Question

Let $G={\mathbb{Z}}_4\times {\mathbb{Z}}_4$and let $N$ be the cyclic subgroup generated by $\left(3,2\right)$. Show that $G/N\cong {\mathbb{Z}}_4$.

Short answer

The group$G/N\cong {\mathbb{Z}}_4$ .

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 group or factor group of $G$ by $N$.

Quotient group G/N

Let $G={\mathbb{Z}}_4\times {\mathbb{Z}}_4$ be the group of order 16 and N be the subgroup generated by $\left(3,2\right)$ which are $\left(3,2\right)$, $\left(3,2\right)+\left(3,2\right)=\left(6,4\right)\equiv \left(2,0\right)$,$\left(3,2\right)+\left(3,2\right)+\left(3,2\right)=\left(9,6\right)\equiv \left(1,2\right)$ , $\left(3,2\right)+\left(3,2\right)+\left(3,2\right)+\left(3,2\right)=\left(12,8\right)\equiv \left(0,0\right)$ in $G$.

Thus, the subgroup generated by $\left(3,2\right)$are $\left\{\left(0,0\right)\left(2,0\right)\left(1,2\right)\left(3,2\right)\right\}$which is of order 4. Then the order of the quotient group$G/N$ is 4.

G/N≅ℤ4

Recall the theorem 8.7. Let p be a positive prime integer. Every order of group p is isomorphic to ${\mathbb{Z}}_p$.

Here, the order of $G/N$ is a positive integer 4; then by the theorem,$G/N\cong {\mathbb{Z}}_4$.

Therefore, the group $G/N\cong {\mathbb{Z}}_4$.