$G$ is a group and $f:G \rightarrow G$ is a function defined as $f(a)=a^{-1}$ where $a^{-1}$ is the inverse of $a$ under the group operation. Prove that $f$ is an isomorphism if and only if $G$ is abelian.
I understand that I have to prove $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. How might I do that?
Reference: Fraleigh p. 49 Question 4.40 in A First Course in Abstract Algebra
Best Answer
First note that it is a bijection of $G$ onto $G$ no matter what. So this boils down to $G$ Abelian if and onbly if $f$ is a homorphism.
If $G$ is Abelian, I think you can show that $f$ is a homomorphism.
Now if $f$ is a homomorphism $$ ab=(a^{-1})^{-1}(b^{-1})^{-1}=(b^{-1}a^{-1})^{-1}=(f(b)f(a))^{-1}=(f(ba))^{-1}=((ba)^{-1})^{-1}=ba. $$