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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 \(g_1, g_2 \in G\). 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.

\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 \(g \in G\). 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 \(n {\mathbb Z}\).