A homomorphism between groups $(G, \cdot)$ and $(H, \circ)$ is a map $\phi :G \rightarrow H$ such that \begin{equation*}\phi( g_1 \cdot g_2 ) = \phi( g_1 ) \circ \phi( g_2 )\end{equation*} for $g_1, g_2 \in G$. The range of $\phi$ in $H$ is called the homomorphic image of $\phi$.

Two groups are related in the strongest possible way if they are isomorphic; however, a weaker relationship may exist between two groups. For example, the symmetric group $S_n$ and the group ${\mathbb Z}_2$ are related by the fact that $S_n$ can be divided into even and odd permutations that exhibit a group structure like that ${\mathbb Z}_2$, as shown in the following multiplication table.

We use homomorphisms to study relationships such as the one we have just described.

##### Example11.1

Let $G$ be a group and $g \in G$. Define a map $\phi : {\mathbb Z} \rightarrow G$ by $\phi( n ) = g^n$. Then $\phi$ is a group homomorphism, since \begin{equation*}\phi( m + n ) = g^{ m + n} = g^m g^n = \phi( m ) \phi( n ).\end{equation*} This homomorphism maps ${\mathbb Z}$ onto the cyclic subgroup of $G$ generated by $g$.

##### Example11.2

Let $G = GL_2( {\mathbb R })$. If \begin{equation*}A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\end{equation*} is in $G$, then the determinant is nonzero; that is, $\det(A) = ad - bc \neq 0$. Also, for any two elements $A$ and $B$ in $G$, $\det(AB) = \det(A) \det(B)$. Using the determinant, we can define a homomorphism $\phi : GL_2( {\mathbb R }) \rightarrow {\mathbb R}^\ast$ by $A \mapsto \det(A)$.

##### Example11.3

Recall that the circle group ${ \mathbb T}$ consists of all complex numbers $z$ such that $|z|=1$. We can define a homomorphism $\phi$ from the additive group of real numbers ${\mathbb R}$ to ${\mathbb T}$ by $\phi : \theta \mapsto \cos \theta + i \sin \theta$. Indeed, \begin{align*} \phi( \alpha + \beta ) & = \cos( \alpha + \beta ) + i \sin( \alpha + \beta )\\ & = (\cos \alpha \cos \beta - \sin \alpha \sin \beta) + i( \sin \alpha \cos \beta + \cos \alpha \sin \beta )\\ & = (\cos \alpha + i \sin \alpha )(\cos \beta + i \sin \beta)\\ & = \phi( \alpha ) \phi( \beta ). \end{align*} Geometrically, we are simply wrapping the real line around the circle in a group-theoretic fashion.

The following proposition lists some basic properties of group homomorphisms.

Let $\phi : G \rightarrow H$ be a group homomorphism and suppose that $e$ is the identity of $H$. By Proposition 11.4, $\phi^{-1} ( \{ e \} )$ is a subgroup of $G$. This subgroup is called the kernel of $\phi$ and will be denoted by $\ker \phi$. In fact, this subgroup is a normal subgroup of $G$ since the trivial subgroup is normal in $H$. We state this result in the following theorem, which says that with every homomorphism of groups we can naturally associate a normal subgroup.

##### Example11.6

Let us examine the homomorphism $\phi : GL_2( {\mathbb R }) \rightarrow {\mathbb R}^\ast$ defined by $A \mapsto \det( A )$. Since 1 is the identity of ${\mathbb R}^\ast$, the kernel of this homomorphism is all $2 \times 2$ matrices having determinant one. That is, $\ker \phi = SL_2( {\mathbb R })$.

##### Example11.7

The kernel of the group homomorphism $\phi : {\mathbb R} \rightarrow {\mathbb C}^\ast$ defined by $\phi( \theta ) = \cos \theta + i \sin \theta$ is $\{ 2 \pi n : n \in {\mathbb Z} \}$. Notice that $\ker \phi \cong {\mathbb Z}$.

##### Example11.8

Suppose that we wish to determine all possible homomorphisms $\phi$ from ${\mathbb Z}_7$ to ${\mathbb Z}_{12}$. Since the kernel of $\phi$ must be a subgroup of ${\mathbb Z}_7$, there are only two possible kernels, $\{ 0 \}$ and all of ${\mathbb Z}_7$. The image of a subgroup of ${\mathbb Z}_7$ must be a subgroup of ${\mathbb Z}_{12}$. Hence, there is no injective homomorphism; otherwise, ${\mathbb Z}_{12}$ would have a subgroup of order 7, which is impossible. Consequently, the only possible homomorphism from ${\mathbb Z}_7$ to ${\mathbb Z}_{12}$ is the one mapping all elements to zero.

##### Example11.9

Let $G$ be a group. Suppose that $g \in G$ and $\phi$ is the homomorphism from ${\mathbb Z}$ to $G$ given by $\phi( n ) = g^n$. If the order of $g$ is infinite, then the kernel of this homomorphism is $\{ 0 \}$ since $\phi$ maps ${\mathbb Z}$ onto the cyclic subgroup of $G$ generated by $g$. However, if the order of $g$ is finite, say $n$, then the kernel of $\phi$ is $n {\mathbb Z}$.