Please help me start this problem on homomorphisms. Let $G$ be a group. Prove that the following are equivalent:
(a) $G$ is Abelian
(b) $f$ : $G\rightarrow G$ defined by $f(x) = x^{-1}$ is a homomorphism.
(c) $f : G \rightarrow G$ defined by $f(x) = x^2$ is a homomorphism.
The trouble is proving they are equivalent. Does that meant a implies b, b implies c, c implies b, b implies a, a implies c, c implies a?

For example assume (a), then we need to check $f(x)=x^{-1}$ is a homomorphism.
Let $x,y \in G$ then $f(xy)=(xy)^{-1}$. Since $G$ is abelian (we are assuming (a)) then $(xy)^{-1}=x^{-1}y^{-1}$ so $f(xy)=(xy)^{-1}=x^{-1}y^{-1}=f(x)f(y)$. So that (b) holds
Now assume (b) and we need to prove (c).
Let $x,y \in G$, then since $(b)$ holds we have that $f(xy)=f(x)f(y)$, that is:
$x^{-1}y^{-1}=(xy)^{-1}$
On the other hand it is always the case that $(xy)^{-1}=y^{-1}x^{-1}$ therefore:
$(yx)^{-1}=(xy)^{-1}$
Hence $yx=xy$.
Thus $f(xy)=(xy)^{2}=(xy)(xy)=x(yx)y=x(xy)y=x^{2}y^{2}$ so $f$ is a homomorphism.
Now just show $c$ implies $a$.
