Question

If ${\mathbf{\left(}\mathbf{ab}\mathbf{\right)}}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}{\mathit{b}}^{\mathbf{2}}$ for all$\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$, prove that $\mathit{G}$is abelian.

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 thatG is an abelian

Consider ${\left(ab\right)}^2=a^2{b}^2$where$a,b$be arbitrary elements in$G$ .

Multiply by $a^{-1}$from the left side of the given equality and by $b^{-1}$from the right side of the equality then we have,

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

Thus,

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

Therefore,

$ba=ab$

Hence Proved.