\hfill \thepage} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Problems and Comments For Section 2} \vspace{1pt} \textbf{Problems: }2.1, 2.5, 2.7, 2.8 Problem 2.13 requires some thought. It is optional. A product term is defined \textit{recursively } as follows: \begin{enumerate} \item $p=a$ is a product where $a$ is any element of the group $\mathbf{G.}$ \item If $p_{1}$and $p_{2}$ are products, then $p=(p_{1}\cdot p_{2})$ is a product. \item All products are obtained that way. \end{enumerate} \vspace{1pt} For any two elements $a,b$ one has that $p=(a\cdot b)$ is a product. For three elements $a,b,c$ there are two possibilities to form a product of these elements without changing the order: $((a\cdot (b\cdot c))$ and $% ((a\cdot b)\cdot c).$ By associativity, both products have the same value. \vspace{1pt} The \textit{complexity} of a product is the number of dots in it. $p=a$ has complexity $0.$ If $p$ has complexity $n$ and $q$ complexity $m$ then $% (p\cdot q)$ has complexity $n+m+1.$ A product is a string of parenthesis (left and right), elements of $\mathbf{G}$ and dots. However such strings have to be constructed according to the rules 1. and 2. They have to be% \textit{\ well-formed. }\newline If, read from left to right, the elements in a product are $% a_{1},a_{2},\ldots ,a_{n}$ then the product $p=p(a_{1},a_{2},\ldots ,a_{n})$ has complexity $n-1.$Given a list $\ a_{1},a_{2},\ldots ,a_{n}$ of elements, the \textit{normal product} $\ n(a_{1},a_{2},\ldots ,a_{n})\ $of these elements is defined recursively by \begin{enumerate} \item $n(a_{n})=a_{n}$ \item $n(a_{1},a_{2},\ldots ,a_{n})=(a_{1}\cdot n(a_{2},\ldots ,a_{n}))$ \end{enumerate} \vspace{1pt} The claim of 2.13 now is% \[ p(a_{1},\ldots ,a_{n})=n(a_{1},\ldots ,a_{n}) \] for any product $p(a_{1},\ldots ,a_{n}).$Prove this by induction over the complexity of the product. 2.12 is a preparation for the general proof. Notice that for 2.12 and 2.13, only associativity is used. \textbf{Question:} Why did I say "dots", and not products, in the definition of complexity? \vspace{1pt} \textbf{Comments} A group is most often defined as an algebraic system with three operations: $% \cdot ,^{-1},e.$ Here $\cdot $ is binary, $^{-1}$ is unary and $e$ is nullary. The axioms for a group then are all \textit{equations}: \begin{enumerate} \item $a\cdot (b\cdot c)=(a\cdot b)\cdot c$ \item $a\cdot e=e\cdot a=a$ \item $a\cdot a^{-1}=a^{-1}\cdot a=e$ \end{enumerate} An algebraic system $\mathbf{A}=(A,\cdot )\ $with only one binary associative operation is called a\textit{\ semi group}. A semi group with an identity (or unit) is called a \textit{monoid}. Thus a group is a monoid where every element has an inverse. Notice that under this definition, a group cannot be empty. It must at least have one element, the identity. \vspace{1pt} With this convention, the notation for the additive group of integers is $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion =(Z,+,-,0).$ Here addition is the binary group operation. The multiplicative group of non-zero real numbers is $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{\ast }=(R\backslash \{0\},\cdot ,^{-1},1).$ For any set $S$, $Map(S)=\{f|\;f:S\rightarrow S\}$ is the set of maps $f$ from $S$ to $S.\ $It is a semigroup. Here the operation is composition of maps. The identity map $id:S\rightarrow S,x\mapsto x$ is the identity. \[ Map(S)=(\{f|\;f:S\rightarrow S\},\circ ,id) \] is the prototype of a monoid. It is not a group unless $S$ has only one element. \end{document}