Question

Prove that $\left\{\left(\begin{array}{lll}1& 2& 3\\ 2& 1& 3\end{array}\right),\left(\begin{array}{lll}1& 2& 3\\ 1& 2& 3\end{array}\right)\right\}$is a subgroup of${\mathit{S}}_{\mathbf{3}}$but not normal.

Short answer

It is proved that $\left\{\left(\begin{array}{lll}1& 2& 3\\ 2& 1& 3\end{array}\right),\left(\begin{array}{lll}1& 2& 3\\ 1& 2& 3\end{array}\right)\right\}$is a subgroup of${\mathrm{S}}_3$but not normal.

Step-by-step solution

Prove that given set is subgroup 

Assume that the given subset is $\mathrm{H}=\left\{\left(\begin{array}{lll}1& 2& 3\\ 2& 1& 3\end{array}\right),\left(\begin{array}{lll}1& 2& 3\\ 1& 2& 3\end{array}\right)\right\}$

Prove is a subgroup as:

$\left(\begin{array}{lll}1& 2& 3\\ 2& 1& 3\end{array}\right)\left(\begin{array}{lll}1& 2& 3\\ 2& 1& 3\end{array}\right)=\left(\begin{array}{lll}1& 2& 3\\ 1& 2& 3\end{array}\right)$

Hence, it is proved that $\mathrm{H}$is a subgroup.

Prove that H is not normal

Consider the element $\left(\begin{array}{lll}1& 2& 3\\ 1& 3& 2\end{array}\right)\in {\mathrm{S}}_3$this implies that:

$$\begin{gathered} \left(\begin{array}{lll}1& 2& 3\\ 1& 3& 2\end{array}\right)\left(\begin{array}{lll}1& 2& 3\\ 2& 1& 3\end{array}\right)\left(\begin{array}{lll}1& 2& 3\\ 1& 3& 2\end{array}\right)=\left(\begin{array}{lll}1& 2& 3\\ 3& 1& 2\end{array}\right) \\ \in \mathrm{H} \end{gathered}$$

This implies that $\mathrm{H}$is not normal.

Hence, it is proved that$\left\{\left(\begin{array}{lll}1& 2& 3\\ 2& 1& 3\end{array}\right),\left(\begin{array}{lll}1& 2& 3\\ 1& 2& 3\end{array}\right)\right\}$is a subgroup of${\mathrm{S}}_3$but not normal.