Question

Write out the operation table of $\mathit{G}\mathbf{/}\mathit{M}$, using the four cosets $\mathit{M}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$,$\mathit{M}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$ , $\mathit{M}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$, $\mathit{M}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$ .

Short answer

The operation table of $G/M$ is shown below:

	$M+(0,0)$	$M+(1,0)$	$M+(0,1)$	$M+(1,1)$
$M+(0,0)$	$M+(0,0)$	$M+(1,0)$	$M+(0,1)$	$M+(1,1)$
$M+(1,0)$	$M+(1,0)$	$M+(0,0)$	$M+(1,1)$	$M+(0,1)$
$M+(0,1)$	$M+(0,1)$	$M+(1,1)$	$M+(0,0)$	$M+(1,0)$
$M+(1,1)$	$M+(1,1)$	$M+(0,1)$	$M+(1,0)$	$M+(0,0)$

Step-by-step solution

Normal subgroup and Quotient group

Let $N$ be the normal subgroup of $G$. Then

1. $G/N$ is a group under the operation defined by $\left(Na\right)\left(Nc\right)=Nac$.

2. If $G$ is finite, then the order of $G/N$ is $\left|G\right|/\left|N\right|$.

3. If $G$ is an abelian group, then so is $G/N$.

The group $G/N$ is called the quotient groupor factor group of $G$ by $N$.

Group G and their subgroups 

Let $\mathrm{G}={\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$ be the group of order 8 and M be the subgroup generated by $(0,2)$.

The elements in M are $\left\{(0,2)(0,0)\right\}$, which is of order 2; then the quotient group $G/M$has order 4.

Operation table of  G/M

The operation table $G/M$ of four cosets is shown below:

	$M+(0,0)$	$M+(1,0)$	$M+(0,1)$	$M+(1,1)$
$M+(0,0)$	$M+(0,0)$	$M+(1,0)$	$M+(0,1)$	$M+(1,1)$
$M+(1,0)$	$M+(1,0)$	$M+(0,0)$	$M+(1,1)$	$M+(0,1)$
$M+(0,1)$	$M+(0,1)$	$M+(1,1)$	$M+(0,0)$	$M+(1,0)$
$M+(1,1)$	$M+(1,1)$	$M+(0,1)$	$M+(1,0)$	$M+(0,0)$

Therefore, the required operation table of $G/M$ is made.