Question

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

$\mathit{f}\mathbf{:}\mathbf{\mathbb{Z}}\mathbf{\to }{\mathbf{\mathbb{Z}}}_{\mathbf{2}}\mathbf{\times }{\mathbf{\mathbb{Z}}}_{\mathbf{4}}$, whererole="math" localid="1654590570026" $\mathit{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{(}{\mathbf{\left[}\mathbf{a}\mathbf{\right]}}_{\mathbf{2}}\mathbf{,}{\mathbf{\left[}\mathbf{a}\mathbf{\right]}}_{\mathbf{4}}\mathbf{)}$.

Short answer

It is proved that, $f$is a homomorphism and Ker $f=\{4a:a\in \mathbb{Z}\}$ .

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 role="math" localid="1654590767873" $f:G\to H$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="1654590877126" $G$onto an identity element in $H$ by the homomorphism role="math" localid="1654590892791" $f$.

Let $\mathrm{f}:\mathrm{\mathbb{Z}}\to {\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$be the function defined as, $f\left(a\right)=({\left[a\right]}_2,{\left[a\right]}_4)$.

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

Let $a,b\in \mathrm{\mathbb{Z}}$.

Then, to prove homomorphism we will show that $f(a-b)=f\left(a\right)-f\left(b\right)$ as follows:

$\begin{array}{c}f(a-b)=({[a-b]}_2,{[a-b]}_4)\\ =({\left[a\right]}_2-{\left[b\right]}_2,{\left[a\right]}_4-{\left[b\right]}_4)\\ =({\left[a\right]}_2,{\left[a\right]}_4)-({\left[b\right]}_2,{\left[b\right]}_4)\\ =f\left(a\right)-f\left(b\right)\end{array}$

$\therefore f(a-b)=f\left(b\right)-f\left(b\right)$

Hence, $f$ is a homomorphism.

To find kernel of f

We find kernel of $f$ as:

Then, Ker$f$$=\{a\in \mathbb{Z}:a\equiv 0\mathrm{mod}2anda\equiv 0\mathrm{mod}4\}$

Here, $a$ is divisible by 2 as well as 4, for $\mathrm{a}\in \mathrm{\mathbb{Z}}$.

Then, $a$ must be multiple of 4.

Hence, ker$f=\{4a:a\in \mathbb{Z}\}$.

Thus, role="math" localid="1654591839396" $f$is a homomorphism and Kerrole="math" localid="1654591843699" $f=\{4a:a\in \mathbb{Z}\}$ .