\part{Cyclic Groups}

\chapter{Cyclic Groups}



Sometimes it is interesting to see how many distinct elements can be formed by repeating the operation on some specific element; we call this the order of the element.



\section{Cyclic Groups}

\begin{definition}
A \emph{cyclic group} $(G,\cdot)$ is a group such that there exists some element $\gamma$ that can generate all other elements of the group under the operation. That is, all elements are of the form $g^n$. We say that such a $\gamma$ \emph{generates} $G$.
	\[ G \text{ is cyclic } \iff  \forall g \in G ( \exists n \in \mathbb{Z} [ \gamma^n = g   ] ) \]
\end{definition}


If the group operator is clear, one may write the set of elements generated by the element $g$ as the following.
\[ \langle \gamma \rangle = \{ \gamma^{n} : n \in \mathbb{Z} \} \].

One may see that this is indeed forms its own group, and furthermore, a cyclic one.
\begin{proposition}
\[(G,\cdot) \land \gamma \in G\]
\[  ( \langle \gamma \rangle, \cdot) \leq (G,\cdot )  \]
\[ ( \langle \gamma \rangle, \cdot) \text{ is cyclic} \]
\end{proposition}

This is how we construct cyclic subgroups, and this gives us a reason to think of 'orders of elements' as 'orders of the cyclic subgroup they generate'.
\begin{definition}
The \emph{order} of a group element $g$ is the cardinality of $\langle g \rangle$.
The order of $g$ is denoted as $\mathrm{ord}(g)$.
\[ \mathrm{ord}(g) = | \langle g \rangle | \]
\end{definition}





\begin{definition}
Let $(G,\cdot)$ be a cyclic group. A \emph{generator of a cyclic group} is an element $\gamma $ that generates $G$. The order of a generator is the order of the group that it forms.
\[ \gamma \text{ is a generator of } G \iff G = \langle \gamma \rangle \]
\end{definition}




\section{Isomorphisms of Cyclic Groups}


Finite cyclic groups of the same order are all isomorphic by the isomorphism $f(\gamma^n)=\eta^n$ (where $G=\langle \gamma \rangle, H = \langle \eta \rangle$, hence we define a notation to represent the unique cyclic group of order $n$ (unique up to an isomorphism). 

\begin{example}
The \emph{cyclic group of order $n$} $\mathrm{Cyc}(n)$
Cyclic group of $n$ elements.
$\mathrm{Cyc}(n)= \langle a : a^n = 1 \rangle$
\begin{itemize}
	\item It is Abelian
	\item It has order $n$
\end{itemize}
\end{example}



Given a cyclig group generated by $g$, the law $g^{a}\cdot g^{b} = g^{a+b}$ in retrospect looks very much like a homomorphism to some $\mathbb{Z} / n \mathbb{Z}$ (where $n$ is the order of $g$); this turns out to be a fully fledged isomorphism.

So cyclic groups are isomorphic to additive groups of integers!

\begin{proposition}
\[ \mathrm{Cyc}(n) \cong \mathbb{Z} / n\mathbb{Z} \]
\end{proposition}


\section{Properties of Cyclic Groups}

The isomorphic equivalence $\mathrm{Cyc}(n) \cong \mathbb{Z} / n\mathbb{Z}$ really means that cylic groups are actually all about additive integer groups modulo $n$. This permits us to invoke techniques from number theory to fully realize the properties of cyclic groups.

The following lemma helps us understand more about the cyclic nature of $\mathbb{Z}/n\mathbb{Z}$
\begin{lemma}
	Let $m \in \mathbb{Z} /n \mathbb{Z}$, then the order of $m$ is $\frac{\lcm (m,n)}{m}$, or equivalently, $\frac{n}{\gcd(m,n)}$.
\end{lemma}

\begin{corollary}
Let $m \in \mathbb{Z} /n \mathbb{Z}$, then $m$ generates $\mathbb{Z} / n\mathbb{Z}$ iff $m$ is coprime to $n$.
\end{corollary}



\begin{proposition}
There are $\varphi(n)$ generators of $\mathrm{Cyc}(n)$.
\end{proposition}



\begin{proposition}
Groups of prime order are cyclic.
\[ |G| \text{ is prime } \implies G \cong \mathrm{Cyc}(|G|) \]
\end{proposition}


Now to prove some propositions on the subgroups of cyclic groups.

\begin{proposition}
Subgroups of cyclic groups are cyclic.
\end{proposition}
The following proposition is proved by switching from $\mathrm{Cyc}(n)$ to $\mathbb{Z} /n\mathbb{Z}$ showing that all elements in subgroups of $\mathbb{Z} / n\mathbb{Z}$ must be multiples (repeated operation) of the smallest nonzero number in that subgroup (if such a smallest nonzero number is in the group, which is true for all but the trivial group). This is a good example of leveraging the properties of integers, bringing their nice properties to facilitate proofs.




\begin{proposition}
If $d$ divides the order of a cyclic group, then there is a unique cyclic subgroup of order $d$.
\[  d | n \implies \exists! H \leq \mathrm{Cyc}(n) [|H|=d]\]
\end{proposition}
Again leveraging number theory, existence follows by noting that $\frac{|G|}/d$ generates a group isomorphic to $\mathbb{Z} / d \mathbb{Z}$ (which is cyclic). Then the proof is finished by showing that any other generator of order $d$ is a multiple of our $\frac{n}{d}$.


Just as modular groups can be used to prove stuff about cyclic groups, the theory of cyclic groups can conversely prove propositions in number theory.

\begin{proposition}
\[ n = \sum_{d : d|n} \varphi(d) \]
\end{proposition}


Recall our beloved Lagrange's theorem; it's pretty cool that the order of subgroups divide the order of the group. However, if a number divides the order of a group, does that mean a subgroup with an order of that number be made? This is the converse to Lagrange's theorem, and it is not true in general, however it turns out that this is true for prime numbers and these groups happen to be cyclic!

\begin{proposition}[Cauchy's theorem]
Let $G$ be a finite group. If a prime number $p$ divides the order of $G$, then there exists a cyclic subgroup of $G$ with order $p$.
\end{proposition}
