Question

Which one of the following subsets of $S_3$ is a subgroup of $S_3$?

• A. $\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\}$
• B. $\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right)\}$
• C. $\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right)\}$
• D. $\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)\}$

Answer 1

The correct option is B $\{\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right)\}$

Option C does not contain an identity, so it can't be a subgroup of $S_3$

Since, $\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right)\in A,B,D$

$\therefore$ A,B, D has identity element

Since, they have two elements in which one is identity, so other need to be an inverse of itself.

In A,

$\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\cdot \left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)=\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)\ne I$

$\therefore$ inverse does not exist

$\therefore A$ is not a subgroup of $S_3$

In B,

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

$\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right)$ is an inverse of itself.

$\therefore$ Inverse exist

$\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right)\cdot \left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right)=\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\in B$

$\therefore B$ satisfies closure property

$\therefore B$ is a subgroup of $S_3$

In D,

$\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)\cdot \left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)=\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\ne I$

$\therefore$ inverse does not exist

$\therefore D$ is not a subgroup of $S_3$

Hence, only option B is correct.