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?
 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 $a$.
A: It means if you take any two of them they will imply each other.  So (a) implies (b) and (b) implies (a); (b) implies (c) and (c) implies (b); etc.  To prove that they are equivalent you only need to construct a chain of implications that allow you to move from one to the other.  For instance if you showed (a) implies (b), (b) implies (c), and (c) implies (a) you'd be done.  To show that (b) implies (a) you would only observe that (b) implies (c) and (c) implies (a).
A: $(a) \Rightarrow (b)$:
$f(xy) = (xy)^{-1} = y^{-1}x^{-1} = x^{-1}y^{-1} = f(x)f(y)$
So $f$ is indeed a homomorphism.
$(b) \Rightarrow (a)$, or really not$(a) \Rightarrow$ not$(b)$:
Let $x$ and $y$ be two elements which don't commute. Then
$f(xy) =  (xy)^{-1} = y^{-1}x^{-1}$ but $f(x)f(y) = x^{-1}y^{-1}$
Since $x$ and $y$ don't commute, their inverses can't commute either, since we would then have
$xy = ((xy)^{-1})^{-1} = (y^{-1}x^{-1})^{-1} = (x^{-1}y^{-1})^{-1} = ((yx)^{-1})^{-1} = yx$
$(a) \Rightarrow (c)$:
$f(xy) = (xy)(xy) = xyxy = x^2y^2 = f(x)f(y)$
Not$(a) \Rightarrow$ not$(c)$:
Let $x$ and $y$ be two elements that don't commute. We then have
$f(xy) = (xy)(xy) = xyxy$ but $f(x)f(y) = x^2y^2 = xxyy$
These cannot be equal to each other, since we would then have the contradiction
$xyxy = xxyy \Rightarrow x^{-1}xyxyy^{-1} = x^{-1}xxyyy^{-1} \Rightarrow yx = xy$
Hence both $(b)$ and $(c)$ are equivalent to $(a)$.
