Question

If $f=\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 3& 2& 4& 5& 1& 6& \end{array}\right)\in S_6$. then $f^{-1}$ is

• A. $\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 2& 5& 1& 3& 4& 6& \end{array}\right)$.
• B. $\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 5& 2& 1& 3& 4& 6& \end{array}\right)$.
• C. $\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 4& 2& 1& 3& 5& 6& \end{array}\right)$.
• D. $\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 3& 2& 1& 5& 4& 6& \end{array}\right)$.

Answer 1

The correct option is B $\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 5& 2& 1& 3& 4& 6& \end{array}\right)$.

Given $f=\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 3& 2& 4& 5& 1& 6& \end{array}\right)\in S_6$.

We have to find $f^{-1}$.

That is, we need to find inverse permutation.

• From $f$ we see that, $1\to 3\to 4\to 5\to 1$. Therefore in $f^{-1},1\to 5$
• From $f$ we see that, $2\to 2$. Therefore in $f^{-1},2\to 2$
• From $f$ we see that, $3\to 4\to 5\to 1\to 3$. Therefore in $f^{-1},3\to 1$
• From $f$ we see that, $4\to 5\to 1\to 3\to 4$. Therefore in $f^{-1},4\to 3$
• From $f$ we see that, $5\to 1\to 3\to 4\to 5$. Therefore in $f^{-1},5\to 4$
• From $f$ we see that, $6\to 6$. Therefore in $f^{-1},6\to 6$
  

Thus $f^{-1}=\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& \\ 5& 2& 1& 3& 4& 6& \end{array}\right)$.