Question

Prove that$\mathit{G}$is abelian if and only if ${\mathbf{\left(}\mathbf{ab}\mathbf{\right)}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}{\mathit{b}}^{\mathbf{-}\mathbf{1}}$ for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$.

Short answer

It is proved that $G$ is abelian.

Step-by-step solution

Write the basic properties of G

If $G$is a group and $a,b\in G$then,

1. ${\left(ab\right)}^{-1}=b^{-1}{a}^{-1}$;
2. ${\left(a^{-1}\right)}^{-1}=a$

Show that G is an abelian

Consider ${\left(ab\right)}^{-1}=a^{-1}{b}^{-1}$ where \[a,b\] be arbitrary elements in \[G\].

As we know that,

\[\left( ab \right)\left( {{a}^{-1}}{{b}^{-1}} \right)=e\]

Multiply both sides by \[b\] and \[a\] on the right, then we have,

\[ab=ba\]

Thus, \[G\] must be abelian.

If \[G\] is abelian, we have,

\[{{\left( ab \right)}^{-1}}={{b}^{-1}}{{a}^{-1}}={{a}^{-1}}{{b}^{-1}}\]

Therefore,

\[{{\left( ab \right)}^{-1}}={{a}^{-1}}{{b}^{-1}}\]

Hence Proved.