Let $G$ be a group and $N\leq G$. Recall the equivalence relation $\sim$ on $G$ defined by

$$a\sim b\ \mbox{if}\ ab^{-1}\in N$$ for any $a,b\in G$. Each equivalence class $[a]$ is identified with right coset $Na$. The quotient set $G/\sim$ is the set of all equivalence classes or equivalently all right cosets of the subgroup $N$ in $G$. A question we can ask is can we give a group structure to the quotient set $G/\sim$? The answer is affirmative if $N$ is a normal subgroup of $G$. This is what we are going to discuss in this lecture.

Let $N\vartriangleleft G$. Denote by $G/N$ the set of all right cosets of $N$ in $G$ i.e.

$$G/N=\{Na:a\in G\}.$$

For any $Na, Nb\in G/N$,

\begin{align*}

(Na)(Nb)&=N(aN)b\\

&=N(Na)b\\

&=N(Nab)\\

&=Nab.

\end{align*}

So, it appears that if $N\vartriangleleft G$, we may define an operation $\cdot$ on $G/N$ naturally from the binary operation on $G$ by the equation \begin{equation}

\label{eq:cosetoper}

Na\cdot Nb=Nab

\end{equation} for any $Na,Nb\in G/N$. But is this operation well-defined? To see that let $Na=Nc$ and $Nb=Nd$. Then $Nac^{-1}=N$ and $Nbd^{-1}=N$. So $ac^{-1}\in N$ and $bd^{-1}\in N$. Now,

\begin{align*}

N(ac)(bd)^{-1}&=N(ac)(d^{-1}b^{-1})\\

&=Na(cd^{-1})b^{-1}\\

&=a(Ncd^{-1})b^{-1}\\

&=aNb^{-1}\\

&=Nab^{-1}\\

&=N.

\end{align*}

That is, $Na\cdot Nb=Nab=Ncd=Nc\cdot Nd$. Hence, right coset operation given by the equation \eqref{eq:cosetoper} is well-defined. Conversely, if $N\leq G$ and right coset operation given by the equation \eqref{eq:cosetoper} is well-defined, then $N$ must be a normal subgroup of $G$. The verification of this is left to readers as an exercise. Furthermore, $Ne=N$ is an identity element in $G/N$ and for each $Na\in G/N$, $Na^{-1}\in G$ and $NaNa^{-1}=N$, hence $(Na)^{-1}=Na^{-1}$. Therefore, $(G/N,\cdot)$ is a group. This group is called a *factor group* or a *quotient group* of $G$ modulo $N$.

*Theorem*. If $N\vartriangleleft G$ then there exists an onto homomorphism (an epimorphism) $\psi:G\longrightarrow G/N$ such that $\ker\psi=N$, Such a homomorphism is called a *natural homomorphism* or a *canonical homomorphism*.

*Proof*. Define $\psi: G\longrightarrow G/N$ by

$$\psi(a)=Na$$

for any $a\in G$. If $a=b$ then $\psi(a)=Na=Nb=\psi(b)$ so $\psi$ is well-defined. Let $Na\in G/N$. Then $a\in G$ and $\psi(a)=Na$, so $\psi$ is onto. $\psi(ab)=Nab=NaNb=\psi(a)\psi(b)$, hence $\psi$ is a homomorphism. To show that $\ker\psi=N$,

\begin{align*}

a\in\ker\psi&\Longleftrightarrow \psi(a)=N\\

&\Longleftrightarrow Na=N\\

&\Longleftrightarrow a\in N.

\end{align*}

$G/N$ is the set of all right coset of $N$ in $G$, so $|G/N|=|G:N|=\frac{|G|}{|N|}$.

*Example*. Since $\mathbb{Z}$ is an abelian group, for any $n\in\mathbb{N}$, $n\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$. So, for any $n\in\mathbb{N}$, $\mathbb{Z}/n\mathbb{Z}$ is a factor group. Each $\mathbb{Z}/n\mathbb{Z}$ is indeed isomorphic to $\mathbb{Z}_n$.

The notion of factor groups can be used to prove an important theorem in group theory called *Cauchy Theorem*. First we study the following theorem as we will also need it (actually its corollary) to prove Cauchy Theorem.

*Theorem*. Let $G$ be a finite group and $a\in G$ with $|a|=n$. Then for any $k\in\mathbb{Z}$,

- $|a^k|=\frac{n}{d}$ where $d=(k,n)$.
- $|a^k|=n$ if and only if $(k,n)=1$.

*Proof*. Since 2 follows from 1, we prove 1 only. Since $d=(k,n)$, $n=n_1d$ and $k=k_1d$ for some $n_1,k_1\in\mathbb{Z}$ such that $(n_1,k_1)=1$.

$$(a^k)^{n_1}=a^{k_1n}=(a^n)^{k_1}=e.$$ So, if we let $|a^k|=s$, then $s|n_1$. On the other hand, $e=|a^k|^s=a^{ks}$, so $n|ks$ which implies that $n_1|k_1s$. Since $(n-1,k_1)=1$, $n_1|s$. Therefore, $s=n_1=\frac{n}{d}$.

*Corollary*. If a finite group $G$ has no nontrivial subgroup, then $G$ must be a cyclic group of a prime order.

*Proof*. Let $a\ne e\in G$. Then $\langle a\rangle\leq G$. Since $G$ has no nontrivial subgroup and $a\ne e$, $\langle a\rangle=G$ i.e. $G$ is a cyclic group. Let $|a|=n$. If $k\ne n$ and $k|n$. Then by the above theorem $|a^k|=\frac{n}{k}$ since $(n,k)=k$. Since $a^k\ne e$ and $G$ has no nontrivial subgroup, $\langle a^k\rangle=G$. This implies that $|a^k|=n$ and so $k=1$. Therefore, $n$ is a prime.

*Theorem*. [Cauchy] If $G$ a finite abelian group and and $p||G|$ where $p$ is a prime, then $G$ has an element of order $p$.

*Proof*. We prove the theorem by induction on $|G|$. If $|G|=1$, then there is no prime that divides $|G|$, so the theorem is vacuously true. Suppose that the statement is true for all abelian groups whose order is less than $|G|$. Let $\{e\}\not\leq N\not\leq G$. If $p||N|$, then by induction hypothesis there exists $a\in N$ such that $|a|=p$. Since $N\subset G$, we are done in this case. Now suppose that $p\not|N|$. Since $G$ is abelian, $N\vartriangleleft G$. Since $p|G$ and $p\not| N$, $p||G/N|<|G|$. So by induction hypothesis, $G/N$ has an element $Na$ of order $p$.

$$(Na)^p=Na^p=N\Longrightarrow a^p\in N$$

but $a\not\in N$ since $Na\ne N$. Let $|N|=m$. Then $e=(a^p)^m=(a^m)^p$. Let $b=a^m$. If $b\ne e$, then $b$ is an element of order $p$. What if $b=a^m=e$? If so, $(Na)^m=N$. Since $|Na|=p$, $p|m=|N|$. but by assumption $p\not||N|$. A contradiction! So, we are done if $G$ has a nontrivial subgroup. What if $G$ does not have a nontrivial subgroup? If so, by corollary above $G$ must be a cyclic group of a prime order. Since $p||G|$, $|G|=p$. In this case, any $a\ne e\in G$ is an element of order $p$.