Prove that a group $G$ is Abelian if and only if $(ab)^{-1}=a^{-1}b^{-1}$ Prove that a group $G$ is Abelian if and only if $(ab)^{-1}=a^{-1}b^{-1} \forall a,b \in G$
Since G is a group then that means that it contains the following properties: Associativity, Identity, and Inverse.
To show that $G$ is abelian, then $ab=ba, \forall a,b \in G$
Would it be something like this? 
$(ab)(ab)^{-1}=a^{-1}b^{-1}(ab)\\
(ab)(ab)^{-1}=e\\
\rightarrow a^{-1}(ab)(ab)^{-1}=a^{-1}e\\ \rightarrow (a^{-1}a)b(ab)^{-1}=a^{-1}
\\ \rightarrow b(ab)^{-1}=a^{-1}
\\ \rightarrow b^{-1}b(ab)^{-1} = b^{-1}a^{-1} \rightarrow (ab)^{-1}=b^{-1}a^{-1} $ Or did I went the wrong direction?
 A: $$ 
ab=ba 
\overset{\small{\text{by inversing}}}{\Longleftrightarrow} 
b^{-1}a^{-1}=a^{-1}b^{-1} 
\Longleftrightarrow 
(ab)^{-1}=a^{-1}b^{-1} 
.
$$
A: This is really a long comment, but let me get you started.
Easy direction: If $G$ is abelian, then $(ab)^{-1}=a^{-1}b^{-1}$.
Hints: Use the fact that $(ab)^{-1}=b^{-1}a^{-1}$ and that $G$ is abelian to prove this.
Harder direction: If $(ab)^{-1}=a^{-1}b^{-1}$ for all $a,b$, then $G$ is abelian.
Hints: Start with the two equations $(ab)^{-1}=a^{-1}b^{-1}$ (given) and $(ab)^{-1}=b^{-1}a^{-1}$ (always true).  Then, the right-hand-sides are equal, so $a^{-1}b^{-1}=b^{-1}a^{-1}$.  Now, perform some manipulations to reach $ab=ba$.
A: Another short proof:
$$
ab=((ab)^{-1})^{-1}=(a^{-1}b^{-1})^{-1}=(b^{-1})^{-1}(a^{-1})^{-1}=ba.
$$
A: For the backwards direction, here is a very short proof: 
The hypothesis is 
$(ab)(a^{-1}b^{-1})=e$, so $\;(ab)a^{-1}=eb=b\;$ and finally $\; ab=ba$.
A: Let $G$ be a group with identity $e$ and let $a,b \in G$ be given.
First assume $G$ is Abelian. We want to prove $(ab)^{-1}=a^{-1}b^{-1}$. Group inverses are unique, so it suffices to show $(ab)(a^{-1}b^{-1})=e$. Since groups are associative, this is equivalent to showing $(a)(b)(a^{-1})(b^{-1})=e$. Since $G$ is Abelian, this can be re-written as $(a)(a^{-1})(b)(b^{-1})=(aa^{-1})(bb^{-1})=(e)(e)=e$ as required.
Next assume $(ab)^{-1}=a^{-1}b^{-1}$. We want to prove $G$ is Abelian; that is $ab=ba$. By assumption, we have $(ab)(a^{-1}b^{-1})=e$. By associativity, we have $(aba^{-1})(b^{-1})=e$. Since group inverses are unique, this implies $aba^{-1}=b$. Multiplying by $a$ on the right, we see $(aba^{-1})a=(b)a$ and so $(ab)(a^{-1}a)=ba$ and finally $(ab)(e)=ab=ba$ as required.
