$(xy)^{-1}=x^{-1}y^{-1}$ if and only if $G$ is Abelian I have to prove the following statement: $(xy)^{-1}=x^{-1}y^{-1}$ $\forall x,y\in G$ if and only if $G$ is Abelian. $G$ is a group and $x,y\in G$. This is what I did:
If $G$ is Abelian, then $(xy)^{-1}=y^{-1}x^{-1}=x^{-1}y^{-1}$ for all $x,y\in G$.
Now suppose that $(xy)^{-1}=x^{-1}y^{-1}$ holds for all $x,y\in G$, we also know that $(xy)^{-1}=y^{-1}x^{-1}$, so $y^{-1}x^{-1}=x^{-1}y^{-1}$ and $G$ has to be Abelian.
Is this proof complete or not? Thank you.   
 A: While your proof does capture the essential idea it leaves out a detail that the professor probably wants you to be explicit about because you're in an introductory class.  The direction $G$ Abelian implies $(xy)^{-1} = x^{-1}y^{-1}$ is perfectly fine, but for the other direction you've shown that $x^{-1}y^{-1} = y^{-1}x^{-1}$ for all $x$ and $y$.  You are supposed to show that $xy = yx$ for all $x$ and $y$.  The general format of the proof should be as follows:


*

*Assume $x, y \in G$.

*Use the assumption to calculate $yx$.

*Conclude that $xy = yx$.

A: Your proof is basically correct; you're one step away from finishing. You end by concluding that $x^{-1}y^{-1}=y^{-1}x^{-1}$. What we really want is $xy=yx$. 
A: try... this .. 
by definition, 'e' as identity
$$(xy)^{-1} = (xy)^{-1}\cdot e \cdot e$$
$$= (xy)^{-1}\cdot (xx^{-1})(yy^{-1})$$ 
 $$= (xy)^{-1}\cdot (x)(x^{-1}\cdot y)(y^{-1})$$ 
 $$= (xy)^{-1}\cdot (x)(y\cdot  x^{-1} )(y^{-1})$$ 
 $$= (xy)^{-1}\cdot (x y)\cdot  (x^{-1} )(y^{-1})$$ 
 $$= e \cdot x^{-1}  y^{-1}$$ 
$$=  x^{-1} y^{-1}$$ 
NOW the other way.. 
$$xy = xy\cdot e =  xy \cdot (yx)^{-1}(yx) = xy \cdot y^{-1}x^{-1}(yx) = x(y \cdot y^{-1})x^{-1}(yx)=x\cdot e\cdot x^{-1}\cdot yx = yx$$
----->>>>>>>>>in case you need further explanations.. some of the steps are typically justified by somthing like this.. ----below....
by definition, 'e' as identity
$$(xy)^{-1} = (xy)^{-1}\cdot e \cdot e$$
$$= (xy)^{-1}\cdot (xx^{-1})(yy^{-1})$$\tag{by definition of inverses}\
 $$= (xy)^{-1}\cdot (x)(x^{-1}\cdot y)(y^{-1})$$\tag{by associativity of groups}\
 $$= (xy)^{-1}\cdot (x)(y\cdot  x^{-1} )(y^{-1})$$\tag{[important :-)] using abelian  assumption}\
 $$= (xy)^{-1}\cdot (x y)\cdot  (x^{-1} )(y^{-1})$$\tag{associative prop of groups}\
 $$= e \cdot x^{-1}  y^{-1}$$\tag{def of inverses}\
$$=  x^{-1} y^{-1}$$\tag{def of identity}\
