Let be $G$ a group and $a,b \in G$. Prove that $ (ab) ^{-1} = b^{-1}a^{-1}$

Question

I have a proof:

Let $x$ the inverse of $ab$ so $$ x(ab) = 1 $$ $$ xa = b^{-1 }$$ $$ x = b^{-1 }a^{-1} $$

Some of you know another way to do it? Something like: $$ (ab)^{-1} = \cdots =\cdots=b^{-1}a^{-1}$$ ?

Answer 1

\begin{align} (ab)^{-1} = & (ab)^{-1}e \\ = & (ab)^{-1}[aa^{-1}] \\ = & (ab)^{-1}\big[(ae)a^{-1}\big] \\ = & (ab)^{-1}\Big[\big(a[bb^{-1}]\big)a^{-1}\Big] \\ = & (ab)^{-1}\Big[([ab]b^{-1})a^{-1}\Big] \\ = & (ab)^{-1}\Big[[ab]\big(b^{-1}a^{-1}\big)\Big] \\ = & \Big[(ab)^{-1}[ab]\Big]\big(b^{-1}a^{-1}\big) \\ = & e\big(b^{-1}a^{-1}\big) \\ = & b^{-1}a^{-1} \end{align}

Answer 2

$ab.b^{-1}a^{-1}=a.a^{-1}=1$

And $b^{-1}a^{-1}.ab=b^{-1}.b=1$

Thus $(ab)^{-1}=b^{-1}a^{-1}$

Answer 3

$$(ab)^{-1}=(ab)^{-1}((ab)(b^{-1}a^{-1}))=((ab)^{-1}(ab))(b^{-1}a^{-1})=b^{-1}a^{-1}.$$