Merge branch 'master' of ssh://git.betalupi.com:33/Mark/ormc-handouts

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Mark 2023-01-16 08:30:46 -08:00
commit 284107ab48

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@ -6,7 +6,7 @@ A group must have the following properties: \\
\begin{enumerate} \begin{enumerate}
\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$. \item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
\item $\ast$ is associative: $a \ast b = b \ast a$ \item $\ast$ is associative: $(a \ast b) \ast c = a \ast (b \ast c)$
\item There is an \textit{identity} $\overline{0} \in G$, so that $a \ast \overline{0} = a$ for all $a \in G$. \item There is an \textit{identity} $\overline{0} \in G$, so that $a \ast \overline{0} = a$ for all $a \in G$.
\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = \overline{0}$. $b$ is called the \textit{inverse} of $a$. \\ \item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = \overline{0}$. $b$ is called the \textit{inverse} of $a$. \\
This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise. This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise.
@ -42,13 +42,59 @@ Make it one by modifying $\mathbb{R}$. \\
\vfill \vfill
\problem{} \problem{}
Can you construct a group that contains a single element? What is the smallest group we can create?
\begin{solution} \begin{solution}
Let $(G, \circledcirc)$ be our group, where $G = \{\star\}$ and $\circledcirc$ is defined by the identity $\star \circledcirc \star = \star$ Let $(G, \circledcirc)$ be our group, where $G = \{\star\}$ and $\circledcirc$ is defined by the identity $\star \circledcirc \star = \star$
Verifying that the trivial group is a group is trivial. Verifying that the trivial group is a group is trivial.
\end{solution} \end{solution}
\vfill
\problem{}
Let $G$ be the set of all bijections $A \to A$. \\
Let $\circ$ be the usual composition operator. \\
Is $(G, \circ)$ a group?
\vfill
\pagebreak
\problem{}
Show that a group has exactly one identity element.
\vfill
\problem{}
Show that each element in a group has exactly one inverse.
\vfill
\problem{}
Let $(G, \ast)$ be a group and $a, b, c \in G$. Show that...
\begin{itemize}
\item $a \ast b$ and $a \ast c \implies b = c$
\item $b \ast a$ and $c \ast a \implies b = c$
\end{itemize}
What does this mean intuitively?
\vfill
\problem{}
Let $(G, \ast)$ be a finite group (i.e, $G$ has finitely many elements), and let $g \in G$. \\
Show that $\exists~n \in Z^+$ so that $g^n = \overline{0}$ \\
\hint{$g^n = g \ast g \ast ... \ast g$ $n$ times.}
\vspace{2mm}
The smallest such $n$ defines the \textit{order} of $(G, \ast)$.
\vfill
\problem{}
What is the order of 5 in $(\mathbb{Z}/25, +)$? \\
What is the order of 2 in $((\mathbb{Z}/17)^\times, \times)$? \\
\vfill
\problem{}
Show that if $G$ has four elements, $(G, \ast)$ is abelian.
\vfill \vfill
\pagebreak \pagebreak
@ -76,20 +122,12 @@ Recall your tables from \ref{modtables}: \\
\end{tabular} \end{tabular}
\end{center} \end{center}
Convince yourself that $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ are the same group. \\ Look at these tables and convince yourself that $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ are the same group. \\
We say that two such groups are \textit{isomorphic}. We say that two such groups are \textit{isomorphic}.
\vspace{2mm} \vspace{2mm}
Intuitively, this means that $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ have the same algebraic structure. We can translate statements about addition in $\mathbb{Z}/4$ into statements about multiplication in $(\mathbb{Z}/5)^\times$ \\ Intuitively, this means that these two groups have the same algebraic structure. We can translate statements about addition in $\mathbb{Z}/4$ into statements about multiplication in $(\mathbb{Z}/5)^\times$ \\
\problem{}
Show that a group has exactly one identity element.
\vfill
\problem{}
Show that each element in a group has exactly one inverse.
\vfill
\pagebreak \pagebreak