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 welldefined 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 \(G\text{.}\)
Theorem8.3.1
Let \(G\) be a group with identity element \(e\text{,}\) and let \(N\unlhd G\text{.}\) Then \(G/N\) is a group under left coset multiplication (that is, under the operation \((aN)(bN)=abN\) for all \(a,b\in G\)), and \(G/N=(G:N)\) (in particular, if \(G\lt \infty\text{,}\) then \(G/N=G/N\)).
Proof
We already know that since \(N\unlhd G\text{,}\) left coset multiplication on \(G/N\) is welldefined; further, by definition, that \((aN)(bN)=abN\in G/N\) for every \(aN,bN\in G/N\text{.}\)
Associativity of this operation on \(G/N\) follows from the associativity of \(G\)'s operation: indeed, if \(aN\text{,}\) \(bN\text{,}\) \(cN \in
G/N\text{,}\) then
\begin{align*}
((aN)(bN))(cN)\amp =(abN)(cN)\\
\amp =(ab)cN\\
\amp =a(bc)N\\
\amp =(aN)(bcN)\\
\amp =(aN)((bN)(cN)).
\end{align*}
Next, \(N=eN\in G/N\) acts as an identity element since for all \(a\in
G\text{,}\)
\((aN)(N)=aeN=aN\) and \(N(aN)=eaN=aN\text{.}\)
Finally, let \(aN\in G/N\text{.}\) Then \(a^{1}N\in G/N\) with \((aN)(a^{1}N)=aa^{1}N=N\) and \((a^{1}N)(aN)=a^{1}aN=N\text{.}\)
So \(G/N\) is a group under left coset multiplication. Since \((G:N)\) is, by definition, the cardinality of \(G/N\text{,}\) we're done.
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:
Theorem8.3.4
Let \(n,d \in \Z^+\) with \(d\) dividing \(n\text{.}\) Then we have:
\(n\Z\unlhd d\Z\) and \(\langle d\rangle \unlhd \Z_n\text{;}\)
\(d\Z/n\Z\simeq \Z_{n/d}\) (special case: \(\Z/n\Z \simeq \Z_n\)); and
\(Z_n/\langle d\rangle \simeq \Z_d\text{.}\)
Proof
Since \(d\) is a positive divisor of \(n\text{,}\) \(n\Z\) and \(\langle d\rangle\) are clearly subgroups of, respectively, \(d\Z\) and \(\Z_n\text{.}\) Moreover, since \(d\Z\) and \(\Z_n\) are abelian, all of their subgroups are normal.
This follows from the facts that \(d\Z/n\Z=\langle d+n\Z\rangle\text{,}\) hence is cyclic, and that \(d\Z/n\Z=(d\Z:n\Z)=n/d\) (see Example 7.3.5). (Unpacking the statement that \(d\Z/n\Z=\langle d+n\Z\rangle\text{:}\) Notice that \(d\Z=\langle d\rangle\text{,}\) so every element of \(d\Z\) is of the form \(kd\) for some integer \(k\text{.}\) Thus, every element of \(d\Z/n\Z\) is of the form \(kd+n\Z\) for some integer \(k\text{.}\) But for each \(k\in \Z\text{,}\) using the definition of left coset multiplication we have that \(kd+n\Z=k(d+n\Z)\text{.}\) So \(d\Z/n\Z=\langle d+n\Z\rangle\text{.}\))
Similarly, since \(\Z_n=\langle 1\rangle\text{,}\) \(\Z_n/\langle d\rangle =\langle 1+\langle d\rangle \rangle\text{,}\) so is cyclic. Since \((\Z_n:\langle d\rangle )=d\) (again, see Example 7.3.5), \(\Z_n/\langle d\rangle\) is isomorphic to \(\Z_d\text{,}\) as desired. \qedhere
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.
Theorem8.3.6
Let \(G\) be a group and \(N\unlhd G\text{.}\) Then:
If \(G\) is finite, then \(G/N\) is finite.
If \(G\) is abelian, then \(G/N\) is abelian; and
If \(G\) is cyclic, then \(G/N\) is cyclic.
Proof
Clearly this holds, since in this case \(G/N=G/N\text{.}\)
The proof of this statement is left as an exercise for the reader.

Let \(G\) be cyclic. Then there exists \(a\in G\) with \(G=\langle a\rangle\text{.}\) We claim \(G/N=\langle aN\rangle\text{.}\) Indeed:
\begin{equation*}
\langle aN\rangle =\{(aN)^i\,:\,i\in \Z\}=\{a^iN\,:\,i\in \Z\}.
\end{equation*}
But every element of \(G\) is an integer power of \(a\text{,}\) so this set equals \(\{xN\,:x\in G\}\text{,}\) that is, it equals \(G/N\text{.}\)
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.
Theorem8.3.10
{(Correspondence Theorem)} Let \(G\) be a group, and let \(K\unlhd G\text{.}\) Then the subgroups of \(G/K\) are exactly those of the form \(H/K\text{,}\) where \(H\leq G\) with \(K\subseteq H\text{.}\) Moreover, the normal subgroups of \(G/K\) are exactly those of the form \(N/K\text{,}\) where \(N\unlhd G\) with \(K\subseteq N\text{.}\)
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 onetoone. (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.
Theorem8.3.13
Let \(G\) be a group and let \(N\unlhd G\text{.}\) Then the map \(\Psi: G\to G/N\) defined by \(\Psi(g)=gN\) is an epimorphism.
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:
Theorem8.3.15
Let \(G\) be a group and \(N\) a subset of \(G\text{.}\) Then \(N\) is a normal subgroup of \(G\) if and only if \(N\) is the kernel of a homomorphism from \(G\) to some group \(G'\text{.}\)