8.3: Introduction to Factor Groups

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 \(G\text{.}\)

Theorem \(\PageIndex{1}\)

Let \(G\) be a group with identity element \(e\text{,}\) and let \(N\: G\text{.}\) 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 \(N⊴G\), left coset multiplication on \(G/N\) is well-defined; further, by definition, that \((aN)(bN)=abN∈G/N\) for every \(aN,bN∈G/N\).

Associativity of this operation on \(G/N\) follows from the associativity of \(G\)'s operation: indeed, if \(aN, bN, cN∈G/N\), then

\(\begin{array}& ((aN)(bN))(cN)&=(abN)(cN)\\&=(ab)cN\\&=a(bc)N\\&=(aN)(bcN)\\&=(aN)((bN)(cN)). \end{array}\)

Next, \(N=eN∈G/N\) acts as an identity element since for all \(a∈G\),

\((aN)(N)=aeN=aN \text{ and } N(aN)=eaN=aN\).

Finally, let \(aN∈G/N\). Then \(a^{−1}N∈G/N\) with \((aN)(a^{−1}N)=aa^{−1}N=N\) and \((a^{−1}N)(aN)=a^{−1}aN=N\).

So \(G/N\) is a group under left coset multiplication. Since \((G:N)\) is, by definition, the cardinality of \(G/N\), we're done.

Definition: Factor Group

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.

Example \(\PageIndex{1}\)

Let \(G=\mathbb{Z}\) and \(N=3\mathbb{Z}\text{.}\) 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, \(\mathbb{Z}/3\mathbb{Z}\)) 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 \(\mathbb{Z}/3\mathbb{Z}\text{,}\) we obtain the group table for \(\mathbb{Z}_3\text{:}\)

\(+\)	\(0\)	\(1\)	\(2\)
\(0\)	\(0\)	\(1\)	\(2\)
\(1\)	\(1\)	\(2\)	\(0\)
\(2\)	\(2\)	\(0\)	\(1\)

Thus, we see that \(\mathbb{Z}/3\mathbb{Z}\) is isomorphic to \(\mathbb{Z}_3\text{.}\)

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

Theorem \(\PageIndex{2}\)

Let \(n,d \in \mathbb{Z}^+\) with \(d\) dividing \(n\text{.}\) Then we have:

\(n\mathbb{Z}\unlhd d\mathbb{Z}\) and \(\langle d\rangle \unlhd \mathbb{Z}_n\text{;}\)
\(d\mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}_{n/d}\) (special case: \(\mathbb{Z}/n\mathbb{Z} \simeq \mathbb{Z}_n\)); and
\(Z_n/\langle d\rangle \simeq \mathbb{Z}_d\text{.}\)

Proof

Since dd is a positive divisor of \(n\), \(n\mathbb{Z}\) and \(\langle d \rangle\) are clearly subgroups of, respectively, \(d\mathbb{Z}\) and \(\mathbb{Z}n\). Moreover, since \(d\mathbb{Z}\) and \(\mathbb{Z}n\) are abelian, all of their subgroups are normal.
This follows from the facts that \(d\mathbb{Z}/n\mathbb{Z}= \langle d+n\mathbb{Z} \rangle\), hence is cyclic, and that \(|d\mathbb{Z}/n\mathbb{Z}|=(d\mathbb{Z}:n\mathbb{Z})=n/d\) (see Example \(7.3.3\)). (Unpacking the statement that \(d\mathbb{Z}/n\mathbb{Z}= \langle d+n\mathbb{Z} \rangle\): Notice that \(d\mathbb{Z}=\langle d \rangle\), so every element of \(d\mathbb{Z}\) is of the form \(kd\) for some integer \(k\). Thus, every element of \(d\mathbb{Z}/n\mathbb{Z}\) is of the form \(kd+n\mathbb{Z}\) for some integer \(k\). But for each \(k∈\mathbb{Z}\), using the definition of left coset multiplication we have that \(kd+n\mathbb{Z}=k(d+n\mathbb{Z})\). So \(d\mathbb{Z}/n\mathbb{Z}= \langle d+n\mathbb{Z} \rangle\).)
Similarly, since \(\mathbb{Z}_n=\langle 1 \rangle, \mathbb{Z}n/ \langle d \rangle=\langle 1+ \langle d \rangle \rangle\), so is cyclic. Since \((\mathbb{Z}_n: \langle d \rangle)=d\) (again, see Example \(7.3.3\)), \(\mathbb{Z}n/ \langle d \rangle\) is isomorphic to \(\mathbb{Z}d\), as desired.

Q.E.D.

Example \(\PageIndex{2}\)

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 \(\mathbb{Z}_2\text{.}\)

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.

Theorem \(\PageIndex{3}\)

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 \(|G/N|=|G|/|N|\).
The proof of this statement is left as an exercise for the reader.
Let \(G\) be cyclic. Then there exists \(a∈G\) with \(G=\langle a \rangle\). We claim \(G/N=\langle aN \rangle\). Indeed:

\(\langle aN \rangle={(aN)i:i∈\mathbb{Z}}={aiN:i∈\mathbb{Z}}\).

But every element of \(G\) is an integer power of \(a\), so this set equals \(\{xN:x∈G\}\), that is, it equals \(G/N\).

Note

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.

Example \(\PageIndex{3}\)

\(S_3\) is nonabelian (and therefore of course noncyclic), but we saw above that \(N=\langle (123)\rangle\) is a normal subgroup of \(S_3\) with \(S_3/N \simeq \mathbb{Z}_2\text{,}\) a cyclic (and therefore of course abelian) group.

Example \(\PageIndex{4}\)

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

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.

Theorem \(\PageIndex{4}\): 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 \(K\subseteq N\text{.}\)

Example \(\PageIndex{5}\)

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

Definition: Epimorphism and Monomorphism

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.

Theorem \(\PageIndex{5}\)

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.

Definition: Canonical Epimorphism

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:

Theorem \(\PageIndex{6}\)

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 \(G'\text{.}\)