%\input{tcilatex} \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amsfonts} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2606} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %TCIDATA{Created=Thursday, July 22, 2004 13:38:11} %TCIDATA{LastRevised=Wednesday, April 04, 2007 14:41:27} %TCIDATA{} %TCIDATA{} %TCIDATA{Language=American English} %TCIDATA{CSTFile=Math.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %F=36,\PARA{038

\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}[Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section*{Problems and Comments for Section 5} \vspace{1pt} \textbf{Problems: }5.1, 5.2, 5.7, 5.18 (a), 5.19, 5.20, 5.21 \vspace{1pt}\textbf{Comments: }For a given algebra $(A,(f_{i})_{i\in I})$, a subset $C~$is called closed if it is closed under all the operations: If for example $f_{i}$ is a binary operation, say $\star ,$ and if $a,b\in C$ then also $a\star b\in C.$ Also a closed subset must contain all constants, which are the values of the nullary operations. For a group $\mathbf{G}=(G,\cdot ,^{-1},e)$ this means that a subset $C~$is closed if \begin{enumerate} \item If $a,b\in C$ then $a\cdot b\in C.$ \item If $a\in C$ then $a^{-1}\in C.$ \item $e\in C.$ \end{enumerate} One easily shows that the intersection of closed subsets is closed. The argument is the following. Assume that the $C_{t},t\in T,$ are closed subsets of the algebra $A.$ Let:% $C=\dbigcap\limits_{t\in T}C_{t}$ We wish to show that $C$ is closed. So assume, that $a,b\in C.$Then $a,b\in C_{t}$ for all $t\in T.$ Because each $C_{t}$ is closed we have that $a\star b\in C_{t}$ for all $t\in T.$ Hence $a\star b\in C.$ Of course, the whole algebra $A$ is closed. The empty set $\emptyset$ is closed only if the algebra does not have constant operations. It is clear that a closed subset of a group is a group where the group operations are restricted to the closed subset. This is so because the group axioms are equations. For example, associativity $a\cdot (b\cdot c)=(a\cdot b)\cdot c$ holds for all group elements, in particular then for elements of the closed subset. Thus it is appropriate to call a closed subset of a group a subgroup. A closed subset of an arbitrary algebra is called a subalgebra. For any subset $S$ of an algebra $A,$ the intersection of all closed subsets of $A$ which contain $S$ is a subalgebra. It is called the subalgebra generated by $S~$and it is of course then the smallest subalgebra that contains $S$ and is denoted $.$ For a group $G$, and $S=\{a\},$ the subgroup generated by $\{a\}$ is then the cyclic group generated by $a.$ \end{document}