Is Abstract Algebra the Key to Unlocking Mathematical Concepts?

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The discussion centers on the creation and review of Abstract Algebra notes, specifically focusing on the definitions and properties of binary operations and groups. The author seeks feedback to ensure accuracy and clarity, emphasizing the importance of correct concepts in mathematics. A small correction was noted regarding the inclusion of zero in the set of integers modulo n, clarifying that \mathbb{Z}_n includes 0 along with 1 to n-1. The author expresses gratitude for the feedback and inquires about the cost of converting their web notes into LaTeX format. Overall, the thread highlights collaboration in learning and refining mathematical concepts.
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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?)
 

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Hi dalcde! :smile:

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!
 
Thanks!
 
huh?
 
how much to put my web notes into tex?
 
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