Abstract Algebra: Theory and Applications

Let us examine the homomorphism $\varphi :G{L}_2(\mathbb{R})\to {\mathbb{R}}^{\ast }$ 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, $\ker \varphi =S{L}_2(\mathbb{R})\text{.}$

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

Suppose that we wish to determine all possible homomorphisms $\varphi$ from ${\mathbb{Z}}_7$ to ${\mathbb{Z}}_{12}\text{.}$ Since the kernel of $\varphi$ 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.

Let $G$ be a group. Suppose that $g\in G$ and $\varphi$ is the homomorphism from $\mathbb{Z}$ to $G$ given by $\varphi (n)=g^n\text{.}$ If the order of $g$ is infinite, then the kernel of this homomorphism is $\{0\}$ since $\varphi$ 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 $\varphi$ is $n\mathbb{Z}\text{.}$