Question

Question: Prove that a group G is abelian if and only if the function $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{G}$given by $\mathit{f}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{-}\mathbf{1}}$is a homomorphism of groups. In this case, show that f is an isomorphism.

Short answer

It has been proved that is abelian if and only if the function$f:G\to G$ given by$f\left(x\right)=x^{-1}$ is a homomorphism of groups.

Step-by-step solution

Step-By-Step SolutionStep 1: Suppose that is abelian

Consider$f:G\to G$given by $f\left(x\right)=x^{-1}$.

Let $x,y\in G$then$x^{-1},y^{-1}\in G$

Now, Since Gis abelian

$$\begin{gathered} f\left(x·y\right)={\left(x·y\right)}^{-1} \\ =y^{-1}·x^{-1} \\ =x^{-1}·y^{-1} \\ =f\left(x\right)·f\left(y\right) \end{gathered}$$

Hence, f is homomorphism.

Since the inverse exists, it is clear that f is an isomorphism as well.

Suppose that  f is a homomorphism

Consider $f:G\to G$given by $f\left(x\right)=x^{-1}$is a homomorphism.

Let ,

$x,y\in G$

$$\begin{gathered} xy=f\left(x^{-1}\right)f\left(y^{-1}\right) \\ =f\left(x^{-1}{y}^{-1}\right) \\ =f\left({\left(yx\right)}^{-1}\right) \\ =yx \end{gathered}$$

Hence, G is abelian

Conclusion

Hence, G is abelian if and only if the function $f:G\to G$given by $f\left(x\right)=x^{-1}$is a homomorphism of groups.