Question

Let $\mathit{G}\mathbf{=}{\mathit{\mathbb{Z}}}_{\mathbf{6}}\mathbf{\times }{\mathit{\mathbb{Z}}}_{\mathbf{2}}$ and let $\mathit{N}$ be the cyclic subgroup$\mathbf{⟨}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{⟩}$ . Describe the quotient group $\mathit{G}\mathbf{/}\mathit{N}$.

Short answer

The elements in group $G/N$ are$\{(0,1)N,N\}$ .

Step-by-step solution

Normal subgroup and Quotient group

Let $N$ be the normal subgroup of role="math" localid="1654496927729" $G$. 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 role="math" localid="1654496938591" $G$ is an abelian group, then so is $G/N$.

The group$G/N$ is called thequotient group or factor group of $G$ by $N$.

Quotient group G/N

Let $\mathrm{G}={\mathrm{\mathbb{Z}}}_6\times {\mathrm{\mathbb{Z}}}_2$ be the group of order 12 and N be the subgroup generated by $(1,1)$which are $(1,1)$, $2(1,1)=(2,2)\equiv (2,0)$, $3(1,1)=(3,3)\equiv (3,1)$,$4(1,1)=(4,4)\equiv (4,0)$, $5(1,1)=(5,5)\equiv (5,1)$, $6(1,1)=(6,6)\equiv (0,0)$ in $G$.

Thus, the elements in the subgroup generated by $(1,1)$are$\left\{(1,1)(2,0)(3,1)(4,0)(5,1)(0,0)\right\}$ which is of order 6. Then the order of the quotient group $G/N$ is 2.

Elements in G/N

The element $(0,1)$generates the quotient group then $G/N=\{(0,1)N,N\}$, which is of order 2.

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

Here, the order of role="math" localid="1654497413114" $G/N$is a positive integer 2; then by the theorem, $G/N\cong {\mathbb{Z}}_2$.

Therefore, the elements in group $G/N$are $\{(0,1)N,N\}$ and $G/N\cong {\mathbb{Z}}_2$.