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\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}
~~