Question

Prove that ${\mathbb{Z}}_{12}\cong {\mathbb{Z}}_3\times {\mathbb{Z}}_4$. [Consider , given by $f\left(a\right)=\left({\left[a\right]}_3,{\left[a\right]}_4\right)$.]

Short answer

It is proved that, the group ${\mathbb{Z}}_{12}$ is isomorphic to ${\mathbb{Z}}_3\times {\mathbb{Z}}_4$.

Step-by-step solution

First Isomorphism Theorem

Let$f:G\to H$ be a surjective homomorphism of groups with kernel K. Then the quotient group $G/K$ is isomorphic to H.

 f is surjective

Define a map $f:\mathbb{Z}\to {\mathbb{Z}}_3\times {\mathbb{Z}}_4$ by $f\left(a\right)=\left({\left[a\right]}_3,{\left[a\right]}_4\right)$.

Let $\left({\left[a\right]}_3,{\left[b\right]}_4\right)\in {\mathbb{Z}}_3\times {\mathbb{Z}}_4$.

Then, there exists an element $4a+9b$ in $\mathbb{Z}$ such that .

$f\left(4a+9b\right)=\left({\left[4a+9b\right]}_3,{\left[4a+9b\right]}_4\right)$

Simplify $f\left(4a+9b\right)=\left({\left[4a+9b\right]}_3,{\left[4a+9b\right]}_4\right)$ as:

$\begin{array}{c}f\left(4a+9b\right)=\left({\left[4a+9b\right]}_3,{\left[4a+9b\right]}_4\right)\\ =\left({\left[a\right]}_3,{\left[a\right]}_4\right)\end{array}$

Thus, f is surjective.

 f is a homomorphism

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

Then, prove f is a homomorphism as follows:

$\begin{array}{c}f\left(x-y\right)=\left({\left[x-y\right]}_3,{\left[x-y\right]}_4\right)\\ =\left({\left[x\right]}_3,{\left[x\right]}_4\right)-\left({\left[y\right]}_3,{\left[y\right]}_4\right)\\ =f\left(x\right)-f\left(y\right)\end{array}$

Therefore, f is a homomorphism.

Kernel  f 

The kernel f will be the set $Ker\left(f\right)=\left\{z\in \mathbb{Z}:f\left(z\right)=\left(0,0\right)\text{\hspace{0.17em}in\hspace{0.17em}}{\mathbb{Z}}_3\times {\mathbb{Z}}_4\right\}$ that can be written as, $Ker\left(f\right)=\left\{z\in \mathbb{Z}:\left({\left[z\right]}_3,{\left[z\right]}_4\right)=\left(0,0\right)\text{\hspace{0.17em}in\hspace{0.17em}}{\mathbb{Z}}_3\times {\mathbb{Z}}_4\right\}$.

Here, $\ker f$ consists of all multiples of 12 implies $\ker f=12\mathbb{Z}$.

Since f is a surjective homomorphism with the kernel $12\mathbb{Z}$,byFirst Isomorphism Theorem, ${\mathbb{Z}}_{12}\cong {\mathbb{Z}}_3\times {\mathbb{Z}}_4$.

Therefore, ${\mathbb{Z}}_{12}\cong {\mathbb{Z}}_3\times {\mathbb{Z}}_4$.