Merge branch 'master' of ssh://git.betalupi.com:33/Mark/ormc-handouts
This commit is contained in:
commit
284107ab48
@ -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
|
||||||
|
|
||||||
|
Loading…
x
Reference in New Issue
Block a user