end{smallmatrix})
 83	\neq
 84	(\begin{smallmatrix}1 & 2 & 3\\ 2 & 1 & 3\end{smallmatrix})
 85	$$
 86
 87	so, $\alpha \cdot \beta \neq \beta \cdot \alpha$.
 88\end{solution}
 89
 90\begin{solution}{2.26}\
 91
 92	We want to prove that $f: G \rightarrow H$ is a \emph{monomorphism} iff $\ker f=\{e\}$.\\
 93	We know that $f$ is a \emph{monomorphism} (\emph{injective}) iff $\forall a, b \in G$, $f(a) = f(b) \Rightarrow a = b$.\\
 94	Let $a, b \in G$ such that $f(a)=f(b)$. Then
 95	$$f(a) f(b)^{-1} = f(b) (f(b))^{-1} = e$$
 96	$$f(a) f(b^{-1}) = e$$
 97	$$f(ab^{-1}) = e$$
 98	as $\ker f = \{e\}$, then we see that $ab^{-1}=e$, so $a=b$. Thus $f$ is a \emph{monomorphism}.
 99\end{solution}
100
101\end{document}