Question

Let $G$ be the subgroup $〈3〉$ of $\mathbb{Z}$and let $N$be the subgroup $〈15〉$. Find the order of $6+N$ in the group $G/N$.

Short answer

The order of $6+N$ in the group $G/N$ is 5.

Step-by-step solution

Normal subgroup and Quotient group

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

1. If$G/N$ 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 3 of  ℤ

The elements in the group $\mathbb{Z}$ are $\left\{-\infty ,\dots ,+\infty \right\}$ , and the elements in the cyclic subgroup $〈3〉$ of $\mathbb{Z}$ are the multiples of 3.

Let $N$ be the subgroup $〈15〉$ that contains all the multiples of 15.

Order of 6+N in the group  G/N

Let $k$ be the order of $6+N$ in $G/N$; then$k\left(6+N\right)=6k+N$

Since $6k+N\subset N$implies that $6k\in N$,

.

Let $k=5$ then $30\in N$ which is divisible by 15.

Therefore, the order of $6+N$ in the group $G/N$ is 5.