Question

In Exercises 1-9, verify that the given function is a homomorphism and find its kernel.$\mathbf{h}\mathbf{:}\mathbf{ℂ}\mathbf{\to }\mathbf{ℂ}$where$\mathit{h}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{4}}$ .

Short answer

It is proved that,$h$ is a homomorphism and ker $h=\{1,-1,i,-i\}$.

Step-by-step solution

To show f is a homomorphism

Definition of Group Homomorphism

Let $(G,\ast )$and $(G\text{'},\ast \text{'})$be any two groups. A function role="math" localid="1654584633323" $f:G\to G\text{'}$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 of$f$is defined by the set $\{a\in G:f\left(a\right)=e_H\}$, where$e_H$is an identity element.

It is the mapping from elements in $G$onto an identity element in $H$by the homomorphism $f$.

Let $\mathrm{h}:\mathrm{ℂ}\to \mathrm{ℂ}$be the function defined as$h\left(x\right)=x^4$.

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

Let$x,y\in ℂ$.

Then, we will show that $h\left(x{y}^{-1}\right)=h\left(x\right)h{\left(y\right)}^{-1}$as follows:

$$\begin{gathered} h\left(x{y}^{-1}\right)={\left(x{y}^{-1}\right)}^4 \\ \begin{array}{c}=x^4{y}^{-4}\\ =x^4{\left(y^4\right)}^{-1}\\ =h\left(x\right)h{\left(y\right)}^{-1}\end{array} \end{gathered}$$

$\therefore h\left(x{y}^{-1}\right)=h\left(x\right)h{\left(y\right)}^{-1}$

Hence, $h$is a homomorphism.

To find kernel of h

Now, we find kernel of $h$.

Then, ker $h=\{x\in ℂ:x^4=1\}$.

This implies that, only elements $\mathrm{x}\in \mathrm{ℂ}$such that $x^4=1$are $1,-1,i$and $-i$

since ${\left(1\right)}^4={(-1)}^4=1$and ${\left(\mathrm{i}\right)}^4={(-\mathrm{i})}^4=1$.

Hence, ker.$h=\{1,-1,i,-i\}$

Thus, $h$is a homomorphism and ker $h=\{1,-1,i,-i\}$.