158 lines
4.3 KiB
TeX
158 lines
4.3 KiB
TeX
\section{Groups}
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\definition{}
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A \textit{group} $G = (S, \ast)$ consists of a set $S$ and a binary operator $\ast$. \par
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By definition, a group always has the following properties:
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\begin{enumerate}
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\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
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\item $\ast$ is associative: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
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\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
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\item Any $a \in G$ has an \textit{inverse} $a^{-1} \in G$ that satisfies $a \ast a^{-1} = a^{-1} \ast a = e$. \par
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\end{enumerate}
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\note[Note]{
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Commutativity is \textit{not} a required property of a group! \\
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In most cases, $a \ast b \neq b \ast a$.
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}
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\problem{}
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Is $(\znz{5}, +)$ a group? \par
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How about $(\znz{5}, -)$? \par
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\hint{In this problem, $+$ and $-$ work just as you'd expect.}
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\vfill
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\problem{}
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What is the smallest possible group?
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\begin{solution}
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Let $(G, \ast)$ be our group, where $G = \{e\}$ and $\ast$ is defined by the identity $e \ast e = e$
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Verifying that the trivial group is a group is trivial.
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\end{solution}
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\vfill
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\problem{}
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How many distinct groups have two elements? \par
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\hint{
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Two groups are \say{the same} if the elements of one can be renamed to get the other. \\
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A group is fully defined by its multiplication table.
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}
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\vfill
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\pagebreak
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%\problem{}<firstcross>
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%Is $(\znz{17}, \times)$ a group? \par
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%How should we modify $\znz{17}$ to make it one?
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%\problem{}<secondcross>
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%Is $(\znz{6}, \times)$ a group? \par
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%How should we modify $\znz{6}$ to make it one? \par
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%\hint{
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% Be careful, this isn't as easy as \ref{firstcross}. \\
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% Which elements aren't invertible?
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%}
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%\definition{}
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%Building on problems \ref{num:firstcross} and \ref{num:secondcross}, we'll define $(\znz{n})^\times$ as the multiplicative
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%group of integers mod $n$. \par
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%Specifically, $(\znz{n})^\times$ is the set of all integers coprime to $n$. \par
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%\vspace{2mm}
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%For example, $(\znz{6})^\times = \{1, 5\}$ \par
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%and $(\znz{15})^\times = \{1, 2, 4, 7, 8, 11, 13, 14\}$ \par
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%\vspace{2mm}
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%Note that $0$ is the identity in $\znz{n}$ and $1$ is the identity in $(\znz{n})^\times$\hspace{-1.5ex}. \par
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%\note[Note]{
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% Also, notice that we've omitted the operations $+$ and $\times$ in the two groups above. \\
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% These operations are implicitly \say{attached} to $\znz{n}$ and $(\znz{n})^\times$\hspace{-1.5ex}, \\
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% and we rarely write them for the sake of cleaner notation.
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%}
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\vfill
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\definition{}
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Let $G$ be a group, $a$ an element of $G$, and $n \in \mathbb{Z}^+$. \par
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$a^n$ is the defined as $a \ast a \ast ... \ast a$, repeated $n$ times.
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\vspace{1mm}
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Note that this is \textbf{not} \say{normal} exponentiation! \par
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If our group's operator is $+$ (for example, $\znz{5}$), $a^n = a + ... + a$, \par
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which you'll recognize as multipication.
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\vspace{1mm}
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Beware of this odd notation. By convention, we use \say{multiplicative} notation
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when working with groups---so, $a \ast b$ may also be written as $ab$,
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and $a \ast a \ast a$ may be written as $a^3$.
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\vspace{1mm}
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Again, remember that $a^n$ simply means \say{$\ast$ $a$ with itself $n$ times,} \par
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regardless of the specific operator our group uses.
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\problem{}
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Let $a$ be an element of a finite group. \par
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Show that there is a positive integer $n$ so that $a^n = e$. \par
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\vspace{2mm}
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The smallest such $n$ defines the \textit{order} of $g$.
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\vfill
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\problem{}
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Find the order of 5 in $(\znz{25}, +)$. \par
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%Find the order of 2 in $((\znz{17})^\times, \times)$. \par
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Find the order of 2 in $(\znz{7}, +)$. \par
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\vfill
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\pagebreak
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\definition{}
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Let $G$ be a group. \par
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We say a $g \in G$ is a \textit{generator} of $G$
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if every element in $G$ can be written as some power of $g$.
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\vspace{2mm}
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If $G$ has a generator, we say $G$ is \textit{cyclic.}
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\problem{}
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Find a generator of $\znz{7}$. Then, find a generator of $(\znz{7})^\times$
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\vfill
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\definition{}
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Let $G$ be a group. \par
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The \textit{order} of $G$ is the number of elements in $G$. \par
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We'll write this as $|G|$, using the same notation we use with sets. \par
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\note[Note]{
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Don't confuse the order of an \textbf{element}
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with the order of a \textbf{group}!
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}
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\problem{}
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Let $G$ be a cyclic group, and let $g$ be any generator in $G$. \par
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Show that $\text{ord}(g) = |G|$. \par
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\hint{Contradiction.}
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\vfill
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\pagebreak |