Theorem 1   If $a$ and $b$ are arbitrary elements of the group $(G,\,*)$ , then the inverse of $a*b$ is

(1)

Proof. Let the neutral element of the group, which may be proved unique, be $e$ . Using only the group postulates we obtain $$(a*b)*(b^{-1}*a^{-1}) = a*(b*(b^{-1}*a^{-1})) = a*((b*b^{-1})*a^{-1}) = a*(e*a^{-1}) = a*a^{-1} = e,$$ $$(b^{-1}*a^{-1})*(a*b) = b^{-1}*(a^{-1}*(a*b)) = b^{-1}*((a^{-1}*a)*b) = b^{-1}*(e*b) = b^{-1}*b = e,$$ Q.E.D.

Note. The formula (1) may be by induction extended to the form $$(a_1*\cdots*a_n)^{-1} = a_n^{-1}*\cdots*a_1^{-1}.$$