Question

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

Short answer

It is proved that, $f$is a homomorphism and Ker$f=\{1,-1\}$ .

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 $H$in by the homomorphism $f$.

Let $\mathrm{f}:{\mathrm{\mathbb{Q}}}^{\ast }\to {\mathrm{\mathbb{Q}}}^{\ast \ast }$be the function defined by $f\left(x\right)=\left|x\right|$.

Let $\mathrm{x},\mathrm{y}\in {\mathrm{\mathbb{Q}}}^{\ast }$.

Then, find $f\left(x{y}^{-1}\right)$as:

$\begin{array}{c}f\left(x{y}^{-1}\right)=\left|x{y}^{-1}\right|\\ =\left|x\right|\left|y^{-1}\right|\\ =\left|x\right|{\left|y\right|}^{-1}\\ =f\left(x\right)f{\left(y\right)}^{-1}\end{array}$

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

Hence,$f$ is a homomorphism.

To find kernel of f

Let us find kernel of $f$.

Then, Ker $f=\{x\in {\mathbb{Q}}^{\ast }:\left|x\right|=1\}$.

Then, only $\mathrm{x}\in {\mathrm{\mathbb{Q}}}^{\ast }$such that role="math" localid="1654582900949" $\left|x\right|=1$are 1 and$-1$, since $\left|1\right|=1$and$\left|1\right|=1$.

Thus, we get an identity element 1.

Hence, Ker $f=\{1,-1\}$.

Therefore,$f$ is a homomorphism and Ker $f=\{1,-1\}$.