\chapter{Quotient Groups}


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. 



\section{Normal subgroups}


Recall that left cosets of a subgroup form an equivalence relation over a group, essentially \emph{dividing} the group into a bunch of sets that partition the group. Imagine we want to synthetically form group that acts just like $G$, except all elements of the same left coset are treated as the same element; can we always make such a group? The right approach is to treat the left cosets as the elements of this group themselves. Now the question becomes this; can we make a group of these equivalences classes, where composition $*$ acts as such?
\[xH * yH = (xy)H\]

Let's define a function from $G$ to $G / H$ that shows

\begin{definition}[Quotient map]
The \emph{Quotient map} $\pi$ is the following function.
\[ \pi : G / H \to G \]
\[\pi(x)=xH\]
\end{definition}


Using the definition of our quotient map, we can restate the condition we want to the following.

\[xH * yH = (xy)H \implies f(x)f(y)=f(xy)\]

But this is the condition of a homomorphism! So if our quotient map is an epimorphism (since we want our group to contain all left cosets), then such a group can be created. One can immediately see that the quotient map is surjective, so it remains to see that the homomorphism condition is satisfied.


Unfortunately the quotient map isn't always a homomorphism. To see why, imagine we have $aH=bH$ (i.e $a,b$ are in the same left coset), if the quotient map was a homomorphism then we have $(ag)H=(bg)H$, since $a,b$ should 'act' the same, but this is not true in general. 

For what subgroups is this true? For such subgroups, the quotient map would be a homomorpism and our constructed group would be valid. The answer lies in \emph{normal subgroups}.

%We have discovered that cosets formed from subgroups essentially \emph{divides} a groups elements into equally sized classes, by Lagrange's theorem. 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 $g_1 H * g_2 H = (g_1 \circ g_2)H$.

%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 $ (ag)H=(bg)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.



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









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

Assume $N$ is a normal subgroup and we have $aN=bN$, let's show that $(ag)H=(bg)H$
\[agh_1=ah_2 g = bh_3 g = bgh_4\]
\[\implies (ag)H=(bg)H\]

If we begin with the assumption that $aH=bH\implies agH=bgH$, we can prove that the group must be a normal subgroup as such.
\[gh_1= a^-1 a gh_1 = a^-1 bgh_2 = h_3 g h_2\]
\[gh_1= h_3 g h_2\]
\[g(h_1 h^{-1}_2)= h_3 g\]


As mathematicians, we would like to be familiar with sufficient conditions to form a normal subgroup so that we can involve them when we start studying quotient groups.

\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}

\begin{corollary}
Centers are normal subgroups.
\[ Z(G) \lhd G \]
\end{corollary}

\begin{proposition}
Kernels are normal subgroups.
\[ \mathrm{ker}(f) \lhd G \]
\[ f : G \to H \text{ is a homomorphism } \implies \mathrm{ker}(f) \lhd G \]
\end{proposition}




\section{Quotient groups}


We can now introduce the quotient group; a group of cosets.



\begin{definition}[Quotient group]
Let $(G,\circ)$ be a group and $N \lhd G$ be a normal subgroup, \emph{the quotient group of $G$ by $N$} is the group $(G / N, *)$ of unique left cosets of $N$ with $*$-composition defined in the following way.
\begin{itemize}
	\item $G / N$ is the set of left cosets of $N$ in $G$
	\item $  (gN) * (kN) = (gk)N $
\end{itemize}

\end{definition}


We should make sure that $*$-composition is well defined; we didn't introduce normal subgroups for nothing!

For the sake of simplicity, we'll often use the equivalence class notation $[g]=gN$ when the normal subgroup is familiar and omit the $*$-composition symbol. Here's an example below.

\[ gN * kN = [g] * [k] = [g][k] \]




When is a quotient group is Abelian? For quotient group $G / N$, it is when $g^-1 h^-1 gh \in N$.

$G / Z(G)$ is an abelian subgroup iff it is isomorphic to the trivial group.

And of course, we know the folloing from our previous discussion.
\begin{proposition}
Let $G / N$ be a quotient group, then the quotient map $\pi : G \to G / N$ is an epimorphism.
\end{proposition}


\subsection{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} $( \mathbb{T} , \circ)$, 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}

\subsection{First isomorphism theorem}


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

\begin{proposition}
Given a homomorphism $f : G \to H$, its kernel is a normal subgroup of $G$.
\[\mathrm{ker}(f) \lhd G\]
\end{proposition}


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





One way of interpreting this lemma is that kernel cosets correspond to sets of elements with the same homomorphism mappings. In other words, elements in the same left coset of the kernel map to the same $H$. To eliminate this redundancy, one can consider a map from the quotient group which bundles these identically treated elements as oe left coset; this gives us a monomorphism from the quotient group to $H$; This is the essence of the \emph{first isomorphism theorem}.

There are a few ways this theorem is traditionally posed. We will derive a general version of the theorem and prove the rest of the theorem as corollaries.

\begin{theorem}[First isomorphism theorem (group theory)]
Let $\varphi : G \to H$ be a homomorphism. Then there exists a unique epimorphism $k : G / \mathrm{ker}(\varphi) \to H$ defined as
\[ k : g \mathrm{ker}(f) \mapsto f(g)\]
Since $k$ is an epimorphism, we have $\mathrm{Im}(f) \cong G / \mathrm{ker}(f)$
%\begin{tikzcd}[column sep=small]
%	G \arrow{r}{f} \arrow{d}{\pi} & \mathrm{Im}(f) \\
%G/\mathrm{ker}(f) \arrow{ru}{k}
%\end{tikzcd}
% https://q.uiver.app/#q=WzAsMyxbMCwwLCJHIl0sWzEsMCwiSCJdLFswLDEsIkcgLyBcXG1hdGhybXtrZXJ9KGYpIl0sWzIsMSwiayIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDEsImYiXSxbMCwyLCJcXHBpIiwyXV0= 
\[\begin{tikzcd} G & H \\ {G / \mathrm{ker}(f)} \arrow["f", from=1-1, to=1-2] \arrow["\pi"', from=1-1, to=2-1] \arrow["k"', dashed, from=2-1, to=1-2] \end{tikzcd}\]

\end{theorem}




If the image of the original homomorphism $\varphi$ is the whole of $H$ (i.e $\varphi$ is surjective, an epimorphism), our epimorphism gets promoted to an isomorphism, and we have the following instantly.  

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


With Lagrange's theorem, we can now obtain an interesting way to view the order of finite groups through the perspective of homomorphims.

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