Question

In Exercises 1-9, verify that the given function is a homomorphism and find its kernel.

$\mathbf{f}:\mathrm{ℂ}\to \mathrm{ℝ}$, where$\mathbf{f}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{bi}\mathbf{)}\mathbf{=}\mathbf{b}$.

Short answer

It is proved that,$f$ is a homomorphism and Ker$f=\left\{0\right\}$ .

Step-by-step solution

To show f is a homomorphism

Definition of Group Homomorphism

Let $(G,\ast )$and $(\mathrm{G}\text{'},\ast \text{'})$be any two groups. A function $f:G\to G\text{'}$ is said to be a group homomorphism if $f(a\ast b)=f\left(a\right)\ast \text{'}f\left(b\right),\forall a,b\in G$.

Definition of Kernel of a Function

Let $f:G\to H$be a homomorphism of groups. Then the kernel ofrole="math" localid="1654576673880" $f$is defined by the setrole="math" localid="1654576678261" $\{a\in G:f\left(a\right)=e_H\}$, whererole="math" localid="1654576682737" $e_H$is an identity element.

It is the mapping from elements in role="math" localid="1654576686084" $G$onto an identity element in role="math" localid="1654576690041" $H$by the homomorphism role="math" localid="1654576694351" $f$.

Let role="math" localid="1654576749815" $\mathrm{f}:\mathrm{ℂ}\to \mathrm{ℝ}$be the function defined by $f(a+bi)=b$.

We have to show that $f$is a homomorphism.

Let $\mathrm{a}+\mathrm{bi},\mathrm{c}+\mathrm{di}\in \mathrm{ℂ}$.

Then, find$\mathrm{f}((\mathrm{a}+\mathrm{bi})+(-\mathrm{c}-\mathrm{di}))$ as:

$$\begin{gathered} f((a+bi)+(-c-di))=f\left(\right(a-c)+i(b-d\left)\right) \\ =b-d \\ =f(a+ib)+f(-c-id) \end{gathered}$$

$\therefore f((a+ib)+(-c-id))=f(a+ib)+f(-c-id)$.

Hence, is a homomorphism.

To find kernel of f

In order to find kernel of $f$, we claim that Ker $\mathrm{f}=\{\mathrm{a}:\mathrm{a}\in \mathrm{ℝ}\}$.

If $\mathrm{a}+\mathrm{bi}\in \mathrm{ℂ}$ such that $f(a+ib)=0$$\Rightarrow b=0$.

On the other hand, for any$\mathrm{a}\in \mathrm{ℝ}$, we have $\mathrm{f}\left(\mathrm{a}\right)=0$.

This implies that,Ker f as:

$\begin{array}{c}\text{Ker\hspace{0.17em}}f=\{a:a\in ℝ\}\\ =\left\{0\right\}\end{array}$

Hence, $f$is a homomorphism and Ker$f=\left\{0\right\}$ .