Although it is not evident at first, factor groups correspond exactly to homomorphic images, and we can use factor groups to study homomorphisms. We already know that with every group homomorphism $\phi: G \rightarrow H$ we can associate a normal subgroup of $G$, $\ker \phi$. The converse is also true; that is, every normal subgroup of a group $G$ gives rise to homomorphism of groups.

Let $H$ be a normal subgroup of $G$. Define the natural or canonical homomorphism \begin{equation*}\phi : G \rightarrow G/H\end{equation*} by \begin{equation*}\phi(g) = gH.\end{equation*} This is indeed a homomorphism, since \begin{equation*}\phi( g_1 g_2 ) = g_1 g_2 H = g_1 H g_2 H = \phi( g_1) \phi( g_2 ).\end{equation*} The kernel of this homomorphism is $H$. The following theorems describe the relationships between group homomorphisms, normal subgroups, and factor groups.

##### Proof

Mathematicians often use diagrams called commutative diagrams to describe such theorems. The following diagram “commutes” since $\psi = \eta \phi$.

##### Example11.11

Let $G$ be a cyclic group with generator $g$. Define a map $\phi : {\mathbb Z} \rightarrow G$ by $n \mapsto g^n$. This map is a surjective homomorphism since \begin{equation*}\phi( m + n) = g^{m+n} = g^m g^n = \phi(m) \phi(n).\end{equation*} Clearly $\phi$ is onto. If $|g| = m$, then $g^m = e$. Hence, $\ker \phi = m {\mathbb Z}$ and ${\mathbb Z} / \ker \phi = {\mathbb Z} / m {\mathbb Z} \cong G$. On the other hand, if the order of $g$ is infinite, then $\ker \phi = 0$ and $\phi$ is an isomorphism of $G$ and ${\mathbb Z}$. Hence, two cyclic groups are isomorphic exactly when they have the same order. Up to isomorphism, the only cyclic groups are ${\mathbb Z}$ and ${\mathbb Z}_n$.

##### Proof

Notice that in the course of the proof of Theorem 11.13, we have also proved the following theorem.

##### Example11.15

By the Third Isomorphism Theorem, \begin{equation*}{\mathbb Z} / m {\mathbb Z} \cong ({\mathbb Z}/ mn {\mathbb Z})/ (m {\mathbb Z}/ mn {\mathbb Z}).\end{equation*} Since $| {\mathbb Z} / mn {\mathbb Z} | = mn$ and $|{\mathbb Z} / m{\mathbb Z}| = m$, we have $| m {\mathbb Z} / mn {\mathbb Z}| = n$.