
## Section9.3Normal Subgroups and Factor Groups

### Subsection9.3.1Normal Subgroups

A subgroup $H$ of a group $G$ is normal in G if $gH = Hg$ for all $g \in G\text{.}$ That is, a normal subgroup of a group $G$ is one in which the right and left cosets are precisely the same.

###### Example9.1

Let $G$ be an abelian group. Every subgroup $H$ of $G$ is a normal subgroup. Since $gh = hg$ for all $g \in G$ and $h \in H\text{,}$ it will always be the case that $gH = Hg\text{.}$

###### Example9.2

Let $H$ be the subgroup of $S_3$ consisting of elements $(1)$ and $(12)\text{.}$ Since

\begin{equation*} (123) H = \{ (123), (13) \} \quad \text{and} \quad H (123) = \{ (123), (23) \}, \end{equation*}

$H$ cannot be a normal subgroup of $S_3\text{.}$ However, the subgroup $N\text{,}$ consisting of the permutations $(1)\text{,}$ $(123)\text{,}$ and $(132)\text{,}$ is normal since the cosets of $N$ are

\begin{gather*} N = \{ (1), (123), (132) \}\\ (12) N = N (12) = \{ (12), (13), (23) \}. \end{gather*}

The following theorem is fundamental to our understanding of normal subgroups.

(1) $\Rightarrow$ (2). Since $N$ is normal in $G\text{,}$ $gN = Ng$ for all $g \in G\text{.}$ Hence, for a given $g \in G$ and $n \in N\text{,}$ there exists an $n'$ in $N$ such that $g n = n' g\text{.}$ Therefore, $gng^{-1} = n' \in N$ or $gNg^{-1} \subset N\text{.}$

(2) $\Rightarrow$ (3). Let $g \in G\text{.}$ Since $gNg^{-1} \subset N\text{,}$ we need only show $N \subset gNg^{-1}\text{.}$ For $n \in N\text{,}$ $g^{-1}ng=g^{-1}n(g^{-1})^{-1} \in N\text{.}$ Hence, $g^{-1}ng = n'$ for some $n' \in N\text{.}$ Therefore, $n = g n' g^{-1}$ is in $g N g^{-1}\text{.}$

(3) $\Rightarrow$ (1). Suppose that $gNg^{-1} = N$ for all $g \in G\text{.}$ Then for any $n \in N$ there exists an $n' \in N$ such that $gng^{-1} = n'\text{.}$ Consequently, $gn = n' g$ or $gN \subset Ng\text{.}$ Similarly, $Ng \subset gN\text{.}$

### Subsection9.3.2Factor Groups

If $N$ is a normal subgroup of a group $G\text{,}$ then the cosets of $N$ in $G$ form a group $G/N$ under the operation $(aN) (bN) = abN\text{.}$ This group is called the factor or quotient group of $G$ and $N\text{.}$ Our first task is to prove that $G/N$ is indeed a group.

The group operation on $G/N$ is $(a N ) (b N)= a b N\text{.}$ This operation must be shown to be well-defined; that is, group multiplication must be independent of the choice of coset representative. Let $aN = bN$ and $cN = dN\text{.}$ We must show that

\begin{equation*} (aN) (cN) = acN = bd N = (b N)(d N). \end{equation*}

Then $a = b n_1$ and $c = d n_2$ for some $n_1$ and $n_2$ in $N\text{.}$ Hence,

\begin{align*} acN & = b n_1 d n_2 N\\ & = b n_1 d N\\ & = b n_1 N d\\ & = b N d\\ & = b d N. \end{align*}

The remainder of the theorem is easy: $eN = N$ is the identity and $g^{-1} N$ is the inverse of $gN\text{.}$ The order of $G/N$ is, of course, the number of cosets of $N$ in $G\text{.}$

It is very important to remember that the elements in a factor group are sets of elements in the original group.

###### Example9.5

Consider the normal subgroup of $S_3\text{,}$ $N = \{ (1), (123), (132) \}\text{.}$ The cosets of $N$ in $S_3$ are $N$ and $(12) N\text{.}$ The factor group $S_3 / N$ has the following multiplication table.

\begin{equation*} \begin{array}{c|cc} & N & (12) N \\ \hline N & N & (12) N \\ (12) N & (12) N & N \end{array} \end{equation*}

This group is isomorphic to ${\mathbb Z}_2\text{.}$ At first, multiplying cosets seems both complicated and strange; however, notice that $S_3 / N$ is a smaller group. The factor group displays a certain amount of information about $S_3\text{.}$ Actually, $N = A_3\text{,}$ the group of even permutations, and $(12) N = \{ (12), (13), (23) \}$ is the set of odd permutations. The information captured in $G/N$ is parity; that is, multiplying two even or two odd permutations results in an even permutation, whereas multiplying an odd permutation by an even permutation yields an odd permutation.

###### Example9.6

Consider the normal subgroup $3 {\mathbb Z}$ of ${\mathbb Z}\text{.}$ The cosets of $3 {\mathbb Z}$ in ${\mathbb Z}$ are

\begin{align*} 0 + 3 {\mathbb Z} & = \{ \ldots, -3, 0, 3, 6, \ldots \}\\ 1 + 3 {\mathbb Z} & = \{ \ldots, -2, 1, 4, 7, \ldots \}\\ 2 + 3 {\mathbb Z} & = \{ \ldots, -1, 2, 5, 8, \ldots \}. \end{align*}

The group ${\mathbb Z}/ 3 {\mathbb Z}$ is given by the multiplication table below.

\begin{equation*} \begin{array}{c|ccc} + & 0 + 3{\mathbb Z} & 1 + 3{\mathbb Z} & 2 + 3{\mathbb Z} \\\hline 0 + 3{\mathbb Z} & 0 + 3{\mathbb Z} & 1 + 3{\mathbb Z} & 2 + 3{\mathbb Z} \\ 1 + 3{\mathbb Z} & 1 + 3{\mathbb Z} & 2 + 3{\mathbb Z} & 0 + 3{\mathbb Z} \\ 2 + 3{\mathbb Z} & 2 + 3{\mathbb Z} & 0 + 3{\mathbb Z} & 1 + 3{\mathbb Z} \end{array} \end{equation*}

In general, the subgroup $n {\mathbb Z}$ of ${\mathbb Z}$ is normal. The cosets of ${\mathbb Z } / n {\mathbb Z}$ are

\begin{gather*} n {\mathbb Z}\\ 1 + n {\mathbb Z}\\ 2 + n {\mathbb Z}\\ \vdots\\ (n-1) + n {\mathbb Z}. \end{gather*}

The sum of the cosets $k + {\mathbb Z}$ and $l + {\mathbb Z}$ is $k+l + {\mathbb Z}\text{.}$ Notice that we have written our cosets additively, because the group operation is integer addition.

###### Example9.7

Consider the dihedral group $D_n\text{,}$ generated by the two elements $r$ and $s\text{,}$ satisfying the relations

\begin{align*} r^n & = \identity\\ s^2 & = \identity\\ srs & = r^{-1}. \end{align*}

The element $r$ actually generates the cyclic subgroup of rotations, $R_n\text{,}$ of $D_n\text{.}$ Since $srs^{-1} = srs = r^{-1} \in R_n\text{,}$ the group of rotations is a normal subgroup of $D_n\text{;}$ therefore, $D_n / R_n$ is a group. Since there are exactly two elements in this group, it must be isomorphic to ${\mathbb Z}_2\text{.}$