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\(\def\Z{\mathbb{Z}} \def\zn{\mathbb{Z}_n} \def\znc{\mathbb{Z}_n^\times} \def\R{\mathbb{R}} \def\Q{\mathbb{Q}} \def\C{\mathbb{C}} \def\N{\mathbb{N}} \def\M{\mathbb{M}} \def\G{\mathcal{G}} \def\0{\mathbf 0} \def\Gdot{\langle G, \cdot\,\rangle} \def\phibar{\overline{\phi}} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\Ker}{Ker} \def\siml{\sim_L} \def\simr{\sim_R} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

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 \(G\text{.}\)


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.


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:


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.


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.


\(S_3\) 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.


\(\Z\) 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.


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.


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.


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: