Crypto edits

Advanced/Cryptography/parts/2 groups.tex

@@ -29,7 +29,8 @@ Is $(\mathbb{Z}_5, -)$ a group? \par
 
 
 \problem{}
-Show that $(\mathbb{R}, \times)$ is not a group, then make it one by modifying $\mathbb{R}$. \par
+Show that $(\mathbb{R}, \times)$ is not a group,
+then find a subset $S$ of $\mathbb{R}$ so that $(S, \times)$ is a group.
 
 \begin{solution}
 	$(\mathbb{R}, \times)$ is not a group because $0$ has no inverse. \par
@@ -58,8 +59,8 @@ What is the smallest group we can create?
 
 \problem{}
 Let $(G, \ast)$ be a group with finitely many elements, and let $a \in G$. \par
-Show that $\exists n \in \mathbb{Z}^+$ so that $a^n = e$ \par
-\hint{$a^n = a \ast a \ast ... \ast a$ repeated $n$ times.}
+Show that there exists an $n$ in $\mathbb{Z}^+$ so that $a^n = e$ \par
+\hint{$a^n \coloneqq a \ast a \ast ... \ast a$, with $a$ repeated $n$ times.}
 
 \vspace{2mm}
 
@@ -77,8 +78,9 @@ What is the order of 2 in $(\mathbb{Z}_{17}^\times, \times)$? \par
 \theorem{}
 Let $p$ be a prime number. \par
 In any group $(\mathbb{Z}_p^\times, \ast)$ there exists a $g \in \mathbb{Z}_p^\times$ where...
-\begin{itemize}
-	\item The order of $g$ is $p - 1$
+
+\begin{itemize}[itemsep=1mm]
+	\item The order of $g$ is $p - 1$, and
 	\item $\{a^0,~ a^1,~ ...,~ a^{p - 2}\} = \mathbb{Z}_n^\times$
 \end{itemize}
 We call such a $g$ a \textit{generator}, since its powers generate every other element in the group.