# Section10.1Factor Groups and Normal Subgroups¶ permalink

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

##### Example10.1

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

##### Example10.2

Let $H$ be the subgroup of $S_3$ consisting of elements $(1)$ and $(12)$. Since \begin{equation*}(123) H = \{ (123), (13) \} \quad \text{and} \quad H (123) = \{ (123), (23) \},\end{equation*} $H$ cannot be a normal subgroup of $S_3$. However, the subgroup $N$, consisting of the permutations $(1)$, $(123)$, and $(132)$, 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.

##### Proof

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

##### Proof

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

##### Example10.5

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

This group is isomorphic to ${\mathbb Z}_2$. 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 $S_3$. Actually, $N = A_3$, 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.

##### Example10.6

Consider the normal subgroup $3 {\mathbb Z}$ of ${\mathbb Z}$. 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.

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

##### Example10.7

Consider the dihedral group $D_n$, generated by the two elements $r$ and $s$, 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, $R_n$, of $D_n$. Since $srs^{-1} = srs = r^{-1} \in R_n$, the group of rotations is a normal subgroup of $D_n$; therefore, $D_n / R_n$ is a group. Since there are exactly two elements in this group, it must be isomorphic to ${\mathbb Z}_2$.