Question

Let $N$ be the subgroup $\left(4\right)$ of ${\mathbb{Z}}_{20}$. Find the order of $13+N$ in the group ${\mathbb{Z}}_{20}/N$.

Short answer

The order of $13+N$ in the group ${\mathbb{Z}}_{20}/N$ is 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$N$.

Subgroup 4 of  ℤ20

The elements in the group ${\mathbb{Z}}_{20}$ are $\left\{0,1,2,\dots ,19\right\}$ , and the elements in the cyclic subgroup $〈4〉$ of ${\mathbb{Z}}_{20}$ are $\left\{0,4,8,12,16\right\}$ .

Order of 13+N in the group  ℤ20/N

Let $k$ be the order of $13+N$ in ${\mathbb{Z}}_{20}/N$; then $k\left(13+N\right)=13k+N$.

Since $13k+N\subset N$ implies that $13k\in N$,

Let $k=4$. Then $52\in N$, which is divisible by 4.

Therefore, the order of $13+N$ in the group ${\mathbb{Z}}_{20}/N$ is 4.