Question

Find the inverse of each permutation in $S_3$.

Short answer

The inverse of the permutation in$S_3$ are.

$$\begin{gathered} \left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right) \\ \left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right) \end{gathered}$$

Step-by-step solution

Permutation

Let $S$ be a non-empty set. A permutation on $S$ is defined as a bijective mapping

$f:S\to S$. If, $S=\left\{a_1,a_2,.......,a_n\right\}$. Then, the number of bijection from$S$ onto$S$ is $n!$ Let one of the bijection be $f$ which maps $a_i$ to$f\left(a_i\right)$. This is denoted by the symbol,

$\left(\begin{array}{cccc}{a}_1& a_2& ...& a_n\\ f\left(a_1\right)& f\left(a_2\right)& ...& f\left(a_n\right)\\ & & & \end{array}\right)$

Inverse of Permutation

Let$f:S\to S$be a permutation on set $S$. Since, $f$is a bijective mapping, it admits of a unique inverse, s$f^{-1}:S\to S$uch that $f f^{-1}=f^{-1} f=i$. If, by $f$,

$a_i\to a_j$, then, by $f^{-1}$$a_j\to a_i$.

Therefore, if $\left(\begin{array}{cccc}{a}_1& a_2& ...& a_n\\ f\left(a_1\right)& f\left(a_2\right)& ...& f\left(a_n\right)\\ & & & \end{array}\right)$, then role="math" localid="1653674491895" $f^{1=}\left(\begin{array}{cccc}f\left(a_1\right)& f\left(a_2\right)& ...& f\left(a_n\right)\\ \left(a_1\right)& f\left(a_2\right)& ...& f\left(a_n\right)\\ & & & \end{array}\right)$

Conclusion

Given, $S_3=\left\{1,2,3\right\}$, the number of permutations is given by $3!=6$.

Which are,role="math" localid="1653675374421" $$\begin{gathered} \left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right) \\ \left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right) \end{gathered}$$

And their inverses are,$$\begin{gathered} \left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right) \\ \left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right),\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right) \end{gathered}$$