\part{Quotient Groups}


\chapter{Normal subgroups}

Now that some familiarity with the basic theory of groups is established, we can now turn towards some common groups constructions; ways in which different groups can be related to one another to form new groups. 

We have discovered that cosets formed from subgroups essentially \emph{divides} a groups elements into equally sized classes. One may ask if a group on these cosets can be formed with the operator $+$ behaving similar to the original group in the sense that $g_1 H + g_2 H = (g_1 \cdot g_2)H$.

One slightly small caveat is that cosets formed by different elements be the same coset (more specifically, if $b \in aH$, then $aH=bH$), so if we have $aH=bH$,we want $ (a \cdot g)H=(b\cdot g)H$, which is not true in general! So as a precursor to defining these 'groups divided into cosets', we must first address this concern by finding what type of subgroup permits this type of behaviour.


\section{Normal subgroups}



Normal subgroups are what is required to ensure that identical cosets are treated the same by our 'coset operation'.

\begin{definition}
A \emph{normal subgroup}
\[ N \lhd G \iff \forall g \in G [ gNg^{-1} =N ] \]
\end{definition}





Here is a useful sufficient condition for normal subgroups.
\begin{proposition}
If a subgroup is Abelian, then it is normal.
\[ N \leq G \land N \text{ is Abelian } \implies N \lhd G\]
\end{proposition}


This immediately implies the following.
\begin{proposition}
Centers are normal subgroups.
\[ Z(G) \lhd G \]
\end{proposition}

Although kernels are not necessarily Abelian, they are indeed normal.

\begin{proposition}
Kernels are normal subgroups.
\[ \mathrm{ker}(f) \lhd G \]
\end{proposition}




\chapter{Quotient groups}



\begin{definition}
A \emph{quotient group on $G$} is a group of unique left cosets on the normal subgroup $N \lhd G$ with multiplication defined in the following way.
\[ ( G / N , \otimes )\]
\[ (gN) \otimes (kN) = (gk)N\]
\end{definition}


We should make sure that $\otimes$ is a well defined operation; we didn't introduce normal subgroups for nothing!

For the sake of simplicity, we'll often use the equivalence class notation $[g]=gN$ with the original group's multiplication symbol $\cdot$ for elements and multiplication of a quotient group. Here's an example below.

\[ gN\otimes kN = [g] \cdot [k] = [g][k] \]


\section{Quotient map}

There is a simple homomorphism from a group to any of its quotient groups. Enter \emph{the quotient map}; a map from group elements to their left cosets.

\begin{definition}
The \emph{Quotient map} $\pi$ of a normal group $N$ is the following group epimorphism from the original group $G$ to cosets of the normal subgroup $gN$.
\[ \pi : g \mapsto gN \]
\end{definition}


\section{Examples of quotient groups}

\begin{example}
Consider the \emph{additive group of integers modulo n} $(\mathbb{Z} / n\mathbb{Z},+)$
Recall that $n \mathbb{Z} \leq \mathbb{Z}$ and that $\mathbb{Z}$ and its subgroups are Abelian, hence it is indeed well defined.
Since numbers with the same Euclidean remainder when divided by $n$ fall in the same coset, the cosets are $\mathbb{Z} / n\mathbb{Z} = \{n\mathbb{Z}, 1+n\mathbb{Z}, 2+n\mathbb{Z}, ..., (n-1)+n\mathbb{Z}\}$
\end{example}


\begin{example}
The \emph{circle group} $( \mathbb{T} , \cdot)$, where $\mathbb{T} = \{z \in C : |z| =1  \}$ can be represented isomorphically as a quotient group.
One can show that $\mathbb{T} \cong \mathbb{R} / 2\pi \mathbb{Z}$ by a result called the \emph{first isomorphism thorem}.
\end{example}

\section{First isomorphism theorem}


We can create quotient groups given any normal subgroup. Remember that kernels and images of homomorphisms actually form subgroups themselves; that's pretty neat. However can one make quotient groups with the kernel?


\begin{lemma}
\[ f : G \to H \text{ is a homomorphism } \implies \mathrm{ker}(f) \lhd G \]
\end{lemma}

This means we can make quotient groups using kernels. So the cosets on the kernel form their own group. Experimenting with the cosets of the kernel brings the following lemma.

\begin{lemma}
\[ f : G \to H \text{ is a homomorphism } \implies [ g_1 \mathrm{ker}(f) = g_2 \mathrm{ker}(f) \iff f(g_1)=f(g_2) ] \]
\end{lemma}

One way of interpreting this lemma is that kernel cosets correspond to sets of elements with the same homomorphism mappings. This is the essence of the \emph{first isomorphism theorem}.

\begin{theorem}[First isomorphism theorem for groups]
Let $f : G \to H$ be a homomorphism . Then $\mathrm{Im}(f) \cong G / \mathrm{ker}(f)$ the following isomorphism $k$.
\[ k : g \mathrm{ker}(f) \mapsto f(g)\]
\begin{tikzcd}[column sep=small]
	G \arrow{r}{f} \arrow{d}{\pi} & \mathrm{Im}(f) \\
G/\mathrm{ker}(f) \arrow{ru}{k}
\end{tikzcd}
\end{theorem}

\begin{corollary}
\[ f : G \to H \text{ is an epimorphism } \implies G / \mathrm{ker}(f) \cong H \]
\end{corollary}


\begin{corollary}
\[ f : G \to H \text{ is a homomorphism } \implies |\mathrm{Im}(f)||\mathrm{Ker}(f)|=|G| \]
\end{corollary}


