<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\section{Groups}

\definition{}
A \textit{group} $G = (S, \ast)$ consists of a set $S$ and a binary operator $\ast$. \par
By definition, a group always has 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: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
	\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
	\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
\end{enumerate}

\note[Note]{
	Commutativity is \textit{not} a required property of a group! \\
	In most cases, $a \ast b \neq b \ast a$.
}



\problem{}
Is $(\znz{5}, +)$ a group? \par
How about $(\znz{5}, -)$? \par
\hint{In this problem, $+$ and $-$ work just as you'd expect.}

\vfill

\problem{}
What is the smallest possible group?

\begin{solution}
	Let $(G, \ast)$ be our group, where $G = \{e\}$ and $\ast$ is defined by the identity $e \ast e = e$

	Verifying that the trivial group is a group is trivial.
\end{solution}

\vfill

\problem{}
How many distinct groups have two elements? \par
\hint{
	Two groups are \say{the same} if the elements of one can be renamed to get the other. \\
	A group is fully defined by its multiplication table.
}


\vfill
\pagebreak



%\problem{}&lt;firstcross&gt;
%Is $(\znz{17}, \times)$ a group? \par
%How should we modify $\znz{17}$ to make it one?

%\problem{}&lt;secondcross&gt;
%Is $(\znz{6}, \times)$ a group? \par
%How should we modify $\znz{6}$ to make it one? \par
%\hint{
%	Be careful, this isn't as easy as \ref{firstcross}. \\
%	Which elements aren't invertible?
%}


%\definition{}
%Building on problems \ref{num:firstcross} and \ref{num:secondcross}, we'll define $(\znz{n})^\times$ as the multiplicative
%group of integers mod $n$. \par

%Specifically, $(\znz{n})^\times$ is the set of all integers coprime to $n$. \par

%\vspace{2mm}

%For example, $(\znz{6})^\times = \{1, 5\}$ \par
%and $(\znz{15})^\times = \{1, 2, 4, 7, 8, 11, 13, 14\}$ \par

%\vspace{2mm}

%Note that $0$ is the identity in $\znz{n}$ and $1$ is the identity in $(\znz{n})^\times$\hspace{-1.5ex}. \par
%\note[Note]{
%	Also, notice that we've omitted the operations $+$ and $\times$ in the two groups above. \\
%	These operations are implicitly \say{attached} to $\znz{n}$ and $(\znz{n})^\times$\hspace{-1.5ex}, \\
%	and we rarely write them for the sake of cleaner notation.
%}


\vfill


\definition{}
Let $G$ be a group, $a$ an element of $G$, and $n \in \mathbb{Z}^+$. \par
$a^n$ is the defined as $a \ast a \ast ... \ast a$, repeated $n$ times.

\vspace{1mm}

Note that this is \textbf{not} \say{normal} exponentiation! \par
If our group's operator is $+$ (for example, $\znz{5}$), $a^n = a + ... + a$, \par
which you'll recognize as multipication.

\vspace{1mm}

Beware of this odd notation. By convention, we use \say{multiplicative} notation
when working with groups---so, $a \ast b$ may also be written as $ab$,
and $a \ast a \ast a$ may be written as $a^3$.

\vspace{1mm}

Again, remember that $a^n$ simply means \say{$\ast$ $a$ with itself $n$ times,} \par
regardless of the specific operator our group uses.

\problem{}
Let $a$ be an element of a finite group. \par
Show that there is a positive integer $n$ so that $a^n = e$. \par

\vspace{2mm}

The smallest such $n$ defines the \textit{order} of $g$.

\vfill

\problem{}
Find the order of 5 in $(\znz{25}, +)$. \par
%Find the order of 2 in $((\znz{17})^\times, \times)$. \par
Find the order of 2 in $(\znz{7}, +)$. \par

\vfill
\pagebreak


\definition{}
Let $G$ be a group. \par
We say a $g \in G$ is a \textit{generator} of $G$
if every element in $G$ can be written as some power of $g$.

\vspace{2mm}

If $G$ has a generator, we say $G$ is \textit{cyclic.}

\problem{}
Find a generator of $\znz{7}$. Then, find a generator of $(\znz{7})^\times$

\vfill

\definition{}
Let $G$ be a group. \par
The \textit{order} of $G$ is the number of elements in $G$. \par
We'll write this as $|G|$, using the same notation we use with sets. \par
\note[Note]{
	Don't confuse the order of an \textbf{element}
	with the order of a \textbf{group}!
}

\problem{}
Let $G$ be a cyclic group, and let $g$ be any generator in $G$. \par
Show that $\text{ord}(g) = |G|$. \par
\hint{Contradiction.}

\vfill
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