Is Abstract Algebra the Key to Unlocking Mathematical Concepts?

[dalcde]

I have started to write Abstract Algebra notes as I am learning them, and typing them with LaTex afterwards. I have just done a bit but I want some of you to help and see if I have got any thing wrong (having the wrong concept in your mind can have terrible consequences) or anything else to make them more comprehensible. Please have a look.

I will post the LaTex code below in case there is formatting problems (I'm not really an expert of LaTex)

\documentclass[a4paper,10pt]{article}

\usepackage[utf8x]{inputenc}
\usepackage{amssymb}

\title{Abstract Algebra}
\author{}
\date{}

\pdfinfo{%
/Title (Abstract Algebra)
/Author ()
/Creator ()
/Producer ()
/Subject ()
/Keywords ()
}

\begin{document}
\maketitle
\tableofcontents
\pagebreak
\section{Groups}

\subsection{Binary Operators}
\textbf{Definition.} A \textbf{binary operation} on a set $S$ is a function $f:S\times S\rightarrow S$.\\
They are usually denoted with infix operators, e.g.
$$s\cdot t, s\ast t, etc.$$
A binary operation, $\ast$ is always closed, i.e.
$$\forall s,t\in S: s\ast t\in S$$
\textbf{Definition.} A binary operation $\ast$ is
\begin{enumerate}
\item \textbf{Associative} if $a\ast(b\ast c)=(a\ast b)\ast c$
\item \textbf{Communative} if $a\ast b=b\ast a$
\end{enumerate}
\textbf{Example.} $+$, $-$, $\times$ are binary operations in $\mathbb{R}$\\
A binary operation can also be defined by a table:\\
\begin{tabular}{|c|c|c|}
\hline
$\ast$ & a & b\\
\hline
a & b & a\\
\hline
b & a & b\\
\hline
\end{tabular}\\
i.e. $a \ast b=b$, $a\ast b=a$\\
$b \ast a=a$, $b\ast b=b$\\
It is communative:
$$a\ast b=b\ast a=a$$
It is also associative (which takes some time to prove).

\subsection{Groups}
\textbf{Definition.} A \textbf{group} is a set $G$ with a binary operator that $\ast$ satisfy
\begin{enumerate}
\item $\forall a, b, c: a\ast(b\ast c)=(a\ast b)\ast c$ (Associativity)
\item $\exists e\in G: \forall a\in G: e\ast a=a\ast e=a$ (Identity)
\item $\forall a\in G: \exists a'\in G: a\ast a'=a'\ast a=e$ (Inverse)
\end{enumerate}
\textbf{Definition.} A group is \textbf{abelian} iff it is communtative.\\
\textbf{Definition.} The \textbf{order} of a group $G$, denoted by $|G|$,m is the number of elements in it.\\
A finite group is a group with finite order.\\
An infinite group is a group with infinite order.\\
\textbf{Example.} $\mathbb{Z}$ with addition is a group, as
\begin{enumerate}
\item Addition is associative
\item 0 is the identity
\item The inverse of any integer $a$ is $-a$
\end{enumerate}
\textbf{Example.} Define $\ast$ on the reals to be
$$a\ast b=a+b+3$$
We shall show that this makes a group
\begin{enumerate}
\item $a\ast(b\ast c)=a\ast(b+c+3)=a+(b+c+3)+3=a+b+c+6$\\
$(a\ast b)\ast c=(a+b+3)\ast c=(a+b+3)+c+3=a+b+c+6$\\
Therefore it is associative.
\item Let $e$ be the identity. Hence $e\ast a = a$, $e+a+3=a$, $e=-3$
\item For all $A$, there should be an inverse $a'$.
\begin{eqnarray*}
a\ast a'&=&-3\\
a+a'+3=-3\\
a=-a-6\\
\end{eqnarray*}
So there exists an invers for all $a$ since subtraction (and negation) is well defined in the reals
\end{enumerate}
\textbf{Definition.} $\mathbb{Z}_n$ is the group (and later ring) of integers modulo $n$, containing 1, 2, ... $n-1$.\\
Operations are defined as the normal operations (addition or multiplication) with the answers modulo $n$\\
\textbf{Example.} $\mathbb{Z}_3$ is a group with the following table:\\
\begin{tabular}{|c|c|c|c|}
\hline
+ & 0 & 1 & 2\\
\hline
0 & 0 & 1 & 2\\
\hline
1 & 1 & 2 & 0\\
\hline
2 & 2 & 0 & 1\\
\hline
\end{tabular}

\pagebreak
\section{Glossary of Definitions}
\textbf{Definition.} A \textbf{binary operation} on a set $S$ is a function $f:S\times S\rightarrow S$.\\
\textbf{Definition.} A binary operation $\ast$ is
\begin{enumerate}
\item \textbf{Associative} if $a\ast(b\ast c)=(a\ast b)\ast c$
\item \textbf{Communative} if $a\ast b=b\ast a$
\end{enumerate}
\textbf{Definition.} A \textbf{group} is a set $G$ with a binary operator that $\ast$ satisfy
\begin{enumerate}
\item $\forall a, b, c: a\ast(b\ast c)=(a\ast b)\ast c$ (Associativity)
\item $\exists e\in G: \forall a\in G: e\ast a=a\ast e=a$ (Identity)
\item $\forall a\in G: \exists a'\in G: a\ast a'=a'\ast a=e$ (Inverse)
\end{enumerate}
or (in words)
\begin{enumerate}
\item For all $a, b$ and $c$ in $G$, $a\ast(b\ast c)=(a\ast b)\ast c$ (Associativity)
\item There exists an $e$ in $G$, called the identity element, such that for all $a$, $e\ast a=a\ast e=a$ (Identity)
\item For any $a$, there is an inverse element, $a'$, in $G$ such that $a\ast a'=a'\ast a=e$ (Inverse)
\end{enumerate}
\textbf{Definition.} A group is \textbf{abelian} iff it is communtative.\\
\textbf{Definition.} The \textbf{order} of a group $G$, denoted by $|G|$,m is the number of elements in it.\\
\textbf{Definition.} $\mathbb{Z}_n$ is the group (and later ring) of integers modulo $n$, containing 1, 2, ... $n-1$.\\
Operations are defined as the normal operations (addition or multiplication) with the answers modulo $n$\\
\end{document}

EDIT: Sorry for the formatting of the LaTex Code presented above. The forum seems to automatically put some of the code into a box (or is it just me?)

 

[micromass]

Hi dalcde!

I only found one small mistake. You said that $\mathbb{Z}_n$ contains 1,2,...,n-1. This is correct, but it also contains 0! Thus $\mathbb{Z}_n=\{0,1,...,n-1\}$.

The rest looks very good!

 

[dalcde]

Thanks!

 

[mathwonk]

huh?

 

[mathwonk]

how much to put my web notes into tex?