Question

Let $\mathit{a}\mathbf{=}\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)$and$\mathit{b}\mathbf{=}\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)$in${\mathit{S}}^{\mathbf{3}}$. Verify That,${\left(ab\right)}^{\mathbf{-}\mathbf{1}}\mathbf{\ne }{\mathit{a}}^{\mathbf{-}\mathbf{1}}{\mathit{b}}^{\mathbf{-}\mathbf{1}}$

Short answer

It is verified that,${\left(ab\right)}^{-1}\ne a^{-1}{b}^{-1}$.

Step-by-step solution

Composition of permutations

Let $\mathit{f}\mathbf{:}\mathit{S}\mathbf{\to }\mathit{S}$and $\mathit{g}\mathbf{:}\mathit{S}\mathbf{\to }\mathit{S}$be two permutations on $\mathit{S}$. The compositions $\mathit{f}\mathit{g}$and $\mathit{g}\mathit{f}$are defined as the compositions $\mathit{f}\mathbf{\circ }\mathit{g}$and $\mathit{g}\mathbf{\circ }\mathit{f}$.

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

Then,

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

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

Inverse of a permutation

Let $\mathit{f}\mathbf{:}\mathit{S}\mathbf{\to }\mathit{S}$be a permutation on set $\mathit{S}$. Since $\mathit{f}$is a bijective mapping, it admits of a unique inverse role="math" localid="1653654411377" $\mathbf{}{\mathit{f}}^{\mathbf{1}}\mathbf{:}\mathit{S}\mathbf{\to }\mathit{S}\mathbf{}$, Such that role="math" localid="1653654433231" $\mathit{f}{\mathit{f}}^{\mathbf{1}}\mathbf{=}{\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathit{f}\mathbf{=}\mathit{i}$. If by $\mathit{f}$role="math" localid="1653654388889" ${\mathit{a}}_{\mathbf{i}}\mathbf{\to }{\mathit{a}}_{\mathbf{j}}$then by role="math" localid="1653654399820" ${\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathit{a}}_{\mathbf{j}}\mathbf{\to }{\mathit{a}}_{\mathbf{i}}\mathbf{}$.

Therefore, if $\mathit{f}\mathbf{=}\left(\begin{array}{cccc}{a}_1& a_2& ...& a_n\\ f\left(a_1\right)& f\left(a_2\right)& ...& f\left(a_n\right)\end{array}\right)$then,

${\mathit{f}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\left(\begin{array}{cccc}f\left(a_1\right)& f\left(a_2\right)& ...& f\left(a_n\right)\\ a_1& a_2& ...& a_n\end{array}\right)$

Proof

It is given that, $a=\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right),b=\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)$then,

role="math" localid="1653655058254" $$\begin{gathered} ab=\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)\left(\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\end{array}\right)=\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right) \\ {\left(\mathrm{ab}\right)}^{-1}=\left(\begin{array}{ccc}1& 2& 3\\ 3& 2& 1\end{array}\right) \\ Again,a^{-1}=\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right)and,b^{-1}=\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right) \\ \Rightarrow a^{-1}{b}^{-1}=\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\end{array}\right) \end{gathered}$$

Hence, it is proved that,${\left(ab\right)}^{-1}\ne a^{-1}{b}^{-1}$