\documentclass[12pt]{article}
\usepackage{amsmath, amssymb, amsthm}
\usepackage[letterpaper,margin=1in]{geometry}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{theorem}{Theorem}

\begin{document}

\title{Normal Subgroups and Homomorphisms}\author{}\date{}\maketitle

We make frequent use of the multiplication of subgroups of a group $G$. If $S$
and $T$ are subgroups, let $S T =\{s t|s \in S, t \in T\}$. The multiplication
of subsets is associative:

\begin{lemma}
  \label{multassoc}If $S, T$ and $U$ are subsets of $G$ then
  \[ (S T) U = S (T U) . \]
\end{lemma}

\begin{proof}
  This is clear since both sets consist of all products $s t u$ with $s \in
  S$, $t \in T$ and $u \in U$.
\end{proof}

As a special case of the multiplication of subsets, if $T =\{t\}$ consists of
a single element we will write $S t$ instead of $S\{t\}$.

A subgroup $K$ of a group $G$ is {\textit{normal}} if $x K x^{- 1} = K$ for all
$x \in G$. Let $G$ and $H$ be groups and let $\phi : G \longrightarrow H$ be a
homomorphism. Then the {\textit{kernel}} $\ker (\phi)$ of $\phi$ is the subgroup
of $G$ consisting of all elements $g$ such that $\phi (g) = 1$.

Not every subgroup is normal. For example if $G = S_3$, then the subgroup
$\left\langle (12) \right\rangle$ generated by the 2-cycle $(12)$ is not
normal. On the other hand, the subgroup $K = \left\langle (123) \right\rangle$
generated by the 3-cycle $(123)$ is normal, since $S_3$ has only one subgroup
of order three, so $g K g^{- 1} = K$ for any $g$.

\begin{proposition}
  The kernel of a homomorphism is a normal subgroup.
\end{proposition}

\begin{proof}
  Let $\phi : G \longrightarrow H$ be a homomorphism and let $K = \ker
  (\phi)$. To show that $K$ is normal, we must show that if $k \in \text{} K$
  and $x \in G$ then $g k g^{- 1} \in K$. Indeed, we have
  \[ \phi (g k g^{- 1}) = \phi (g) \phi (k) \phi (g^{- 1}) = \phi (g) \cdot 1
     \cdot \phi (g)^{- 1} = 1 \]
  because $k \in \ker (\phi)$ so $\phi (k) = 1$. Therefore $g k g^{- 1} \in K$
  and so $K$ is normal.
\end{proof}

Since the kernel of a homomorphism is normal, we may ask the converse question
of whether given a normal subgroup $N$ of $G$ it is always possible to find a
homomorphism $\phi : G \longrightarrow H$ for some group $H$ that has $N$ as
its kernel. The answer is affirmative, as we shall see.

If $N$ is any subgroup of $G$ (normal or not) then for $x \in G$ the set $N x$
is called a {\textit{right coset}}. Similarly $x N$ is called a {\textit{left
coset}}.

\begin{lemma}
  \label{equalordisjoint}
  Let $N$ be any subgroup of $G$. Then two right cosets of a subgroup $N$ are
  either equal or disjoint.
\end{lemma}

\begin{proof}
  Suppose that the cosets $N x$ and $N y$ are not disjoint. Then there exists
  some element $z \in N x \cap N y$. We may write $z = n x$ for some $n \in
  N$. Then $N z = N n x = N x$ where we have used the fact that $N$ is a
  group, so $N = N n$. Similarly $N z = N y$ and so $N x = N y$.
\end{proof}

For Lemma~\ref{equalordisjoint} we did not assume that $N$ a normal subgroup,
but we will assume it next.

\begin{lemma}
  \label{multcoset}Let $N$ be a normal subgroup of $G$.
  \begin{itemize}
    \item[(i)] Every right coset $N x$ equals the left coset $x N$.
    \item[(ii)] If $x, y \in G$ then
      \begin{equation}
       \label{keyformula} N x \cdot N y = N x y,
      \end{equation}
     so the product of two cosets is a coset.
  \end{itemize}
\end{lemma}

\begin{proof}
  To prove (i), since $N$ is normal we have $N = x N x^{- 1}$. Multiplying
  this on the right by $x$ gives $N x = x N$.
  
  To prove (ii), we can obtain equation (\ref{keyformula}) as follows:
  \[ N x N y = N N x y = N x y. \]
  Here the first step uses $xN=Nx$ from~(i).
\end{proof}

\begin{theorem}
  Let $N$ be a normal subgroup of $G$. Then the set $G / N$ of right cosets of
  $N$ is a group whose identity element is $N = N 1$. The map $\phi : G
  \longrightarrow G / N$ defined by $\phi (x) = N x$ is a homomorphism with
  kernel $N$.
\end{theorem}

\begin{proof}
  By Lemma~\ref{multcoset} the product of two cosets is a coset. Let us check
  the group axioms. The multiplication is associative by
  Lemma~\ref{multassoc}. To check that $G/N$ has an identity element,
  note that $N = N \cdot 1$ is itself a coset, and by
  (\ref{keyformula}) we have $N \cdot N x = N x \cdot N = N x$. Finally, taking $x$ and
  $y$ to be inverses in (\ref{keyformula}) shows that $N x^{- 1}$ is a
  multiplicative inverse to $N x$ and so $G / N$ is a group.
  
  Now $\phi : G \longrightarrow G / N$ defined by $\phi (x) = N x$ is a
  homomorphism by (\ref{keyformula}). We have only to check that its kernel is
  $N$. Indeed $x$ is in the kernel if and only if $\phi (x) = N$, and $N x =
  N$ is equivalent to $x \in N$, as required.
\end{proof}

\end{document}
