Question

In Exercises 1-9, verify that the given function is a homomorphism and find its kernel. $\mathbf{f}\mathbf{:}{\mathbf{\mathbb{Z}}}_{\mathbf{12}}\mathbf{\to }{\mathbf{\mathbb{Z}}}_{\mathbf{12}}$, whererole="math" localid="1654587769060" $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathit{x}$.

Short answer

It is proved that,$f$ is a homomorphism and Ker$f=\{0,4,8\}$ .

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$\mathrm{f}:{\mathrm{\mathbb{Z}}}_{12}\to {\mathrm{\mathbb{Z}}}_{12}$be the function defined as,$f\left(x\right)=3x$.

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

Let $\mathrm{x},\mathrm{y}\in {\mathrm{\mathbb{Z}}}_{12}$.

Then to prove homomorphism we will show that localid="1654588598160" $f(x-y)=f\left(x\right)-f\left(y\right)$as follows:

$\begin{array}{c}f(x-y)=3(x-y)\\ =3x-3y\\ =f\left(x\right)-f\left(y\right)\end{array}$

$\therefore f(x-y)=f\left(x\right)-f\left(y\right)$

Hence,$f$is a homomorphism.

To find kernel of f

We find kernel of$f$ as:

Kerrole="math" localid="1654588450227" $f$$=\{x\in {\mathbb{Z}}_{12}:3x\equiv 0\mathrm{mod}12\}$

Here,$3x$ is divisible by 12 for $x\in {\mathbb{Z}}_{12}$.

Then, the values of$x$ has to be one of $0,4,and8$.

Hence, ker $f=\{0,4,8\}$.

Thus, $f$is a homomorphism and Ker $f=\{0,4,8\}$.