Abelian group and morphism equivalent statement

Question

Exercise

Show that the following statements are equivalent:

$(i) \space G \space \text{is abelian.}$

$(ii) \space \text{the map f: G} \to \text{G defined as} \space f(x)=x^{-1} \space \text{is a group morphism.}$

$(iii) \space \text{the map f: G} \to \text{G defined as} \space f(x)=x^2 \space \text{is a group morphism.}$

I could show $(i)$ if and only if $(ii)$, and $(i)$ implies $(iii)$. I got stuck with the implication $(iii) \implies (i)$, so I would appreciate any hints. Also, I wanted to know if there is a way to show $(ii)$ if and only if $(iii)$ without using $(i)$

Answer 1

$f(ab) = abab = f(a) f(b) = aa bb$.

Hence $a^{-1} f(ab) b^{-1} = ba = a^{-1} f(a) f(b) b^{-1} = ab$.

Answer 2

If $x\to x^{-1}$ is homomorphism so $(ab)^{-1}=a^{-1}b^{-1}$ so $$b^{-1}a^{-1}=a^{-1}b^{-1}$$ which you wanted to omit it. Now $$b=aba^{-1},~~~ab=a^2ba^{-1}$$ and so $a^2b^2=a^2b(aba^{-1})=a^2(bab)a^{-1}=(a^2ba^{-1})(a^2ba^{-1})=(ab)^2$. Therefore $x\to x^{2}$ is group homomorphism. If you do the other side iii to ii, you'll see that we cannot escape from i. In fact i makes a strong relation with them.