f(a) = inverse of a is an isomorphism iff a group G is Abelian 
$G$ 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   
 A: 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
$$
ab=(a^{-1})^{-1}(b^{-1})^{-1}=(b^{-1}a^{-1})^{-1}=(f(b)f(a))^{-1}=(f(ba))^{-1}=((ba)^{-1})^{-1}=ba.
$$
A: Hint: No, we always have that $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. (One of the directions of) what you have to prove is that, if $G$ is abelian,
$$b^{-1}a^{-1}=(ab)^{-1}=\underset{\substack{\text{what it means for $f$}\\\text{to be a homomorphism}}}{\fbox{$f(ab)=f(a)f(b)$}}=a^{-1}b^{-1}$$
A: HINT: You have to prove two things:


*

*If $G$ is Abelian, then $f(ab)=f(a)f(b)$ for all $a,b\in G$, which means that $(ab)^{-1}=a^{-1}b^{-1}$ for all $a,b\in G$.

*If $f(ab)=f(a)f(b)$ for all $a,b\in G$, i.e., if $(ab)^{-1}=a^{-1}b^{-1}$ for all $a,b\in G$, then $G$ is Abelian.
You need just one basic fact for both: that in any group $(ab)^{-1}=b^{-1}a^{-1}$.
A: $f\colon G → G$ defined by $f(x)=x^-1$
$f$ isomorphism $⇒$ f is a 1-1 homomorphism
$∀ x,y∈G, xy = ((xy)^{-1})^{-1} = ((y^{-1})(x^{-1}))^{-1}
 = (f(y)(x))^{-1} = f(yx)^{-1}$  (∵$f$ is a homomorphism)
$= ((yx)^{-1})^{-1}$ (by mapping defining)
$= yx (∵ (a^{-1})^{-1} = a) 
∴ xy = yx ⇒ G$ is abelian
