Question

$N=\left\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\right\}$is a normal subgroup of $S_3$ by example 9 of section 8.2. Show that $S_3/N\cong {\mathbb{Z}}_2$ .

Short answer

The group $S_3/N\cong {\mathbb{Z}}_2$.

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 of S3

The elements in the group $S_3$ are$\left\{\left(\right),\left(12\right),\left(13\right),\left(23\right),\left(123\right),\left(132\right)\right\}$.

Given that $N=\left\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\right\}$ is the normal subgroup of $S_3$.

Here, the order of group $S_3$ is 6, and the subgroup $N$ is 3, such that the order of the quotient subgroup $S_3/N$ is 2.

  S3/N≅ℤ2

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 $S_3/N$ is a positive prime integer 2; then by the theorem, $S_3/N\cong {\mathbb{Z}}_2$.

Therefore, the group $S_3/N\cong {\mathbb{Z}}_2$.