# Section8.3Introduction to factor groups

We now return to the notion of equipping $G/H\text{,}$ when $H\unlhd G\text{,}$ with a group structure. We have already saw that left coset multiplication on $G/H$ is well-defined when $H\unlhd G$ (Theorem 8.1.2); it turns out that given this, it is very easy to prove that $G/H$ under this operation is a group.

Before we prove this, we introduce a change in our notation: We initiate the convention of frequently using $N\text{,}$ rather than $H\text{,}$ to denote a subgroup of a group $G$ when we know that that subgroup is normal in $G\text{.}$

##### Definition8.3.2

When $G$ is a group and $N\unlhd G\text{,}$ the above group ($G/N$ under left coset multiplication) is called a factor group or quotient group.

##### Example8.3.3

Let $G=\Z$ and $N=3\Z\text{.}$ Then $N$ is normal in $G\text{,}$ since $G$ is abelian, so the set $G/N=\{N,1+N,2+N\}$ is a group under left coset multiplication. Noting that $N=0+N\text{,}$ it is straightforward to see that $G/N$ (that is, $\Z/3\Z$) has the following group table:

 $+$ $0+N$ $1+N$ $2+N$ $0+N$ $0+N$ $1+N$ $2+N$ $1+N$ $1+N$ $2+N$ $0+N$ $2+N$ $2+N$ $0+N$ $1+N$

Clearly, if we ignore all the $+N's$ after the $0$'s, $1$'s, and $2'$ in the above table, and consider $+$ to be addition mod 3, rather than left coset addition in $\Z/3\Z\text{,}$ we obtain the group table for $\Z_3\text{:}$

 $+$ $0$ $1$ $2$ $0$ $0$ $1$ $2$ $1$ $1$ $2$ $0$ $2$ $2$ $0$ $1$

Thus, we see that $\Z/3\Z$ is isomorphic to $\Z_3\text{.}$

This is not a coincidence! In fact, we have the following:

##### Example8.3.5

Let $N=\langle (123)\rangle \leq S_3\text{.}$ Since $(S_3:N)=2\text{,}$ $N$ must be normal in $S_3\text{,}$ so $S_3/N$ is a group under left coset multiplication. Since $|S_3/N|=2\text{,}$ $S_3/N$ must be isomomorphic to $\Z_2\text{.}$

In the above examples, we were able to identify $G/N$ up to isomorphism relatively easily because we could determine that $G/N$ was cyclic. In general, however, it can be quite difficult to identify the group structure of a factor group. We explore some powerful tools we can use in this identification in the next section, but first we note a couple properties of $G$ that are “inherited” by any of its factor groups.

##### Warning8.3.7

However, $G$ can be nonabelian [noncyclic, nonfinite] and yet have a normal subgroup $N$ such that $G/N$ is abelian [cyclic, finite]. (See the examples below.) Intuitively, the idea is that modding out a group by a normal subgroup can introduce abelianness or cyclicity, or finiteness, but not remove one of those characteristics.

##### Example8.3.8

$S_3$ is nonabelian (and therefore of course noncyclic), but we saw above that $N=\langle (123)\rangle$ is a normal subgroup of $S_3$ with $S_3/N \simeq \Z_2\text{,}$ a cyclic (and therefore of course abelian) group.

##### Example8.3.9

$\Z$ is an infinite group, but it has finite factor group $\Z/2\Z\text{.}$

What do the (normal) subgroups of a factor group $G/N$ look like? Well, they come from the (normal) subgroups of $G\text{!}$ We have the following theorem, whose proof is tedious but straightforward.

##### Example8.3.11

Let $N=18\Z$ in $\Z\text{.}$ Since the subgroups of $\Z$ containing $N$ are the sets $d\Z$ where $d$ is a positive divisor of $18\text{,}$ the subgroups of $\Z/N$ are the sets $d\Z/N\text{,}$ where, again, $d$ is a positive divisor of $18\text{.}$

We end this chapter by noting that given any group $G$ and factor group $G/N$ of $G\text{,}$ there is a homomorphism from $G$ to $G/N$ that is onto. Before we define this homomorphism, we provide some more terminology.

##### Definition8.3.12

Let $\phi: G\to G'$ be a homomorphism of groups. Then $\phi$ is can be called an epimorphism if $\phi$ is onto, and a monomorphism if $\phi$ is one-to-one. (Of course, we already know that an epimorphism that is also a monomorphism is called an “isomorphism.”)

We now present the following theorem, whose straightforward proof is left to the reader.

##### Definition8.3.14

We call this map $\Psi$ the canonical epimorphism from $G$ to $G/N$.

Notice that given $N\unlhd G\text{,}$ the kernel of the canonical epimorphism from $G$ to $G/N$ is $N\text{.}$ Thus, putting this fact together with the fact that every kernel of a homomorphism is a normal subgroup of the homomorphism's domain, we have the following: