Section11.1Group 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.

Example11.1.

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 )\text{.} \end{equation*}

This homomorphism maps $${\mathbb Z}$$ onto the cyclic subgroup of $$G$$ generated by $$g\text{.}$$

Example11.2.

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 $$A \mapsto \det(A)\text{.}$$

Example11.3.

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 )\text{.} \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)\text{.} \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'\text{.} \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)\text{,} \end{equation*}

$$\phi(H_1)$$ is a subgroup of $$G_2$$ by Proposition 3.31.

(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\text{,} \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 11.4, $$\phi^{-1} ( \{ e \} )$$ 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.

Example11.6.

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, $$\ker \phi = SL_2( {\mathbb R })\text{.}$$

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 $$\{ 2 \pi n : n \in {\mathbb Z} \}\text{.}$$ Notice that $$\ker \phi \cong {\mathbb Z}\text{.}$$

Example11.8.

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.

Example11.9.

Let $$G$$ be a group. Suppose that $$g \in G$$ and $$\phi$$ is the homomorphism from $${\mathbb Z}$$ to $$G$$ given by $$\phi( n ) = g^n\text{.}$$ 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 $$n {\mathbb Z}\text{.}$$