Question

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

$\mathbf{g}\mathbf{:}{\mathbf{ℝ}}^{\mathbf{\ast }}\mathbf{\to }{\mathbf{\mathbb{Z}}}_{\mathbf{2}}$, where$\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$if $\mathit{x}\mathbf{>}\mathbf{0}$and$\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}$if$\mathit{x}\mathbf{<}\mathbf{0}$.

Short answer

It is proved that,$g$is a homomorphism and ker $\mathrm{g}=\{\mathrm{x}\in {\mathrm{ℝ}}^{\ast }:\mathrm{x}>0\}$.

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 $G$onto an identity element in $H$by the homomorphism $f$.

Let localid="1654579501454" $\mathrm{g}:{\mathrm{ℝ}}^{\ast }\to {\mathrm{\mathbb{Z}}}_2$be the function defined as:

$\mathrm{g}\left(\mathrm{x}\right)=0,$if $\mathrm{x}>0$ and

$\mathrm{g}\left(\mathrm{x}\right)=1$, iflocalid="1654579496050" $\mathrm{x}<0$

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

Let $x,y\in {\mathrm{ℝ}}^{\ast }$.

We show that $\mathrm{g}\left({\mathrm{xy}}^{-1}\right)=\mathrm{g}\left(\mathrm{x}\right)\mathrm{g}{\left(\mathrm{y}\right)}^{-1}$.

Here, we consider four cases as follows:

Case 1: $x<0$and $y<0$

Since$x<0$and$y<0$, therefore$x{y}^{-1}>0$.

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

$$\begin{gathered} g\left(x{y}^{-1}\right)=0\mathrm{mod}2 \\ =2\mathrm{mod}2 \\ =1+1 \\ =g\left(x\right)+g{\left(y\right)}^{-1} \end{gathered}$$

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

Case 2: $x<0$and $y>0$

Since $x<0$and $y>0$, therefore $x{y}^{-1}<0$.

$$\begin{gathered} g\left(x{y}^{-1}\right)=1\mathrm{mod}2 \\ =1+0 \\ =g\left(x\right)+g{\left(y\right)}^{-1} \end{gathered}$$

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

Case 3: $x>0$and $y<0$

Since$x>0$and $y<0$, therefore$x{y}^{-1}<0$.

.$$\begin{gathered} g\left(x{y}^{-1}\right)=1\mathrm{mod}2 \\ =0+1 \\ =g\left(x\right)+g{\left(y\right)}^{-1} \end{gathered}$$

localid="1659489056317" $\therefore g\left(x{y}^{-1}\right)=g\left(x\right)g{\left(y\right)}^{-1}$

Case 4: $x>0$and $y>0$

Since$x>0$and $y>0$, therefore$x{y}^{-1}>0$.

$\begin{array}{c}g\left(x{y}^{-1}\right)=0\mathrm{mod}2\\ =0+0\\ =g\left(x\right)+g{\left(y\right)}^{-1}\end{array}$

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

Hence,$g$is a homomorphism.

To find kernel of f

Now, we find the kernel of$g$ .

Clearly, ker $g==\{x\in {ℝ}^{\ast }:x>0\}$.

Hence,$g$is a homomorphism and ker $g=\{x\in {ℝ}^{\ast }:x>0\}$.