Question

Show that ${\mathbf{\mathbb{Z}}}_{\mathbf{6}}\mathbf{/}\mathit{N}\mathbf{\cong }{\mathbf{\mathbb{Z}}}_{\mathbf{3}}$, where N is the cyclic subgroup $\left\{0,3\right\}$.

Short answer

The group${\mathrm{\mathbb{Z}}}_6/N\cong {\mathrm{\mathbb{Z}}}_3$

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 ℤ6/N

The elements in the group ${\mathrm{\mathbb{Z}}}_6$are $\left\{0,1,2,3,4,5\right\}$, and the elements in the cyclic subgroup N are $\left\{0,3\right\}$.

Here, the order of group ${\mathrm{\mathbb{Z}}}_6$ is 6, and the subgroup M is 2, such that the order of the quotient subgroup ${\mathrm{\mathbb{Z}}}_6/N$is 3.

ℤ6/N≅ℤ3

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

Here, the order of ${\mathrm{\mathbb{Z}}}_6/N$is a positive prime integer 3; then by the theorem, ${\mathrm{\mathbb{Z}}}_6/N\cong {\mathrm{\mathbb{Z}}}_3$.

Therefore, the group ${\mathrm{\mathbb{Z}}}_6/N\cong {\mathrm{\mathbb{Z}}}_3$.