## Section10.1Factor Groups and Normal Subgroups

### SubsectionNormal 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.

#### 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 $$h \in H\text{,}$$ it will always be the case that $$gH = Hg\text{.}$$

#### Example10.2.

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

\begin{equation*} (123) H = \{ (1 \, 2 \, 3), (1 \, 3) \} \quad \text{and} \quad H (1 \, 2 \, 3) = \{ (1 \, 2 \, 3), (2 \, 3) \}\text{,} \end{equation*}

$$H$$ cannot be a normal subgroup of $$S_3\text{.}$$ However, the subgroup $$N\text{,}$$ consisting of the permutations $$(1)\text{,}$$ $$(1 \, 2 \, 3)\text{,}$$ and $$(1 \, 3 \, 2)\text{,}$$ is normal since the cosets of $$N$$ are

\begin{gather*} N = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}\\ (1 \, 2) N = N (1 \, 2) = \{ (1 \, 2), (1 \, 3), (2 \, 3) \}\text{.} \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, $$Ng \subset gN\text{.}$$

### SubsectionFactor 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)\text{.} \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\text{.} \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.

#### Example10.5.

Consider the normal subgroup of $$S_3\text{,}$$ $$N = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}\text{.}$$ 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 & (1 \, 2) N \\ \hline N & N & (1 \, 2) N \\ (1 \, 2) N & (1 \, 2) 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 $$(1 \, 2) N = \{ (1 \, 2), (1 \, 3), (2 \, 3) \}$$ 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 $${\mathbb Z}\text{.}$$ 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 \}\text{.} \end{align*}

The group $${\mathbb Z}/ 3 {\mathbb Z}$$ is given by the Cayley 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}\text{.} \end{gather*}

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

#### Example10.7.

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}\text{.} \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 $${\mathbb Z}_2\text{.}$$