Question

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

Short answer

It isproved 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,

i) ${\left(ab\right)}^{-1}=b^{-1}{a}^{-1}$

ii)${\left(a^{-1}\right)}^{-1}=a$

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 and 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.