Proving that this group is Abelian via inverse property Prove that a group G is Abelian If and Only If 
$\left ( ab \right )^{-1}=a^{-1}b^{-1}$ for all a and b in G
Proving first the If condition:
$\left ( ab \right )\left ( ab \right )^{-1}=\left ( ab \right )\left ( a^{-1}b^{-1} \right )$
we require commutativity here in order to satisfy $\left ( ab \right )\left ( ab \right )^{-1}=e$
so,
$\left ( ab \right )\left ( a^{-1}b^{-1} \right )=ab\left ( b^{-1}a^{-1} \right )=a\left ( bb^{-1} \right )a^{-1}=aea^{-1}=aa^{-1}=e
$
Now, the Only If condition:
Because G is Abelian, we have 
$a\circ b=b\circ a  
\forall a,b \in G$
We have 
$\left ( ab \right )\left (ab  \right )^{-1}=\left ( ab \right )\left ( b^{-1}a^{-1} \right )=a\left ( bb^{-1} \right )a^{-1}=aea^{-1}=aa^{-1}=e$
Is this proof done correctly? My concern is with the 'If' condition.
 A: All the ideas are there. The problem is the way you have written out the proof. For example, the statement "we require commutativity here..." doesn't justify why you need it, and it isn't entirely clear what you're trying to show with the argument that follows.
Here is how I would write out a solution (to the if statement), but I'll stress again that you have all the ideas already!
Suppose that for all $a,b\in G$ we have $(ab)^{-1} = a^{-1}b^{-1}$. Then for any $a,b\in G$, we have $$e = (ab)(ab)^{-1} = (ab)(a^{-1}b^{-1}).$$
Multiplying on the right by $ba$, we see that
$$ba =(ab)(a^{-1}b^{-1})ba = ab,$$so $ab = ba $. Hence $G$ is abelian.
A: *

*If


$ba=(a^{-1}b^{-1})^{-1}=(b^{-1}a^{-1})^{-1}=ab$
The second equality is essential.


*

*Only if


$(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}$
The second equality is essential.
A: For the if condition: 
$$e=ab(ab)^{-1}= (ab)(a^{-1}b^{-1})$$
Now multiply both sides with $ba$. You get
$$eba =ba =ab.$$
For the only if just use the identity:
$$(ab)^{-1}= b^{-1}a^{-1}$$ and now apply commutativity. 
