Question

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

$\mathit{h}\mathbf{:}{\mathbf{ℝ}}^{\mathbf{\ast }}\mathbf{\to }{\mathbf{ℝ}}^{\mathbf{\ast }}$, where $\mathit{h}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{3}}$.

Short answer

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

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 $f:G\to G\text{'}$ 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 role="math" localid="1654581461326" $G$ onto an identity element in role="math" localid="1654581466347" $H$by the homomorphism role="math" localid="1654581474548" $f$.

Let $h:{\mathrm{ℝ}}^{\ast }\to {\mathrm{ℝ}}^{\ast }$be the function defined by $\mathrm{h}\left(\mathrm{x}\right)={\mathrm{x}}^3$.

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

Let $x,y\in {\mathrm{ℝ}}^{\ast }$.

Then we will show that $\mathrm{h}\left({\mathrm{xy}}^{-1}\right)=\mathrm{h}\left(\mathrm{x}\right)\mathrm{h}{\left(\mathrm{y}\right)}^{-1}$.

Prove $\mathrm{h}\left({\mathrm{xy}}^{-1}\right)=\mathrm{h}\left(\mathrm{x}\right)\mathrm{h}{\left(\mathrm{y}\right)}^{-1}$ as:

$$\begin{gathered} h\left(x{y}^{-1}\right)={\left(x{y}^{-1}\right)}^3 \\ =x^3{y}^{-3} \\ =x^3{\left(y^3\right)}^{-1} \\ =h\left(x\right)h{\left(y\right)}^{-1} \end{gathered}$$

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

Hence,$h$ is a homomorphism.

To find kernel of h

Now, we find kernel of $h$.

Then, ker$h=\{x\in {ℝ}^{\ast }:x^3=1\}$.

This implies that, the only element $x\in {\mathrm{ℝ}}^{\ast }$such that $x^3=1$ is 1.

Hence, ker$h=\left\{1\right\}$.

Thus, $h$ is a homomorphism and ker $h=\left\{1\right\}$.