Question

If k|n and$\mathit{f}\mathbf{:}{\mathit{U}}_{\mathbf{n}}\mathbf{\to }{\mathit{U}}_{\mathbf{k}}$ is given by $\mathit{f}\mathbf{\left(}{\mathbf{\left[}\mathbf{x}\mathbf{\right]}}_{\mathbf{n}}\mathbf{\right)}\mathbf{=}{\mathbf{\left[}\mathbf{x}\mathbf{\right]}}_{\mathbf{k}}$, show that$\mathit{f}$ is a homomorphism and find its kernel.

Short answer

It is proved that, f is a homomorphism with $\ker f=\{{\left[x\right]}_n\in U_n:\text{\hspace{0.17em}\hspace{0.17em}}x\equiv 1\left(\mathrm{mod}k\right)\}$.

Step-by-step solution

General form of the kernel

If G and H are groups and f is a group homomorphism from G to H, then$\mathit{k}\mathit{e}\mathit{r}\mathit{f}\mathbf{=}\mathbf{\{}\mathbf{g}\mathbf{\in }\mathbf{G}\mathbf{,}\mathbf{f}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{=}{\mathbf{e}}_{\mathbf{H}}\mathbf{\}}$ .

Proving that f is a homomorphism

Let${\left[x\right]}_n,{\left[y\right]}_n\in U_n$ .

Then prove f is a homomorphism as:

$\begin{array}{c}f\left({\left[x\right]}_n{\left[y\right]}_n\right)=f\left({\left[xy\right]}_n\right)\\ ={\left[xy\right]}_k\\ ={\left[x\right]}_k{\left[y\right]}_k\\ =f\left({\left[x\right]}_n\right)f\left({\left[y\right]}_n\right)\end{array}$

Hence, f is a homomorphism.

Finding kernel for f

By putting the values in general formula for kernel, we get .

$\ker f=\{{\left[x\right]}_n\in U_n:\text{\hspace{0.17em}\hspace{0.17em}}x\equiv 1\left(\mathrm{mod}k\right)\}$

Hence, f is a homomorphism with $\ker f=\{{\left[x\right]}_n\in U_n:\text{\hspace{0.17em}\hspace{0.17em}}x\equiv 1\left(\mathrm{mod}k\right)\}$.