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 $k\in\mathbb{Z}$,

1.  $|a^k|=\frac{n}{d}$ where $d=(k,n)$.
2.  $|a^k|=n$ 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 $G$ is 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 $\{e\}\lneqq N\lneqq G$. 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 $p$.

Remark. It should be noted that Cauchy Theorem still holds even if the group $G$ is not abelian. We will not however prove the generalized version of Cauchy Theorem here.

In here, it was shown that the kernel $K=\ker\varphi$ of a homomorphism $\varphi: G\longrightarrow G’$ is a subgroup of $G$ and that it satisfies the property $aK=Ka$ for all $a\in G$. The kernel of a homomorphism is an example of a particular kind of subgroups of a group called normal subgroups.

Definition. The subgroup $N$ of $G$ is called a normal subgroup of $G$ if $a^{-1}Na\subset N$, $\forall a\in G$. If $N$ is a normal subgroup of $G$, we write $N\vartriangleleft G$.

Note that $\forall a\in G$, $a^{-1}Na\subset N$ implies that $\forall a\in G$, $a^{-1}Na=N$. Hence, we have

Theorem. $N\vartriangleleft G$ if and only if $\forall a\in G$, $aN=Na$.

If $G$ is abelian, every subgroup is a normal subgroup. But the converse need not be true. There are nonabelian groups in which every subgroup is normal. Such nonabelian groups are called Hamiltonian.

Example. Let $\mathbb{Q}_8=\{\pm 1,\pm, i, \pm j, \pm k\}$ where $1,i,j,k$ satisfy the multiplication laws
\begin{align*}
1^2&=1,\ i^2=j^2=k^2=-1,\\
ij&=-ji=k,\ jk=-kj=i,\ ki=-ik=j.
\end{align*}
Then $\mathbb{Q}_8$ forms a nonabelian group. There are four subgroups of $\mathbb{Q}_8$:
\begin{align*}
\langle -1\rangle&=\{1,-1\},\\
\langle i\rangle&=\{1,-1,i,-i\},\\
\langle j\rangle&=\{1,-1,j,-j\},\\
\langle k\rangle&=\{1,-1,k,-k\}.
\end{align*}
All these subgroups are normal subgroups of $\mathbb{Q}_8$, so $\mathbb{Q}_8$ is Hamiltonian. Note that $1,i,j,k$ form a 4-dimensional algebra of quaternions over $\mathbb{R}$ $$\mathbb{H}=\{a1+bi+cj+dk:a,b,c,d\in\mathbb{R}\}.$$ The real algebra of quaternions $\mathbb{H}$ is identified with the 4-dimensional Euclidean space $\mathbb{R}^4$.

Recall that the index $|G:H|$ of a subgroup $H$ in $G$ is the number of right cosets of $H$ in $G$ or equivalently the number of left cosets of $H$ in $G$ as discussed here.

Theorem. Let $G$ be a group and $N\leq G$. If $|G:N|=2$ then $N\vartriangleleft G$.

Proof. Let $a\in G$. If $a\in N$, $aN=Na$. Now suppose that $a\in G\setminus N$. Then
$$N\cap Na=\emptyset=N\cap aN.$$
Furthermore, since $G=N\cup Na=N\cup aN$, $Na=Na$.

Example. Recall that $|S_n|=n!$ and $|A_n|=\frac{n!}{2}$. Since $|S_n:A_n|=2$, $A_n\vartriangleleft S_n$.

Remark. The converse of the above theorem need not be true, namely the index $|G:N|$ of a normal subgroup $N$ in $G$ is not necessarily 2. The 4th dihedral group (we introduced it here)
$$D_4=\{1,\sigma,\sigma^2,\sigma^3,\tau,\sigma\tau,\sigma^2\tau,\sigma^3\tau\}$$
has the following nontrivial subgroups:
\begin{align*}
&\{1,\sigma^2,\sigma\tau,\sigma^2\tau\},\ \{1,\sigma,\sigma^2,\sigma^3\},\ \{1,\sigma^2,\tau,\sigma^2\tau\},\\
&\{1,\sigma^3\tau\},\ \{1,\sigma\tau\},\ \{1,\sigma^2\},\ \{1,\tau\},\ \{1,\sigma^2\tau\}.
\end{align*}
The subgroups of order 4 are normal subgroups of $D_4$ because their indices are 2. The center of $D_4$ is $\{1,\sigma^2\}$, so it is also a normal subgroup of $D_4$. However, $|D_4:Z(D_4)|=4$. The rest of nontrivial subgroups are not normal subgroups of $D_4$.

Remark. $H\vartriangleleft N, N\vartriangleleft G\not\Longrightarrow H\vartriangleleft G$. In the previous remark, $H=\{1,\tau\}$ is a normal subgroup of $\{1,\sigma^2,\tau,\sigma^2\tau\}$ but $H\not\vartriangleleft D_4$.

Example. The center $Z(G)$ of a group $G$
$$Z(G)=\{x\in G: \forall g\in G, xg=gx\}$$ is a normal subgroup of $G$.

Example. If $N\leq Z(G)$ then $N\vartriangleleft G$.

Example. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\vartriangleleft G$ as seen here.

Example. Let $\mathbb{R}^\ast=\mathbb{R}\setminus\{0\}$. Define $f: \mathrm{GL}(2,\mathbb{R})\longrightarrow\mathbb{R}^\ast$ by
$$f(A)=\det A,\ \mbox{for any}\ A\in\mathrm{GL}(2,\mathbb{R}).$$ Then $f$ is a homomorphism because for any $A,B\in\mathrm{GL}(2,\mathbb{R})$, $f(AB)=\det(AB)=(\det A)(\det B)=f(A)f(B)$. $f$ is also onto: for any $r\in\mathbb{R}^\ast$ let $A=\begin{pmatrix}
r & 0\\
0 & 1 \end{pmatrix}$. Then $A\in\mathrm{GL}(2,\mathbb{R})$ and $f(A)=r$. For any $A\in\mathrm{GL}(2,\mathbb{R})$, $A\in\ker f$ if and only if $\det A=1$ and so $\ker f=\mathrm{SL}(2,\mathbb{R})$. Hence, $\mathrm{SL}(2,\mathbb{R})\vartriangleleft\mathrm{GL}(2,\mathbb{R})$.

In this lecture, we study maps from a group into another group that carry over the same group structure i.e preserve group operations. Such maps are called homomorphisms. It turns out that it suffices to require a homomorphism to preserve a binary operation only. Then the unary operation (inverses) and the nullary operation (the identity element) are automatically preserved.

Definition. Let $(G,\ast)$ and $(G’,\sharp)$ be two groups. A map $\varphi: G\longrightarrow G’$ is called a homomorphism if $\varphi(a\ast b)=\varphi(a)\sharp\varphi(b)$.

Example. Define $\varphi: (\mathbb{R}^+,\cdot)\longrightarrow(\mathbb{R},+)$ by
$$\varphi(x)=\log_{10}x,\ \forall x\in\mathbb{R}^+.$$
Then $\varphi$ is a homomorphism. In addition, $\varphi$ is one-to-one and onto. A homomorphism which is one-to-one and onto is called an isomorphism. If there is an isomorphism from a group onto another group, they are the same group meaning we do not distinguish them.

Example. The map $\varphi: \mathbb{Z}\longrightarrow\mathbb{Z}_n$ defined by
$$\varphi(x)=[x],\ \forall\ x\in\mathbb{Z}$$
is a homomorphism called the canonical or the natural homomorphism. The canonical homomorphism is onto.

Example. Let $G$ be a group and $A(G)$ the set of all bijective (i.e. one-to-one and onto) maps of $G$ onto itself. Recall that $A(G)$ forms a group with composition. Let $a\in G$. Define $T_a: G\longrightarrow G$ by
$$T_ax=ax\ \forall\ x\in G.$$ Hence $\forall a\in G$, $T_a\in A(G)$. Furthermore, $\forall a,b\in G$, $T_aT_b=T_{ab}$: $\forall x\in G$,
$$T_aT_b(x)=T_a(T_bx)=T_a(bx)=a(bx)=(ab)x=T_{ab}x.$$
Define $\varphi: G\longrightarrow A(G)$ by
$$\varphi(a)=T_a,\ \forall\ a\in G.$$ Then $\varphi$ is a homomorphism. In addition, it is one-to-one. If $|G|=n$, then $|A(G)|=n!$, so $\varphi$ cannot be onto.

Definition. A homomorphism which is one-to-one (injective) is called a monomorphism. A homomorphism which is onto (surjective) is called an epimorphism. A homomorphism which is one-to-one and onto (bijective) is called an isomorphism as mentioned above. $G$ is said to be isomorphic to $G’$ if there exists an isomorphism from $G$ onto $G’$. If $G$ is isomorphic to $G’$, we write $G\cong G’$. An isomorphism from a group $G$ onto itself is called an automorphism.

Theorem. [Cayley’s Theorem] Every group $G$ is isomorphic to some subgroup of $A(S)$, for an appropriate set $S$.

We in fact proved this theorem in the previous example by taking $S$ to be the group $G$ itself. But there may be other choices for $S$. If $G$ is finite, $S$ may be taken to be a finite set in which case $A(S)$ is $S_n$, the group of permutations of $\{1,2,\cdots,n\}$. In this case, Cayley’s Theorem is stated as: A finite group can be represented as a group of permutations.

Example. Let $G$ be a group and $a\in G$ be fixed. The map $\varphi: G\longrightarrow G$ defined by
$$\varphi(x)=a^{-1}xa,\ \forall x\in G$$
is an automorphism called the inner automorphism of $G$ induced by $a$.

Example. Define $\varphi: (\mathbb{R},+)\longrightarrow (\mathbb{C}\setminus\{0\},\cdot)$ by
$$\varphi(x)=e^{ix},\ \forall x\in\mathbb{R}.$$
While $\varphi$ is a homomorphism, it is neither one-to-one nor onto.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then

(a) $\varphi(e)=e’$.

(b) $\varphi(a^{-1})=\varphi(a)^{-1}$.

Proof. (a) Let $a\in G$. Then $ae=a$. Since $\varphi$ is a homomorphism, $$\varphi(ae)=\varphi(a)\varphi(e)=\varphi(a)=\varphi(a)e’.$$ So by the cancellation law, we obatin $\varphi(e)=e’$.

(b) Let $a\in G$. Then
\begin{align*}
\varphi(a)\varphi(a^{-1})&=\varphi(aa^{-1})\ (\mbox{because $\varphi$ is a homomorphism})\\
&=\varphi(e)\\
&=e’\ (\mbox{by part (a)})
\end{align*}
Hence, $\varphi(a)^{-1}=\varphi(a^{-1})$.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\varphi(G)\leq G’$.

In here, the kernel of a function was introduced. Let $\varphi: G\longrightarrow G’$ be a homomorphism and $K=\varphi^{-1}(e’)=\{a\in G: \varphi(a)=e’\}$. Then $\ker\varphi$ and $K=\varphi^{-1}(e’)$, the preimage of $e’$ under $\varphi$ are closely related as:
\begin{align*}
(a,b)\in\ker\varphi&\Longleftrightarrow\varphi(a)=\varphi(b)\\
&\Longleftrightarrow\varphi(ab^{-1})=e’\\
&\Longleftrightarrow ab^{-1}\in K.
\end{align*}
So, in group theory we define $K=\varphi^{-1}(e’)$ to be $\ker\varphi$, the kernel of $\varphi$.

Theorem. $\varphi: G\longrightarrow G’$ is one-to-one if and only if $\ker\varphi=\{e\}$.

Proof. ($\Rightarrow$) Suppose that $\varphi$ is one-to-one and let $a\in\ker\varphi$. Then $\varphi(a)=\varphi(e)=e’$. Since $\varphi$ is one-to-one, $a=e$.

($\Leftarrow$) Suppose that $\ker\varphi=\{e\}$ and let $\varphi(a)=\varphi(b)$. Then
\begin{align*}
\varphi(ab^{-1})=e’&\Longrightarrow ab^{-1}\in\ker\varphi\\
&\Longrightarrow ab^{-1}=e\\
&\Longrightarrow a=b.
\end{align*}
So, $\varphi$ is one-to-one.

Theorem. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\leq G$.

Theorem. Let $K=\ker\varphi$. Then $\forall a\in G$, $aK=Ka$.

Proof. It suffices to show (why?) that $\forall a\in G$, $a^{-1}Ka\subset K$. Let $a\in G$. Then $\forall k\in K$,
$$\varphi(a^{-1}ka)=\varphi(a)^{-1}\varphi(k)\varphi(a)=e’\Longrightarrow a^{-1}ka\in K.$$