
Abelian Group Proof
Prove that a group G is Abelian if and only if (ab)^1 = (a^1)(b^1) for all a and b in G.
I know how to do the first part of the proof.
If G is Abelian then (ab)^1 = (a^1)(b^1).
(ab)^1 = (b^1)(a^1)=(a^1)(b^1), since G is abelian.
The second part is where I am snagged.
If (ab)^1 = (a^1)(b^1) then G is abelian.
(ab)^1 = (b^1)(a^1) = (ab)^1 = (a^1)(b^1) Therefore (b^1)(a^1)=(a^1)(b^1). So G must be abelian.
I'm not sure if i can make these statements. It seems like i'm assuming stuff.
Any help is appreciated.

Given that $\displaystyle \left( {ab} \right)^{  1} = a^{  1} b^{  1} $ the following is true:
$\displaystyle \begin{gathered}
\left( {ab} \right)^{  1} b = a^{  1} \hfill \\
\left( {ab} \right)^{  1} ba = e \hfill \\
ba = ab \hfill \\
\end{gathered} $