Question

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

Short answer

It is proved that, $g$is a homomorphism and ker $g=\{(a,1):a\in \mathbb{Q}\}$.

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 $\mathrm{g}:\mathrm{\mathbb{Q}}\times \mathrm{\mathbb{Z}}\to \mathrm{\mathbb{Z}}$be the function defined as $g\left((x,y)\right)=y$.

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

Let $x=(p,q),y=(r,s)\in \mathbb{Q}\times \mathbb{Z}$.

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

$\begin{array}{c}g\left(x{y}^{-1}\right)=g\left((p,q){(r,s)}^{-1}\right)\\ =g(p-r,q-s)\\ =q-s\\ =g\left((p,q)\right)-g\left((r,s)\right)\\ =g\left((p,q)\right)g{\left((r,s)\right)}^{-1}\\ =g\left(x\right)g{\left(y\right)}^{-1}\end{array}$

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

Hence, $g$is a homomorphism.

To find kernel of g

Let us find kernel of $g$.

We have, ker$g=\{(x,y)\in \mathbb{Q}\times \mathbb{Z}:y=1\}$

Hence, ker $g=\{(a,1):a\in \mathbb{Q}\}$.

Thus,$g$ is a homomorphism and ker $g=\{(a,1):a\in \mathbb{Q}\}$.