Edited group handout
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		| @ -6,7 +6,7 @@ A group must have the following properties: \\ | ||||
|  | ||||
| \begin{enumerate} | ||||
| 	\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 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. | ||||
| @ -42,13 +42,59 @@ Make it one by modifying $\mathbb{R}$. \\ | ||||
| \vfill | ||||
|  | ||||
| \problem{} | ||||
| Can you construct a group that contains a single element? | ||||
| What is the smallest group we can create? | ||||
|  | ||||
| \begin{solution} | ||||
| 	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. | ||||
| \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 | ||||
| \pagebreak | ||||
| @ -76,20 +122,12 @@ Recall your tables from \ref{modtables}: \\ | ||||
| \end{tabular} | ||||
| \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}. | ||||
|  | ||||
| \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 | ||||
|  | ||||
|  | ||||
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