Question

Let $\mathbf{G}$ be a multiplicative group. Let${\mathbf{G}}^{\mathbf{op}}$ be the set$\mathbf{G}$ equipped with a new operation$\mathbf{*}$ defined by $\mathbf{a}\mathbf{\ast }\mathbf{b}\mathbf{=}\mathbf{ba}$.

(b) Prove that$\mathbf{G}\mathbf{\cong }{\mathbf{G}}^{\mathbf{op}}$

Short answer

It has been proved that $\mathrm{G}\cong {\mathrm{G}}^{\mathrm{op}}$.

Step-by-step solution

Define a mapping

Define a map $f:G\to G^{op}$

By $f\left(x\right)=x^{-1}$.

The map is clearly injective.

It is surjective as well since$f\circ f=\tau$ is the identity map.

Show homomorphism

Let $x,y\in G$,

$\begin{array}{c}\mathrm{f}\left(\mathrm{xy}\right)={\left(\mathrm{xy}\right)}^{-1}\\ ={\mathrm{y}}^{-1}{\mathrm{x}}^{-1}\\ =\mathrm{f}\left(\mathrm{y}\right)\mathrm{f}\left(\mathrm{x}\right)\\ =\mathrm{f}\left(\mathrm{x}\right)\ast \mathrm{f}\left(\mathrm{y}\right)\end{array}$

Hence homomorphism is proved.

Conclusion

Being injective, surjective and homomorphic, $f$is isomorphism.

Thus, $\mathrm{G}\cong {\mathrm{G}}^{\mathrm{op}}$.