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## Section10.3Group Homomorphisms

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 $g_1, g_2 \in G\text{.}$ The range of $\phi$ in $H$ is called the homomorphic image of $\phi\text{.}$

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 ${\mathbb Z}_2\text{,}$ as shown in the following multiplication table.

\begin{equation*} \begin{array}{c|cc} & \text{even} & \text{odd} \\ \hline \text{even} & \text{even} & \text{odd} \\ \text{odd} & \text{odd} & \text{even} \end{array} \end{equation*}

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

###### Example10.1

Let $G$ be a group and $g \in G\text{.}$ Define a map $\phi : {\mathbb Z} \rightarrow G$ by $\phi( n ) = g^n\text{.}$ 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 $g\text{.}$

###### Example10.2

Let $G = GL_2( {\mathbb R })\text{.}$ If

\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{equation*}

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

###### Example10.3

Recall that the circle group ${ \mathbb T}$ consists of all complex numbers $z$ such that $|z|=1\text{.}$ We can define a homomorphism $\phi$ from the additive group of real numbers ${\mathbb R}$ to ${\mathbb T}$ by $\phi : \theta \mapsto \cos \theta + i \sin \theta\text{.}$ 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.

(1) Suppose that $e$ and $e'$ are the identities of $G_1$ and $G_2\text{,}$ respectively; then

\begin{equation*} e' \phi(e) = \phi(e) = \phi(e e) = \phi(e) \phi(e). \end{equation*}

By cancellation, $\phi(e) = e'\text{.}$

(2) This statement follows from the fact that

\begin{equation*} \phi( g^{-1}) \phi(g) = \phi(g^{-1} g) = \phi(e) = e'. \end{equation*}

(3) The set $\phi(H_1)$ is nonempty since the identity of $G_2$ is in $\phi(H_1)\text{.}$ Suppose that $H_1$ is a subgroup of $G_1$ and let $x$ and $y$ be in $\phi(H_1)\text{.}$ There exist elements $a, b \in H_1$ such that $\phi(a) = x$ and $\phi(b)=y\text{.}$ Since

\begin{equation*} xy^{-1} = \phi(a)[ \phi(b)]^{-1} = \phi(a b^{-1} ) \in \phi(H_1), \end{equation*}

$\phi(H_1)$ is a subgroup of $G_2$ by Proposition 3.16.

(4) Let $H_2$ be a subgroup of $G_2$ and define $H_1$ to be $\phi^{-1}(H_2)\text{;}$ that is, $H_1$ is the set of all $g \in G_1$ such that $\phi(g) \in H_2\text{.}$ The identity is in $H_1$ since $\phi(e) = e'\text{.}$ If $a$ and $b$ are in $H_1\text{,}$ then $\phi(ab^{-1}) = \phi(a)[ \phi(b) ]^{-1}$ is in $H_2$ since $H_2$ is a subgroup of $G_2\text{.}$ Therefore, $ab^{-1} \in H_1$ and $H_1$ is a subgroup of $G_1\text{.}$ If $H_2$ is normal in $G_2\text{,}$ we must show that $g^{-1} h g \in H_1$ for $h \in H_1$ and $g \in G_1\text{.}$ But

\begin{equation*} \phi( g^{-1} h g) = [ \phi(g) ]^{-1} \phi( h ) \phi( g ) \in H_2, \end{equation*}

since $H_2$ is a normal subgroup of $G_2\text{.}$ Therefore, $g^{-1}hg \in H_1\text{.}$

Let $\phi : G \rightarrow H$ be a group homomorphism and suppose that $e$ is the identity of $H\text{.}$ By Proposition 10.4, $\phi^{-1} ( \{ e \} )$ is a subgroup of $G\text{.}$ This subgroup is called the kernel of $\phi$ and will be denoted by $\ker \phi\text{.}$ In fact, this subgroup is a normal subgroup of $G$ since the trivial subgroup is normal in $H\text{.}$ We state this result in the following theorem, which says that with every homomorphism of groups we can naturally associate a normal subgroup.

###### Example10.6

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

###### Example10.7

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

###### Example10.8

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

###### Example10.9

Let $G$ be a group. Suppose that $g \in G$ and $\phi$ is the homomorphism from ${\mathbb Z}$ to $G$ given by $\phi( n ) = g^n\text{.}$ 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 $g\text{.}$ However, if the order of $g$ is finite, say $n\text{,}$ then the kernel of $\phi$ is $n {\mathbb Z}\text{.}$